Konstantin Kryzhitsky (1858-1911) was a Ukrainian-born Russian landscape painter.
We have sections
\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}
Matrices
Let \({\cal F}\) denote either the set \(\mathbb{R}\) of real numbers or the set \(\mathbb{C}\) of complex numbers, and let \(m,n\in\mathbb{N}\). The \(m\times n\) matrix in \({\cal F}\) is defined by
\begin{equation}{\label{laeq1}\tag{1}}
A=\left [\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & \cdots & a_{1n}\\
a_{21} & a_{22} & a_{23} & \cdots & a_{2n}\\
\vdots & \vdots & \vdots & \cdots & \vdots\\
a_{m1} & a_{m2} & a_{m3} & \cdots & a_{mn}
\end{array}\right ].
\end{equation}
We call \(a_{ij}\) the \(ij\)–entry or \(ij\)–component of matrix \(A\). The matrix \(A\) has \(m\) rows and \(n\) columns. The \(j\)th column of \(A\) is
\[A_{\cdot j}=\left [\begin{array}{c} a_{1j}\ a_{2j}\\ \vdots \\ a_{mj}\end{array}\right ]\]
and the \(i\)th row of \(A\) is
\[A_{i\cdot}=\left [a_{i1},a_{i2},\cdots ,a_{in}\right ].\]
We also see that \(A_{\cdot j}\) is an \(m\times 1\) matrix and \(A_{i\cdot}\) is an \(1\times n\) matrix. If \(m=n\), then \(A\) is called a square matrix.
We shall now define the addition and scalar multiplication of matrices. Let \(A\) and \(B\) be two \(m\times n\) matrices with the corresponding entries \(a_{ij}\) and \(b_{ij}\) for \(i=1,\cdots ,m\) and \(j=1,\cdots ,n\).
- The addition of \(A+B\) is defined to be an \(m\times n\) matrix \(C\) such that its entry \(c_{ij}\) is equal to \(a_{ij}+b_{ij}\) for \(i=1,\cdots ,m\) and \(j=1,\cdots ,n\).
- Let \(k\) be a real number, and let \(A\) be an \(m\times n\) matrix. The scalar multiplication of \(A\) by \(k\) is defined by \(kA=C\), where the entry \(c_{ij}\) of$C$ is equal to \(k\cdot a_{ij}\) for \(i=1,\cdots ,m\) and \(j=1,\cdots ,n\).
\begin{equation}{\label{laex2}}\tag{2}\mbox{}\end{equation}
Example \ref{laex2}. Let
\[A=\left [\begin{array}{rrr}
1 & -1 & 0\\ 2 & 3 & -6
\end{array}\right ]\mbox{ and }
B=\left [\begin{array}{rrr}
-1 & 1 & 0\\ -2 & -3 & 6
\end{array}\right ].\]
Then \(A+B\) is a zero matrix given by
\[A+B=\left [\begin{array}{rrr}
0 & 0 & 0\\ 0 & 0 & 0
\end{array}\right ].\]
We also have
\[3A=\left [\begin{array}{rrr}
3 & -3 & 0\\
6 & 9 & -18
\end{array}\right ].\]
Let \(A\) be an \(m\times n\) matrix as shown in (\ref{laeq1}). The transpose of matrix \(A\) is defined to be an \(n\times m\) matrix \(B\) with entries \(b_{ij}=a_{ji}\) for \(i=1,\cdots ,m\) and \(j=1,\cdots ,n\). For convenience, we also write \(B\) as \(A^{\top}\) and
\[A^{\top}=\left [\begin{array}{ccccc}
a_{11} & a_{21} & a_{31} & \cdots & a_{m1}\\
a_{12} & a_{22} & a_{32} & \cdots & a_{m2}\\
\vdots & \vdots & \vdots & \cdots & \vdots\\
a_{1n} & a_{2n} & a_{3n} & \cdots & a_{mn}
\end{array}\right ].\]
It is clear to see \((A^{\top})^{\top}=A.\)
Example. Let \(A\) be given in Example \ref{laex2}. We have
\[A^{\top}=\left [\begin{array}{rrr}
1 & 2\\ -1 & 3\\ 0 & -6
\end{array}\right ].\]
The matrix \(A\) is said to be symmetric iwhen \(A=A^{\top}\). This also says that the symmetric matrix must be a square matrix, i.e., \(m=n\). The square \(n\times n\) matrix \(A\) is said to be a diagonal matrix when \(a_{ij}=0\) for all \(i\neq j\) and \(i,j=1,\cdots ,n\). However, it does not means that \(a_{ii}\neq 0\) for all \(i=1,\cdots ,n\). For example, the zero matrix is also a diagonal matrix. Finally, the unit matrix \(I_{n}\) is an \(n\times n\) diagonal matrix such that \(a_{ii}=1\) for \(i=1,\cdots ,n\). It is clear to see
\[AI_{n}=I_{n}A=A.\]
Let \(A\) be an \(m\times n\) matrix and \(B\) be an \(n\times s\) matrix given by
\[A=\left [\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & \cdots & a_{1n}\\
a_{21} & a_{22} & a_{23} & \cdots & a_{2n}\\
\vdots & \vdots & \vdots & \cdots & \vdots\\
a_{m1} & a_{m2} & a_{m3} & \cdots & a_{mn}
\end{array}\right ]\mbox{ and }
B=\left [\begin{array}{ccccc}
b_{11} & b_{12} & b_{13} & \cdots & b_{1s}\\
b_{21} & b_{22} & b_{23} & \cdots & b_{2s}\\
\vdots & \vdots & \vdots & \cdots & \vdots\\
b_{n1} & b_{n2} & b_{n3} & \cdots & b_{ns}
\end{array}\right ].\]
The multiplication \(AB\) is defined to be an \(m\times s\) matrix \(C\) with entries defined by
\[c_{ik}=\sum_{j=1}^{n}a_{ij}\cdot b_{jk}=a_{i1}\cdot b_{1k}+a_{2i}\cdot b_{2k}+\cdots +a_{in}\cdot b_{nk}.\]
Example. Let
\[A=\left [\begin{array}{rrr}
2 & 1 & 5\\ 1 & 3 & 2\end{array}\right ]\mbox{ and }
B=\left [\begin{array}{rr}
3 & 4\\ -1 & 2\\ 2 & 1
\end{array}\right ].\]
The multiplication \(AB\) is given by
\[AB=\left [\begin{array}{rrr}
2 & 1 & 5\\ 1 & 3 & 2\end{array}\right ]\left [\begin{array}{rr}
3 & 4\\ -1 & 2\\ 2 & 1 \end{array}\right ]
=\left [\begin{array}{rr}
15 & 15\\ 4 & 12
\end{array}\right ].\]
Let \(A\), \(B\) and \(C\) be three matrices such that they can be multiplied. It is not hard to show
\[A(B+C)=AB+BC\mbox{ and }(AB)C=A(BC).\]
Let \(A\) and \(B\) be two matrices such that they can be multiplied. Then \(A^{\top}\) and \(B^{\top}\) can be multiplied satisfying
\[(AB)^{\top}=B^{\top}A^{\top}.\]
Let \(A\) be an \(n\times n\) square matrix. The trace of \(A\) is denoted and defined by
\[\mbox{tr}(A)=\sum_{i=1}^{n}a_{ii},\]
which means the sum of all the diagonal elements of \(A\).
Example. Let
\[A=\left [\begin{array}{rrr}
1 & 7 & 3\\ -1 & 5 & 2\\ 2 & 3 & -4
\end{array}\right ].\]
Then, we obtain \(\mbox{tr}(A)=1+5-4=2\). \(\sharp\)
We have the following observations that will be the exercises for the readers.
(i) Let \(A\) be an \(n\times n\) square matrix. Then, we have
\[\mbox{tr}(A)=\mbox{tr}(A^{\top}).\]
(ii) Let \(A\) and \(B\) be matrices such that they can be multiplied. Then, we have
\[\mbox{tr}(AB)=\mbox{tr}(BA).\]
\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}
Elementary Row and Column Operations of Matrices.
We consider three elementary row operations on an \(m\times n\) matrix \(A\) over the scalar field \({\cal F}\):
(1) multiplication of one row of \(A\) by a non-zero scalar \(c\);
(2) replacement of the \(r\)th row of \(A\) by row \(r\) plus \(c\) times row \(s\), where \(c\) is any non-zero scalar and \(r\neq s\);
(3) interchange of two rows of \(A\).
The elementary column operations can be similarly defined.
Definition. If \(A_{1}\) and \(A_{2}\) are two \(m\times n\) matrix over the same scalar field \({\cal F}\), we say that \(A_{2}\) is row-equivalent to \(A_{1}\) if \(A_{2}\) can be obtained from \(A_{1}\) by a finite steps of elementary row operations. \(\sharp\)
\begin{equation}{\label{laex186}\tag{3}}\mbox{}\end{equation}
Example \ref{laex186}. We consider
\[A_{1}=\left [\begin{array}{rrrr}
2 & -1 & 3 & 2\\ 1 & 4 & 0 & -1\\ 2 & 6 & -1 & 5
\end{array}\right ].\]
We shall perform a finite steps of elementary row operations on \(A_{1}\). Now, we have
\begin{align*} & \left [\begin{array}{rrrr} 2 & -1 & 3 & 2\\ 1 & 4 & 0 & -1\\ 2 & 6 & -1 & 5 \end{array}\right ]\stackrel{(2)}{\longrightarrow} \left [\begin{array}{rrrr} 0 & -9 & 3 & 4\\ 1 & 4 & 0 & -1\\ 2 & 6 & -1 & 5 \end{array}\right ]\stackrel{(2)}{\longrightarrow} \left [\begin{array}{rrrr} 0 & -9 & 3 & 4\\ 1 & 4 & 0 & -1\\ 0 & -2 & -1 & 7 \end{array}\right ]\stackrel{(1)}{\longrightarrow}\\ & \left [\begin{array}{rrrr} 0 & -9 & 3 & 4\\ 1 & 4 & 0 & -1\\ 0 & 1 & \frac{1}{2} & -\frac{7}{2} \end{array}\right ]\stackrel{(2)}{\longrightarrow} \left [\begin{array}{rrrr} 0 & -9 & 3 & 4\\ 1 & 0 & -2 & 13\\ 0 & 1 & \frac{1}{2} & -\frac{7}{2} \end{array}\right ]\stackrel{(2)}{\longrightarrow} \left [\begin{array}{rrrr} 0 & 0 & \frac{15}{2} & -\frac{55}{2}\\ 1 & 0 & -2 & 13\\ 0 & 1 & \frac{1}{2} & -\frac{7}{2} \end{array}\right ]\stackrel{(1)}{\longrightarrow}\\ & \left [\begin{array}{rrrr} 0 & 0 & 1 & -\frac{11}{3}\\ 1 & 0 & -2 & 13\\ 0 & 1 & \frac{1}{2} & -\frac{7}{2} \end{array}\right ]\stackrel{(2)}{\longrightarrow} \left [\begin{array}{rrrr} 0 & 0 & 1 & -\frac{11}{3}\\ 1 & 0 & 0 & \frac{17}{3}\\ 0 & 1 & \frac{1}{2} & -\frac{7}{2} \end{array}\right ]\stackrel{(2)}{\longrightarrow} \left [\begin{array}{rrrr} 0 & 0 & 1 & -\frac{11}{3}\\ 1 & 0 & 0 & \frac{17}{3}\\ 0 & 1 & 0 & -\frac{5}{3} \end{array}\right ]. \end{align*}
Therefore, we finally obtain
\[A_{2}=\left [\begin{array}{rrrr}
0 & 0 & 1 & -\frac{11}{3}\\ 1 & 0 & 0 & \frac{17}{3}\\ 0 & 1 & 0 & -\frac{5}{3}
\end{array}\right ].\]
This says that \(A_{2}\) is row-equivalent to \(A_{1}\). \(\sharp\)
Definition. An \(m\times n\) matrix \(A\) over the scalar field \({\cal F}\) is called row-reduced when the following conditions are satisfied.
- The leading (first) non-zero entry in each non-zero row of \(A\) is equal to \(1\).
- Each column of \(A\) containing the leading non-zero entry of some row must have all its other entries to be \(0\).
Example. The matrix \(A_{2}\) as obtained in Example \ref{laex186} is row-reduced. The following two matrices are not row-reduced.
\[\left [\begin{array}{rrr}
3 & 2 & 1\\ 1 & 0 & -3\\ 0 & 0 & 0
\end{array}\right ]\mbox{ and }
\left [\begin{array}{rrrr}
1 & 0 & 0 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 1 & 0
\end{array}\right ].\]
The first matrix fails to satisfy the first condition of the definition, since the leading non-zero entry of the first row is not equal to \(1\). The second matrix satisfies the first condition, but is fails to satisfy the second condition because of its column 3. \(\sharp\)
Now, we can transform each \(m\times n\) matrix over the scalar field \({\cal F}\) to a row-reduced matrix by using the elementary operations. Let \(A\) be an \(m\times n\) matrix over the scalar field \({\cal F}\).
- If every entry in row \(1\) of \(A\) is \(0\), then the first condition is satisfied.
- If row \(1\) has a non-zero entry, let \(k_{1}\) be the smallest positive integer satisfying \(a_{1k_{1}}\neq 0\). We multiply row \(1\) by \(1/a_{1k_{1}}\). Then the first condition is satisfied.
- For each \(i\geq 2\), we multiply row \(1\) by \(-a_{1k_{1}}\), and add it to row \(i\). Then, the leading non-zero entry of row \(1\) occurs in column \(k\) with value \(1\), and every other entry in column is \(0\).
- Now, we consider the matrix that has been obtained above. If each entry in row \(2\) is zero, we do nothing to row \(2\). If row \(1\) has a non-zero entry, then we can similarly multiply row \(2\) by a suitable scalar such that the leading non-zero entry becomes \(1\).
- Since row \(1\) has a non-zero leading entry in column \(k_{1}\), the non-zero leading entry of row \(2\) cannot occur in column \(k_{2}\). Now, we assume that it is in column \(k_{2}\neq k_{1}\). Similarly, we multiply row \(2\) by a suitable scalar, and add it to the other rows. Then, we can obtain that all entries in column \(k_{2}\) are \(0\) except for the leading non-zero entry in row \(2\). We need to notice that these operations will not change the entries of row \(1\) in columns \(1,\cdots ,k\). We also see that if row \(1\) was all entries to be zero, these operations for row \(2\) will not affect row \(1\).
- Continuing this way, in a finite number of steps, we shall arrive at a row-reduced matrix.
The above procedure says that each \(m\times n\) matrix can perform a finite steps of elementary row operations to obtain a row-reduced matrix. Example \ref{laex186} illustrates this observation. Now, we write it as a theorem below.
\begin{equation}{\label{lat188}\tag{4}}\mbox{}\end{equation}
Theorem \ref{lat188}. Every \(m\times n\) matrix over the scalar field \({\cal F}\) is row-equivalent to a row-reduced matrix. \(\sharp\)
Definition. An \(m\times n\) matrix \(A\) over the scalar field \({\cal F}\) is called row-reduced echelon matrix when the following conditions are satisfied.
- \(A\) is row-reduced
- Every row of \(A\) which has all its entries \(0\) occurs below every row which has a non-zero entry.
- If rows \(1,2,\cdots ,r\) are the non-zero rows of \(A\), and if the leading non-zero entry of row \(i\) occurs in column \(k_{i}\), then \(k_{1}<k_{2}<\cdots <k_{r}\). \(\sharp\)
We define
\[\delta_{ij}=\left\{\begin{array}{ll}1 & \mbox{if \(i=j\)}\\ 0 & \mbox{if \(i\neq j\)}.\end{array}\right .\]
The equivalent description of row-reduced echelon matrix is as follows. Every entry in \(A\) is \(0\), or there exists a positive integer \(r\) with \(1\leq r\leq m\) and \(r\) positive integers \(k_{1},\cdots ,k_{r}\) with \(1\leq k_{i}\leq n\) such that the following conditions are satisfied:
- \(a_{ij}=0\) for \(i>r\) and \(j<k_{i}\);
- \(a_{ik_{j}}=\delta_{ij}\) for \(1\leq i\leq r\) and \(1\leq j\leq r\);
- \(k_{1}<\cdots <k_{r}\).
Example. It is obvious that the \(n\times n\) identity matrix \(I_{n}\) and the \(m\times n\) zero matrix are row-reduced echelon matrix. The following matrix is also a row-reduced echelon matrix
\[\left [\begin{array}{rrrrr}
0 & 1 & -3 & 0 & 6\\ 0 & 0 & 0 & 1 & 2\\ 0 & 0 & 0 & 0 & 0
\end{array}\right ].\]
\begin{equation}{\label{lat197}\tag{5}}\mbox{}\end{equation}
Theorem \ref{lat197}. Every \(m\times n\) matrix over the scalar field \({\cal F}\) is row-equivalent to a row-reduced echelon matrix.
Proof. Theorem \ref{lat188} says that every \(m\times n\) matrix over the scalar field \({\cal F}\) is row-equivalent to a row-reduced matrix. We can perform a finite number of row interchanges on a row-reduced matrix, which can result in a row-reduced echelon form. This completes the proof. \(\blacksquare\)
Definition. An \(n\times n\) elementary matrix is a matrix obtained by performing elementary operations on the \(n\times n\) identity matrix \(I_{n}\). The elementary matrix is
said to be of type 1, 2 or 3 when it is obtained from \(I_{n}\) by performing the elementary operation 1 ,2 , or 3. \(\sharp\)
\begin{equation}{\label{lap176}}\tag{6}\mbox{}\end{equation}
Proposition \ref{lap176}. Let \(A\) be an \(m\times n\) matrix over the scalar field \({\cal F}\). Suppose that \(B\) is obtained from \(A\) by performing one elementary row (resp. column) operation. Then, there exists an \(m\times m\) (resp. \(n\times n\)) elementary matrix \(E\) satisfying \(B=EA\) (resp. \(B=AE\)). As a matter of fact, the elementary matrix \(E\) is obtained from \(I_{m}\) (resp. \(I_{n}\)) by performing the same elementary row (resp. column) operation as that which was performed on \(A\) to obtain \(B\). Conversely, if \(E\) is an \(m\times m\) (resp. \(n\times n\)) elementary matrix, then \(EA\) (resp. \(AE\)) is the matrix obtained from \(A\) by performing the same elementary row (resp. column) operation as that which produces \(E\) from \(I_{m}\) (resp. \(I_{n}\)).
Proof. We shall give a proof for an operation of type (2). The other two types of operations are left as exercises. Given the identity matrix \(I_{m}\), suppose that we perform the operation “replacement of the \(r\)th row of \(A\) by row \(r\) plus \(c\) times row \(s\)”, where \(c\) is any non-zero scalar and \(r\neq s\). Then, the \(ik\)th entry of \(E\) obtained from \(I_{m}\) is
\[e_{ik}=\left\{\begin{array}{ll}\delta_{ik}, & \mbox{if \(i\neq r\)}\\\delta_{rk}+c\cdot\delta_{sk}, & \mbox{if \(i=r\)}.\end{array}\right .\]
Therefore, the \(ij\)th entry of \(EA\) is given by
\[(EA)_{ij}=\sum_{k=1}^{m}e_{ik}a_{kj}=\left\{\begin{array}{ll}a_{ij}, & \mbox{if \(i\neq r\)}\\a_{rj}+c\cdot a_{sj}, & \mbox{if \(i=r\)}.\end{array}\right .\]
This shows that \(EA\) is indeed a matrix obtained from \(A\) by performing type (2) row elementary operation. \(\blacksquare\)
\begin{equation}{\label{lac189}}\tag{7}\mbox{}\end{equation}
Corollary \ref{lac189}. Let \(A\) and \(B\) be \(m\times n\) matrices over the same scalar field \({\cal F}\). Then \(B\) is row-equivalent to \(A\) if and only if \(B=PA\), where \(P\) is a product of \(m\times m\) elementary matrices.
Proof. Suppose that \(B=PA\), where \(P=E_{p}E_{p-1}\cdots E_{1}\) and each \(E_{i}\) is elementary matrix. Then \(E_{1}A\) is row-equivalent to \(A\), and \(E_{2}(E_{1}A)\) is row-equivalent to \(E_{1}A\), which also says that \(E_{2}E_{1}A\) is row-equivalent to \(A\). Continuing in this way, we see that \(PA=E_{p}E_{p-1}\cdots E_{1}A\) is row-equivalent to \(A\). For the converse, suppose that \(B\) is row-equivalent to \(A\). According to Proposition \ref{lap176}, let \(E_{1},E_{2},\cdots ,E_{p}\) be the elementary matrices corresponding to the sequence of elementary row operations which makes \(A\) as \(B\). Then we indeed have \(B=E_{p}E_{p-1}\cdots E_{1}A\). This completes the proof. \(\blacksquare\)
The Inverse of a Matrix.
Let \(A\) be an \(n\times n\) square matrix over the scalar field \({\cal F}\).
- An \(n\times n\) matrix \(B\) is called a left inverse of \(A\) when \(BA=I_{n}\).
- An \(n\times n\) matrix \(B\) is called a right inverse of \(A\) when \(AB=I_{n}\).
- We say that \(A\) is invertible or non-singular when there exists another \(n\times n\) square matrix satisfying \(AB=BA=I_{n}\). In this case, we also write \(B\) as \(A^{-1}\) and is called a two-sided inverse of \(A\). We also call \(A^{-1}\) as the inverse matrix of \(A\).
Proposition. Let \(A\) be an invertible \(n\times n\) matrix. Then its inverse matrix is unique.
Proof. Let \(B\) and \(C\) be \(n\times n\) matrices satisfying
\[AB=BA=AC=CA=I_{n}.\]
We are going to prove \(B=C\). Now, we have
\[B=BI_{n}=B(AC)=(BA)C=I_{n}C=C.\]
This completes the proof. \(\blacksquare\)
\begin{equation}{\label{lap180}}\tag{8}\mbox{}\end{equation}
Proposition \ref{lap180}. We have the following properties.
(i) If \(A\) is an invertible \(n\times n\) matrix, then \(A^{-1}\) is also an invertible \(n\times n\) matrix with \((A^{-1})^{-1}=A\).
(ii) If both \(A\) and \(B\) are invertible \(n\times n\) matrices, then \(AB\) is also an invertible \(n\times n\) matrix with \((AB)^{-1}=B^{-1}A^{-1}\).
(iii) A product of invertible matrices is invertible.
Proof. Part (i) is evident from the definition. Part (ii) follows from the following expression
\[(AB)(B^{-1}A^{-1})=(B^{-1}A^{-1})(AB)=I_{n}.\]
Part (iii) follows immediately from part (ii). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{lap177}}\tag{9}\mbox{}\end{equation}
Proposition \ref{lap177}. Let \(A\) and \(B\) be two \(n\times n\) matrices.
(i) If \(AB\) is invertible, then \(A\) and \(B\) are invertible.
(ii) If \(AB=I_{n}\), then \(A\) and \(B\) are invertible satisfying \(A=B^{-1}\) \((\)equivalently, \(B=A^{-1})\).
Proof. To prove part (i), since \(AB\) is invertible, there exists \(C\) satisfying
\[(AB)C=C(AB)=I_{n},\]
which also says that \((CA)B=I_{n}\). Therefore, \(CA\) is the left-inverse of \(B\). According to Corollary \ref{lac244} that will be proven later, it follows that \(B\) is invertible. We can similarly show that \(A\) is also invertible.
To prove part (ii), since \(AB=I_{n}\), it says that \(AB\) is invertible. According to part (i), we see that \(A\) and \(B\) are invertible. Since \(AB=I_{n}\), by multiplying \(B^{-1}\) on both sides, we obtain \((AB)B^{-1}=I_{n}B^{-1}\), which implies \(A=B^{-1}\). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{lap178}}\tag{10}\mbox{}\end{equation}
Proposition \ref{lap178}. Elementary matrices are invertible. Moreover, its inverse is also an elementary matrix of the same type.
Proof. Let \(E\) be an \(n\times n\) elementary matrix. By definition, \(E\) can be obtained from \(I_{n}\) by elementary row operation. We can also reverse the steps to transform \(E\) back to \(I_{n}\). This means that \(I_{n}\) can be obtained from \(E\) by performing elementary row operations of the same type. Proposition \ref{lap176} says that there exist elementary matrices \(E_{1},\cdots ,E_{p}\) such that \(I_{n}=E_{p}E_{p-1}\cdots E_{1}E\). Let \(\widehat{E}=E_{p}E_{p-1}\cdots E_{1}\). Then \(I_{n}=\widehat{E}E\), and \(\widehat{E}\) is also an elementary matrix, which also says that \(E^{-1}=\widehat{E}\) by part (ii) of Proposition \ref{lap177}. This completes the proof. \(\blacksquare\)
\begin{equation}{\label{lat198}}\tag{11}\mbox{}\end{equation}
Theorem \ref{lat198}. If \(A\) is an \(n\times n\) square matrix, then the following statements are equivalent.
(a) \(A\) is invertible.
(b) \(A\) is row-equivalent to the \(n\times n\) identity matrix \(I_{n}\).
(c) \(A\) is a product of elementary matrices.
Proof. Let \(R\) be a row-reduced echelon matrix which is row-equivalent to \(A\). By Corollary \ref{lac189}, we have
\begin{equation}{\label{eq1}}\tag{12}R=E_{p}E_{p-1}\cdots E_{1}A,\end{equation}
where \(E_{1},E_{2},\cdots ,E_{p}\) are elementary matrices. Since each \(E_{i}\) is invertible by Proposition \ref{lap178}, we also have
\begin{equation}{\label{eq2}}\tag{13}A=E_{1}^{-1}E_{2}^{-1}\cdots E_{p}^{-1}R.\end{equation}
Let \(H=E_{p}E_{p-1}\cdots E_{1}\). By part (iii) of Proposition \ref{lap180}, it follows that \(H\) is invertible and
\[H^{-1}=E_{1}^{-1}E_{2}^{-1}\cdots E_{p}^{-1}.\]
Next, we are going to claim that \(A\) is invertible if and only if \(R\) is invertible. Suppose that \(A\) is invertible. Equality (\ref{eq2}) says that \(I_{n}=A^{-1}H^{-1}R\). Let \(B=A^{-1}H^{-1}\). Then we immediately have \(BR=I_{n}\). Using (\ref{eq1}), we can also obtain
\[RB=HA(A^{-1}H^{-1})=I_{n},\]
which show that \(R\) is invertible with inverse \(B=A^{-1}H^{-1}\). We can similarly show that if \(R\) is invertible, then \(A\) is invertible.
Since \(R\) is a a row-reduced echelon matrix, it means that \(R\) is invertible if and only if each row of \(R\) contains a non-zero entry, which is equivalent to say \(R=I_{n}\). In other words, \(A\) is invertible if and only if \(R=I_{n}\). In this case, we have
\[A=E_{1}^{-1}E_{2}^{-1}\cdots E_{p}^{-1}.\]
These facts show the equivalence of the above three statements. This completes the proof. \(\blacksquare\)
Corollary. Let \(A\) and \(B\) be \(m\times n\) matrices. Then \(B\) is row-equivalent to \(A\) if and only if \(B=PA\), where \(P\) is an invertible \(m\times m\) matrix.
Let \(A\) and \(B\) be \(m\times n\) and \(m\times p\) matrices over the same scalar field \({\cal F}\), respectively. The {\bf augmented matrix} \((A|B)\) is defined to be the \(m\times (n+p)\) matrix \((A,B)\) whose first \(n\) columns are the columns of \(A\), and whose last \(p\) columns are the columns of \(B\). Let \(M\) an another matrix such that the multiplication \(MA\) and \(MB\) can be defined. Then, we can prove
\begin{equation}{\label{laeq201}}\tag{14} M(A|B)=M(A,B)=(MA,MB)=(MA|MB). \end{equation}
The following results will be useful to find the inverse matrix.
\begin{equation}{\label{lac55}}\tag{15}\mbox{}\end{equation}
Proposition \ref{lac55}. We have the following properties.
(i) If \(A\) is an invertible \(n\times n\) matrix, then the \(n\times 2n\) augmented matrix \((A|I_{n})\) can be transformed into \((I_{n}|A^{-1})\) by a finite number of elementary row operations.
(ii) Given an \(n\times n\) matrix \(B\), if \(A\) is an invertible \(n\times n\) matrix such that \((A|I_{n})\) can be transformed into \((I_{n}|B)\) by a finite number of elementary row operations, then \(B\) is the inverse of \(A\).
(iii) If the \(n\times n\) matrix \(A\) is not invertible, then \(A\) can be transformed into a matrix with a row containing all zero entries by a finite number of elementary row operations.
Proof. To prove part (i), Let \(C=(A|I_{n})\). Then, by (\ref{laeq201}), we have
\begin{equation}{\label{laeq202}}\tag{16}
A^{-1}C=A^{-1}(A|I_{n})=(A^{-1}A|AI_{n})=(I_{n}|A^{-1}).
\end{equation}
Since \(A^{-1}\) is invertible, Theorem \ref{lat198} says that \(A^{-1}\) is the product of elementary matrices, which is given by
\[A^{-1}=E_{p}E_{p-1}\cdots E_{1}.\]
Therefore, from (\ref{laeq202}), we obtain
\begin{equation}{\label{laeq203}}\tag{17}
E_{p}E_{p-1}\cdots E_{1}(A|I_{n})=A^{-1}C=(I_{n}|A^{-1}).
\end{equation}
Proposition \ref{lap176} says that multiplying a matrix on the left by an elementary matrix means to transform the matrix by an elementary row operations. Form (\ref{laeq203}), we see that if \(A\) is an invertible \(n\times n\) matrix, then \((A|I_{n})\) can be transformed into \((I_{n}|A^{-1})\) by a finite number of elementary row operations.
To prove part (ii), suppose that the augmented matrix \((A|I_{n})\) can be transformed into \((I_{n}|B)\) by a finite number of elementary row operations. Let \(E_{1},E_{2},\cdots ,E_{p}\) be the elementary matrices corresponding to the elementary row operations according to Proposition \ref{lap176} satisfying
\[E_{p}E_{p-1}\cdots E_{1}(A|I_{n})=(I_{n}|B).\]
Let \(M=E_{p}E_{p-1}\cdots E_{1}\). Then, we also have
\[(MA|M)=M(A|I_{n})=(I_{n}|B),\]
which says that \(MA=I_{n}\) and \(M=B\). Since \(A\) is invertible, it follows \(M=A^{-1}\), i.e. \(B=A^{-1}\).
To prove part (iii), if the \(n\times n\) matrix \(A\) is not invertible, then \(\mbox{rank}(A)<n\) (rank of matrix \(A\) will be defined later; the reader can skip the proof at first reading). Therefore, we cannot transform \((A|I_{n})\) into \((I_{n}|B)\) by a finite number of elementary row operations. Otherwise, we must have \(B=A^{-1}\). In fact, the matrix \(A\) can be transformed into a matrix with a row containing all zero entries. This completes the proof. \(\sharp\)
Example. Let us determine whether the following matrix
\[A=\left [\begin{array}{ccc}
1 & \frac{1}{2} & \frac{1}{3}\\
\frac{1}{2} & \frac{1}{3} & \frac{1}{4}\\
\frac{1}{3} & \frac{1}{4} & \frac{1}{5}
\end{array}\right ]\]
is invertible or not. If it is invertible, we compute its inverse matrix. We shall apply Proposition \ref{lac55} by performing elementary row operations.
\begin{eqnarray*} && \left [\begin{array}{rrr} 1 & \frac{1}{2} & \frac{1}{3}\\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4}\\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \end{array}\right .\left |\left . \begin{array}{rrr} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right .\right ]\longrightarrow\left [\begin{array}{rrr}1 & \frac{1}{2} & \frac{1}{3}\\0 & \frac{1}{12} & \frac{1}{12}\\0 & \frac{1}{12} & \frac{4}{45}
\end{array}\right .\left |\left .\begin{array}{rrr}1 & 0 & 0\\-\frac{1}{2} & 1 & 0\\-\frac{1}{3} & 0 & 1\end{array}\right .\right ]\\ && \longrightarrow\left [\begin{array}{rrr}1 & \frac{1}{2} & \frac{1}{3}\\0 & \frac{1}{12} & \frac{1}{12}\\0 & 0 & \frac{1}{180}\end{array}\right .\left |\left .\begin{array}{rrr}1 & 0 & 0\\-\frac{1}{2} & 1 & 0\\ \frac{1}{3} & -1 & 1\end{array}\right .\right ]\longrightarrow\left [\begin{array}{rrr}1 & \frac{1}{2} & \frac{1}{3}\\0 & 1 & 1\\0 & 0 & 1\end{array}\right .\left |\left .\begin{array}{rrr}1 & 0 & 0\\-6 & 12 & 0\\60 & -180 & 180\end{array}\right .\right ]\\&& \longrightarrow\left [\begin{array}{rrr}1 & \frac{1}{2} & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right .\left |\left .\begin{array}{rrr}-9 & 60 & -60\\-36 & 192 & -180\\60 & -180 & 180\end{array}\right .\right ]\longrightarrow\left [\begin{array}{rrr}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right .\left |\left . \begin{array}{rrr}-9 & -36 & 30\\-36 & 192 & -180\\60 & -180 & 180\end{array}\right .\right ].\end{eqnarray*}
Therefore, \(A\) is invertible with the inverse matrix given by
\[A^{-1}=\left [\begin{array}{rrr}
-9 & -36 & 30\\
-36 & 192 & -180\\
60 & -180 & 180
\end{array}\right ].\]
Example. Let us determine whether the following matrix
\[A=\left [\begin{array}{ccc}
1 & 2 & 1\\
2 & 1 & -1\\
1 & 5 & 4
\end{array}\right ]\]
is invertible or not. If it is invertible, we compute its inverse matrix. We shall apply Proposition \ref{lac55} by performing elementary row operations.
\begin{eqnarray*}\left [\begin{array}{rrr}1 & 2 & 1\\2 & 1 & -1\\1 & 5 & 4\end{array}\right .\left |\left .\begin{array}{rrr}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right .\right ]& \longrightarrow &\left [\begin{array}{rrr}1 & 2 & 1\\0 & -3 & -3\\0 & 3 & 3\end{array}\right .\left |\left .\begin{array}{rrr}1 & 0 & 0\\-2 & 1 & 0\\-1 & 0 & 1\end{array}\right .\right ]\\& \longrightarrow &\left [\begin{array}{rrr}1 & 2 & 1\\0 & -3 & -3\\0 & 0 & 0\end{array}\right .\left |\left .\begin{array}{rrr}1 & 0 & 0\\-2 & 1 & 0\\-3 & 1 & 1\end{array}\right .\right ],\end{eqnarray*}
which is a matrix with a row whose first \(3\) entries are zero. Therefore, the matrix \(A\) is not invertible.$\sharp$
\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}
System of Linear Equations.
We consider the following system of linear equations
\begin{equation}{\label{laeq54}}\tag{18}
\begin{array}{ccc}
a_{11}\cdot x_{1}+a_{12}\cdot x_{2}+\cdots +a_{1n}\cdot x_{n} & = & b_{1}\\
a_{21}\cdot x_{1}+a_{22}\cdot x_{2}+\cdots +a_{2n}\cdot x_{n} & = & b_{2}\\
\vdots & \vdots & \vdots\\
a_{m1}\cdot x_{1}+a_{m2}\cdot x_{2}+\cdots +a_{mn}\cdot x_{n} & = & b_{m},
\end{array}
\end{equation}
where \(a_{ij}\) and \(b_{i}\) are in the scalar field \({\cal F}\) for \(i=1,\cdots ,m\) and \(j=1,\cdots ,n\). If \(b_{i}=0\) for all \(i=1,\cdots ,m\), then the system is called homogeneous.
The basic technique for finding the solutions of a system of linear equations is the technique of elimination. For example, we consider the following homogeneous system
\[\begin{array}{ccccccc}
2x_{1} & – & x_{2} & + & x_{3} & = & 0\\
x_{1} & + & 3x_{2} & + & 4x_{3} & = & 0.
\end{array}\]
In order to eliminate \(x_{1}\), the second equation is multiplied by \(-2\), and add this new equation to the first equation. Then, we obtain \(-7x_{2}-7x_{3}=0\), which also means \(x_{2}=-x_{3}\). Similarly, in order to eliminate \(x_{2}\), the first equation is multiplied by \(3\), and add this new equation to the second equation. Then, we obtain \(7x_{1}+7x_{3}=0\), which also means \(x_{1}=-x_{3}\). Therefore, we conclude that if \((x_{1},x_{2},x_{3})\) is a solution of the linear system, then \(x_{1}=x_{2}=-x_{3}\). This also says that the solutions of the linear systems is a set consisting of the form \((-a,-a,a)\) for \(a\in\mathbb{R}\).
Now, we define
\[A=\left [\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & \cdots & a_{1n}\\
a_{21} & a_{22} & a_{23} & \cdots & a_{2n}\\
\vdots & \vdots & \vdots & \cdots & \vdots\\
a_{m1} & a_{m2} & a_{m3} & \cdots & a_{mn}
\end{array}\right ],
X=\left [\begin{array}{c}
x_{1}\\ x_{2}\\ \vdots\\ x_{n}
\end{array}\right ]\mbox{ and }
B=\left [\begin{array}{c}
b_{1}\\ b_{2}\\ \vdots\\ b_{m}
\end{array}\right ].\]
Then the general system (\ref{laeq54}) can be written as \(AX=B\).
From (\ref{laeq54}), we select \(m\) scalars \(c_{1},\cdots ,c_{m}\), multiply the \(j\)th equation by \(c_{j}\), and then add. We obtain the equation
\[\left (c_{1}a_{11}+\cdots +c_{m}a_{m1}\right )x_{1}+\cdots +\left (c_{1}a_{1n}+\cdots +c_{m}a_{mn}\right )x_{n}=c_{1}b_{1}+\cdots +c_{m}b_{m}.\]
We call such an equation a linear combination of the equations in (\ref{laeq54}). It is obvious that any solution of the entire system (\ref{laeq54}) will also be a solution of this new equation. This is the fundamental idea of the elimination process. If we have another system of linear equations
\begin{equation}{\label{laeq193}}\tag{19}
\begin{array}{ccc}
c_{11}\cdot x_{1}+c_{12}\cdot x_{2}+\cdots +c_{1n}\cdot x_{n} & = & d_{1}\\
c_{21}\cdot x_{1}+c_{22}\cdot x_{2}+\cdots +c_{2n}\cdot x_{n} & = & d_{2}\\
\vdots & \vdots & \vdots\\
c_{p1}\cdot x_{1}+c_{p2}\cdot x_{2}+\cdots +c_{pn}\cdot x_{n} & = & d_{p},
\end{array}
\end{equation}
in which each of the \(p\) equations is a linear combination of the equations in (\ref{laeq54}), then every solution of (\ref{laeq54}) is a solution of this new system (\ref{laeq193}). It may happen that some solutions of (\ref{laeq193}) are not solutions of (\ref{laeq54}). However, if each equation in the original system (\ref{laeq54}) is a linear combination of the equations in the new system (\ref{laeq193}), then we can see that the systems (\ref{laeq54}) and (\ref{laeq193}) have exactly the same solutions. Let us say that two systems of linear equations are equivalent if and only if each equations in each system is a linear combination of the equations in the other system. These observations are shown in the following proposition.
\begin{equation}{\label{lap194}}\tag{20}\mbox{}\end{equation}
Proposition \ref{lap194}. Equivalent systems of linear equations have exactly the same solutions.
\begin{equation}{\label{lap195}}\tag{21}\mbox{}\end{equation}
Proposition \ref{lap195}. If \(A_{1}\) and \(A_{2}\) are row-equivalent \(m\times n\) matrices, the homogeneous systems of linear equations \(A_{1}X={\bf 0}\) and \(A_{2}X={\bf 0}\) have exactly the same solutions.
Proof. Since \(A_{1}\) and \(A_{2}\) are row-equivalent, we can assume that \(A_{1}\) can turn into \(A_{2}\) by a finite sequence of elementary row operations.
\[A_{1}\rightarrow C_{1}\rightarrow C_{2}\rightarrow C_{3}\rightarrow\cdots\rightarrow C_{p}=A_{2}.\]
It suffices to prove that \(C_{j}X={\bf 0}\) and \(C_{j+1}X={\bf 0}\) have the same solutions. Equivalently, we need to claim that one elementary row operations does not alter the set of solutions. Since \(C_{j+1}\) is obtained from \(C_{j}\) by one elementary row operation, we can see that each equation in the system \(C_{j+1}X={\bf 0}\) will be a linear combination of the equations in the system \(C_{j}X={\bf 0}\). On the other hand, since the inverse of elementary row operation is also an elementary row operation, each equation in the system \(C_{j}X={\bf 0}\) will also be a linear combination of the equations in the system \(C_{j+1}X={\bf 0}\). This shows that the systems \(C_{j}X={\bf 0}\) and \(C_{j+1}X={\bf 0}\) are equivalent. By Proposition \ref{lap194}, they have the same solutions, and the proof is complete. \(\blacksquare\)
If \(R\) is an \(m\times n\) row-reduced echelon matrix, we discuss the system \(RX={\bf 0}\). Let rows \(1,2,\cdots ,r\) be the non-zero rows of \(R\). Suppose that the leading non-zero entry of row \(i\) occurs in column \(k_{i}\). Then, the system \(RX={\bf 0}\) consists of \(r\) non-trivial equations. We also see that the unknown \(x_{k_{i}}\) with non-zero coefficient will occur only in the \(i\)th equation. Let \(y_{1},y_{2},\cdots ,y_{n-r}\) denote the \(n-r\) unknowns which are different from \(x_{k_{1}},x_{k_{2}},\cdots ,x_{k_{r}}\). Then, the \(r\) non-trivial equations in \(RX={\bf 0}\) are of the form
\begin{eqnarray} && x_{k_{1}}+\sum_{j=1}^{n-r}c_{1j}y_{j}=0\nonumber\\ && x_{k_{2}}+\sum_{j=1}^{n-r}c_{2j}y_{j}=0\nonumber\\ && \vdots\label{laeq196}\tag{22}\\ && x_{k_{r}}+\sum_{j=1}^{n-r}c_{rj}y_{j}=0.\nonumber\end{eqnarray}
All the solutions to the system \(RX={\bf 0}\) can be obtained by assigning any values to \(y_{1},\cdots ,y_{n-r}\) and computing the values of \(x_{k_{1}},x_{k_{2}},\cdots ,x_{k_{r}}\) according to (\ref{laeq196}). For example, we consider the following matrix
\[\left [\begin{array}{rrrrr}
0 & 1 & -3 & 0 & \frac{1}{2}\\
0 & 0 & 0 & 1 & 2\\
0 & 0 & 0 & 0 & 0
\end{array}\right ].\]
Then \(r=2\), \(k_{1}=2\) and \(k_{2}=4\). The two non-trivial equations are given by
\[x_{2}-3x_{3}+\frac{1}{2}x_{5}=0\mbox{ or }x_{2}=3x_{3}-\frac{1}{2}x_{5}\]
\[x_{4}+2x_{5}=0\mbox{ or }x_{4}=-2x_{5}.\]
Therefore, we can assign any values to \(x_{1},x_{3},x_{5}\), e.g., \(x_{1}=a\), \(x_{3}=b\), \(x_{5}=c\). The solution will be of the form
\[\left (x_{1},x_{2},x_{3},x_{4},x_{5}\right )=\left (a,3b-\frac{1}{2}c,b,-2b,c\right ).\]
\begin{equation}{\label{lap231}}\tag{23}\mbox{}\end{equation}
Proposition \ref{lap231}. Let \(R\) be an \(m\times n\) row-reduced echelon matrix with \(r\) non-zero rows. If \(r<n\), then the system \(RX={\bf 0}\) has a non-trivial solution.
Proof. Since \(r<n\), we can choose some unknowns \(x_{j}\) which is not among the \(r\) unknowns \(x_{k_{1}},x_{k_{2}},\cdots , x_{k_{r}}\). Then, we can construct a solution as shown above such that \(x_{j}=1\). This shows that the system \(RX={\bf 0}\) has a non-trivial solution. \(\sharp\)
\begin{equation}{\label{lap60}}\tag{24}\mbox{}\end{equation}
Proposition \ref{lap60}. If \(A\) is an \(m\times n\) matrix and \(m<n\), then the homogeneous system of linear equations \(AX={\bf 0}\) has a non-trivial solution.
Proof. Let \(R\) be a row-reduced echelon matrix which is row-equivalent to \(A\), which can be guaranteed by Theorem \ref{lat197}. Then, the systems \(AX={\bf 0}\) and \(RX={\bf 0}\) have the same solutions by Proposition \ref{lap195}. If \(r\) is the number of non-zero rows in \(R\), then we have \(r\leq m\). Since \(m<n\), we also have \(r<n\). From Proposition \ref{lap231}, it says that \(RX={\bf 0}\) has a non-trivial solution, which also says that \(AX={\bf 0}\) has a non-trivial solution. This completes the proof. \(\blacksquare\)
The alternate proof of Proposition \ref{lap60} by considering the rank of \(A\) will be given in Proposition \ref{lap205} later.
\begin{equation}{\label{lap69}}\tag{25}\mbox{}\end{equation}
Proposition \ref{lap69}. If \(A\) is an \(n\times n\) square matrix, then \(A\) is row-equivalent to the \(n\times n\) identity matrix if and only if the system of equations \(AX={\bf 0}\) has only the trivial solution \(X=0\).
Proof. If \(A\) is row-equivalent to an identity matrix \(I_{n}\), then \(AX={\bf 0}\) and \(I_{n}X={\bf 0}\) have the same solutions. This shows that \(AX={\bf 0}\) has only the trivial solution \(X=0\). Conversely, suppose that \(AX={\bf 0}\) has only trivial solution \(X={\bf 0}\). Let \(R\) be an \(n\times n\) row -reduced echelon matrix which is row-equivalent to \(A\). Then \(AX={\bf 0}\) and \(RX={\bf 0}\) have the same solutions, i.e., \(RX={\bf 0}\) has only trivial solution. Let \(r\) be the number of non-zero rows in \(R\). Since \(RX={\bf 0}\) has no non-trivial solution, we must have \(r\geq n\) by Proposition \ref{lap231}. This says that \(r=n\). This also means that \(R\) has a leading non-zero entry of \(1\) in each rows of \(R\). Since these \(1\)’s occur in different \(n\) columns, the matrix \(R\) must be the identity matrix \(I_{n}\). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{lat200}}\tag{26}\mbox{}\end{equation}
Theorem \ref{lat200}. For an \(n\times n\) square matrix \(A\), the following statements are equivalent.
(i) \(A\) is invertible.
(ii) The homogeneous system \(AX={\bf 0}\) has only the trivial solution \(X={\bf 0}\).
(iii) The system \(AX=B\) has a solution for each \(B\).
Proof. According to Proposition \ref{lap69}, statement (ii) is equivalent to the fact that \(A\) is row-equivalent to the identity matrix. From Theorem \ref{lat198}, we see that statements (i) and (ii) are equivalent. Next, we want to show that statements (i) and (iii) are equivalent. If \(A\) is invertible, then it is obvious that the solution of \(AX=B\) is \(X=A^{-1}B\). Conversely, suppose that \(AX=B\) has a solution for each given \(B\). Let \(R\) be a row-reduced echelon matrix which is row-equivalent to \(A\), which is always true by Theorem \ref{lat197}. We want to claim that \(R\) is an identity matrix. This is equivalent to showing that the last row of \(R\) is not zero. Let
\[E=\left [\begin{array}{c}
0\\ 0\\ \vdots\\ 0\\ 1
\end{array}\right ].\]
If the system \(RX=E\) has a solution, then the last row of \(R\) cannot be zero. Since \(R=PA\) for some inevrtible matrix \(P\) by Corollary \ref{lac199}, we see that \(RX=E\) if and only if \(AX=P^{-1}E\equiv B\). Since \(AX=P^{-1}E\) has a solution, we conclude that \(R\) is indeed an identity matrix. Theorem \ref{lat198} says that \(A\) is invertible, and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{lac244}}\tag{27}\mbox{}\end{equation}
Corollary \ref{lac244}. An \(n\times n\) square matrix \(A\) with either a left or right inverse is invertible.
Proof. Suppose that \(A\) has a left inverse, i.e., there exists an \(n\times n\) matrix \(B\) satisfying $BA=I_{n}$. If \(AX={\bf 0}\), then
\[X=I_{n}X=(BA)X=B(AX)=B{\bf 0}={\bf 0}.\]
Equivalently, it says that if \(X\neq {\bf 0}\), then \(AX\neq {\bf 0}\). This shows that \(AX={\bf 0}\) has only the trivial solution. Theorem \ref{lat200} says that \(A\) is invertible. Now, suppose that \(A\) has a right inverse, i.e., there exists an \(n\times n\) matrix \(C\) such that \(AC=I_{n}\). This is equivalent to saying that \(C\) has a left inverse, i.e., \(C\) is invertible. Therefore, we also conclude that \(A\) is invertible with inverse \(C\). This completes the proof. \(\blacksquare\)
Corollary. Let \(A=A_{1}A_{2}\cdots A_{r}\), where each \(A_{1},A_{2},\cdots ,A_{r}\) are \(n\times n\) square matrices. Then \(A\) is invertible if and only if each \(A_{i}\) is invertible for \(i=1,\cdots ,r\).
Proof. From part (iii) of Proposition \ref{lap180}, we remain to show that if \(A\) is invertible, then each \(A_{i}\) is invertible for \(i=1,\cdots ,r\). We first claim that \(A_{r}\) is invertible. Suppose that \(X\) is an \(n\times 1\) matrix and \(A_{r}X={\bf 0}\). Then, we have
\[AX=(A_{1}A_{2}\cdots A_{r-1})A_{r}X={\bf 0}.\]
Since \(A\) is invertible, we must have \(X={\bf 0}\) by Theorem \ref{lat200}, which also says that the system \(A_{r}X={\bf 0}\) has no non-trivial solution. Therefore, the matrix \(A_{r}\) must be invertible using Theorem \ref{lat200} again. In this case, we have that
\[A_{1}A_{2}\cdots A_{r-1}=AA^{-1}_{r}\equiv\widehat{A}\]
is invertible. Using the preceding arguments, we can also show that \(A_{r-1}\) is invertible. Continuing in this way, we conclude that each \(A_{i}\) is invertible. This completes the proof. \(\blacksquare\)
\begin{equation}{\label{lat206}}\tag{28}\mbox{}\end{equation}
Theorem \ref{lat206}. Let \(A\) be an \(m\times n\) matrix, and let \(S\) and \(S_{0}\) be set of all solutions of the nonhomogeneous linear system \(AX=B\) and homogeneous linear system \(AX={\bf 0}\), respectively. Then, for any solution \(s\) of \(AX=B\), we have
\[S=S_{0}+\{s\}=\left\{s+s_{0}:s_{0}\in S_{0}\right\}.\]
Proof. If \(t\in S\), then \(At=B\). Therefore, we have
\[A(t-s)=At-As=B-B={\bf 0},\]
which says that \(t-s\in S_{0}\), i.e., \(t\in S_{0}+\{s\}\). Therefore, we obtain the inclusion \(S\subseteq S_{0}+\{s\}\). On the other hand, suppose that \(t\in S_{0}+\{s\}\). Then \(t=s+u\) for some \(u\in S_{0}\). Since
\[At=A(s+u)=As+Au=B+{\bf 0}=B,\]
which says that \(t\in S\). Therefore, we obtain the inclusion \(S_{0}+\{s\}\subseteq S\). This completes the proof. \(\blacksquare\)
Theorem \ref{lat200} shows that \(A\) is invertible if and only if the system \(AX=B\) has a solution for each \(B\). The uniqueness of solution can also be obtained below when \(B\) is fixed.
\begin{equation}{\label{lat207}}\tag{29}\mbox{}\end{equation}
Theorem \ref{lat207}. Let \(AX=B\) be a systems of \(n\) linear equations in \(n\) unknowns. Then \(A\) is invertible if and only if the system \(AX=B\) has a unique solution \(A^{-1}B\).
Proof. Suppose that \(A\) is invertible. Substituting \(X=A^{-1}B\) into the linear system, we obtain
\[AX=A(A^{-1}B)=(AA^{-1})B=I_{n}B=B,\]
which says that \(A^{-1}B\) is a solution. If \(s\) is another solution of the linear system \(AX=B\), then \(As=B\), which says that \(A^{-1}(As)=A^{-1}B\). Therefore, we obtain \(s=A^{-1}B\). This shows that the linear system has one and only one solution \(A^{-1}B\).
Conversely, suppose that the linear system has exactly one solution \(s\). Let \(S_{0}\) denote the set of all solutions for the corresponding homogeneous linear system \(AX={\bf 0}\). Theorem \ref{lat206} says
\[\{s\}=S=S_{0}+\{s\},\]
which implies \(S_{0}=\{{\bf 0}\}\). Therefore, we conclude that \(A\) is invertible by Theorem \ref{lat200}. This completes the proof. \(\blacksquare\)
We remark that, in Theorem \ref{lat200}, the linear system \(AX=B\) must have a solution for each \(B\) in order to guarantee the invertibility of \(A\). However, when \(B\) is fixed, the linear system \(AX=B\) should have a unique solution in order to guarantee the invertibility of \(A\) as shown in Theorem \ref{lat207}.
The elimination process for solving the linear systems is equivalent to the row operations of matrix \(A\).
Example. We consider the following homogeneous linear system
\[\begin{array}{ccccccccc}
2x_{1} & – & x_{2} & + & 3x_{2} & + & 2x_{4} & = & 0\\
x_{1} & + & 4x_{2} &&& – & x_{4} & = & 0\\
2x_{1} & + & 6x_{2} & – & x_{3} & + & 5x_{4} & = & 0.
\end{array}\]
We take
\[A_{1}=\left [\begin{array}{rrrr}
2 & -1 & 3 & 2\\
1 & 4 & 0 & -1\\
2 & 6 & -1 & 5
\end{array}\right ].\]
From Example \ref{laex186}, we can perform a finite steps of elementary row operations on \(A_{1}\) to obtain
\[A_{2}=\left [\begin{array}{rrrr}
0 & 0 & 1 & -\frac{11}{3}\\
1 & 0 & 0 & \frac{17}{3}\\
0 & 1 & 0 & -\frac{5}{3}
\end{array}\right ].\]
The equivalent linear system is given by
\[\begin{array}{ccccccccc}
&&&& x_{3} & – & {\displaystyle \frac{11}{3}x_{4}} & = & 0\\
x_{1} &&&&& + & {\displaystyle \frac{17}{3}x_{4}} & = & 0\\
&& x_{2} &&& – & {\displaystyle \frac{5}{3}x_{4}} & = & 0.
\end{array}\]
The solution has the form of
\[\left (x_{1},x_{2},x_{3},x_{4}\right )=\left (-\frac{17}{3}c,\frac{5}{3}c,\frac{11}{3}c,c\right )\]
for any \(c\in\mathbb{R}\). \(\sharp\)
The elementary row operation can also be used to solve the non-homogeneous linear system \(AX=B\). We form the augmented matrix $A’$ of the system \(AX=B\), where \(A’\) is the \(m\times (n+1)\) matrix whose first \(n\) columns are the columns of \(A\), and whose last column is \(B\).
Example. We consider the following matrix
\[A=\left [\begin{array}{rrr}
1 & -2 & 1\\
2 & 1 & 1\\
0 & 5 & -1
\end{array}\right ]\]
and wish to solve the system \(AX=B\) for some \(b_{1},b_{2},b_{3}\). We shall perform some steps of row operations on the augmented matrix \(A’\) which row-reduces \(A\)
\begin{eqnarray*} && \left [\begin{array}{rrr}1 & -2 & 1\\2 & 1 & 1\\0 & 5 & -1\end{array}\right .\left |\left .\begin{array}{c}b_{1}\\b_{2}\\b_{3}\end{array}\right .\right ]\stackrel{(2)}{\longrightarrow}\left [\begin{array}{rrr}1 & -2 & 1\\0 & 5 & -1\\0 & 5 & -1\end{array}\right .\left |\left .\begin{array}{c}b_{1}\\b_{2}-2b_{1}\\b_{3}\end{array}\right .\right ]\stackrel{(2)}{\longrightarrow}\left [\begin{array}{rrr}1 & -2 & 1\\0 & 5 & -1\\0 & 0 & 0\end{array}\right .\left |\left .\begin{array}{c}b_{1}\\b_{2}-2b_{1}\\b_{3}-b_{2}+2b_{1}\end{array}\right .\right ]\\&& \stackrel{(1)}{\longrightarrow}\left [\begin{array}{rrr}1 & -2 & 1\\0 & 1 & -\frac{1}{5}\\0 & 0 & 0\end{array}\right .\left |\left .\begin{array}{c}b_{1}\\\frac{1}{5}(b_{2}-2b_{1})\\b_{3}-b_{2}+2b_{1}\end{array}\right .\right ]\stackrel{(2)}{\longrightarrow}\left [\begin{array}{rrr}1 & 0 & \frac{3}{5}\\0 & 1 & -\frac{1}{5}\\0 & 0 & 0\end{array}\right .\left |\left .\begin{array}{c}\frac{1}{5}(b_{1}+2b_{2})\\\frac{1}{5}(b_{2}-2b_{1})\\b_{3}-b_{2}+2b_{1}\end{array}\right .\right ].\end{eqnarray*}
Therefore, the non-homogeneous system \(AX=B\) has a solution if and only if \(b_{3}-b_{2}+2b_{1}=0\). In this case, the solution of \(AX=B\) has the form of
\[\left (x_{1},x_{2},x_{3}\right )=\left (-\frac{3}{5}c+\frac{1}{5}(b_{1}+2b_{2}),\frac{1}{5}c+\frac{1}{5}(b_{2}-2b_{1}),c\right )\]
for any \(c\in\mathbb{R}\). \(\sharp\)
\begin{equation}{\label{lap211}}\tag{30}\mbox{}\end{equation}
Proposition \ref{lap211}. Let \(AX=B\) be a system of \(m\) linear equations in \(n\) unknowns, and let \(C\) be an invertible \(m\times m\) matrix. Then, the linear systems \((CA)X=CB\) and \(AX=B\) have the same solutions.
Proof. Let \(S\) and \(S_{C}\) be the solutions sets of the linear systems \(AX=B\) and \((CA)X=CB\), respectively. If \(s\in S\), then \(As=B\), which implies \((CA)s=CB\). This shows that \(s\in S_{C}\) and \(S\subseteq S_{C}\). For the converse, if \(s\in S_{C}\), then \((CA)s=CB\), which implies
\[As=C^{-1}(CAs)=C^{-1}(CB)=B.\]
This shows that \(s\in S\) and \(S_{C}\subseteq S\). Therefore, we obtain \(S=S_{C}\), and the proof is complete. \(\blacksquare\)
Proposition. Let \(AX=B\) be a system of \(m\) linear equations in \(n\) unknowns. If the augmented matrix \((A’|B’)\) is obtained from \((A|B)\) by a finite number of elementary row operations, then the linear systems \(A’X=B’\) and \(AX=B\) have the same solutions.
Proof. Since \((A’|B’)\) is obtained from \((A|B)\) by a finite number of elementary row operations, by Proposition \ref{lap176}, there exist \(p\) \(m\times m\)
elementary matrices \(E_{1},E_{2},\cdots ,E_{p}\) satisfying
\[(A’|B’)=E_{p}E_{p-1}\cdots E_{1}(A|B)\equiv C(A|B)=(CA|CB),\]
where \(C=E_{p}E_{p-1}\cdots E_{1}\). Therefore, we obtain \(A’=CA\) and \(B’=CB\). Since each \(E_{i}\) is invertible, the matrix \(C\) is also invertible. By Proposition \ref{lap211}, we conclude that the linear systems \(A’X=B’\) and \(AX=B\) have the same solutions. This completes the proof. \(\blacksquare\)
Let \((A|B)\) be the augmented matrix of the linear system \(AX=B\). The procedure by performing the elementary row operations to transform \((A|B)\) into a row-reduced echelon matrix is called the Gaussian elimination.
\begin{equation}{\label{lae212}}\tag{31}\mbox{}\end{equation}
Example \ref{lae212}. We consider the following system of linear equations
\[\begin{array}{ccccccccccc}
2x_{1} & + & 3x_{2} & + & x_{3} & + & 4x_{4} & – & 9x_{5} & = & 17\\
x_{1} & + & x_{2} & + & x_{3} & + & x_{4} & – & 3x_{5} & = & 6\\
x_{1} & + & x_{2} & + & x_{3} & + & 2x_{4} & – & 5x_{5} & = & 8\\
2x_{1} & + & 2x_{2} & + & 2x_{3} & + & 3x_{4} & – & 8x_{5} & = & 14
\end{array}.\]
The augmented matrix is given by
\[\left [\begin{array}{rrrrr}
2 & 3 & 1 & 4 & -9\\
1 & 1 & 1 & 1 & -3\\
1 & 1 & 1 & 2 & -5\\
2 & 2 & 2 & 3 & -8
\end{array}\right .\left |\left .\begin{array}{r}
17\\ 6\\ 8\\ 14
\end{array}\right .\right ].\]
Using the Gaussian elimination, we obtain
\begin{eqnarray*}&& \left [\begin{array}{rrrrr}2 & 3 & 1 & 4 & -9\\1 & 1 & 1 & 1 & -3\\1 & 1 & 1 & 2 & -5\\2 & 2 & 2 & 3 & -8\end{array}\right .\left |\left .\begin{array}{r}17\\ 6\\ 8\\ 14\end{array}\right .\right ]\longrightarrow\left [\begin{array}{rrrrr}1 & 1 & 1 & 1 & -3\\2 & 3 & 1 & 4 & -9\\1 & 1 & 1 & 2 & -5\\2 & 2 & 2 & 3 & -8\end{array}\right .\left |\left .\begin{array}{r}6\\ 17\\ 8\\ 14\end{array}\right .\right ]\longrightarrow\\&& \left [\begin{array}{rrrrr}1 & 1 & 1 & 1 & -3\\0 & 1 & -1 & 2 & -3\\0 & 0 & 0 & 1 & -2\\0 & 0 & 0 & 1 & -2\end{array}\right .\left |\left .\begin{array}{r}6\\ 5\\ 2\\ 2\end{array}\right .\right ]\longrightarrow\left [\begin{array}{rrrrr}1 & 1 & 1 & 1 & -3\\0 & 1 & -1 & 2 & -3\\0 & 0 & 0 & 1 & -2\\0 & 0 & 0 & 0 & 0\end{array}\right .\left |\left .\begin{array}{r}6\\ 5\\ 2\\ 0\end{array}\right .\right ]\longrightarrow\\&& \left [\begin{array}{rrrrr}1 & 1 & 1 & 0 & -1\\0 & 1 & -1 & 0 & 1\\0 & 0 & 0 & 1 & -2\\0 & 0 & 0 & 0 & 0\end{array}\right .\left |\left .\begin{array}{r}4\\ 1\\ 2\\ 0\end{array}\right .\right ]\longrightarrow\left [\begin{array}{rrrrr}1 & 0 & 2 & 0 & -2\\0 & 1 & -1 & 0 & 1\\0 & 0 & 0 & 1 & -2\\0 & 0 & 0 & 0 & 0\end{array}\right .\left |\left .\begin{array}{r}3\\ 1\\ 2\\ 0\end{array}\right .\right ]\end{eqnarray*}
The last matrix corresponds to the following system of linear equations
\[\begin{array}{ccccccccccc}
2x_{1} &&& + & 2x_{3} &&& – & 2x_{5} & = & 3\\
&& x_{2} & – & x_{3} &&& + & x_{5} & = & 1\\
&&&&&& x_{4} & – & 2x_{5} & = & 2
\end{array}.\]
In order to solve a system of linear equations for which the augmented matrix is transformed into a row-reduced echelon matrix, we consider the set of variables that appear as leftmost variables in each one of the linear equations of the system. In this example, the set is \(\{x_{1},x_{2},x_{4}\}\). For the remaining variables, we assign the parametric values \(t_{1},t_{2},\cdots\). In this example, we assign \(x_{3}=t_{1}\) and \(x_{5}=t_{2}\). Then, we can solve it for the variables of the set in terms of the assigned values of remaining variables. In this example, we obtain
\begin{eqnarray*}&& x_{1}=-2x_{3}+2x_{5}+3=-2t_{1}+2t_{2}+3\\&& x_{2}=x_{3}-x_{5}+1=t_{1}-t_{2}+1\\&& x_{3}=2x_{5}+2=2t_{2}+2.\end{eqnarray*}
Therefore, the solution has the form of
\[\left [\begin{array}{c}
x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}
\end{array}\right ]=\left [\begin{array}{r}
-2t_{1}+2t_{2}+3\\ t_{1}-t_{2}+1\\ t_{1}\\ 2t_{2}+2\\ t_{2}
\end{array}\right ]=\left [\begin{array}{r}
3\\ 1\\ 0\\ 2\\ 0
\end{array}\right ]+t_{1}\left [\begin{array}{r}
-2\\ 1\\ 1\\ 0\\ 0
\end{array}\right ]+t_{2}\left [\begin{array}{r}
2\\ -1\\ 0\\ 2\\ 1
\end{array}\right ],\]
where \(t_{1},t_{2}\in\mathbb{R}\). \(\sharp\)
Inspired by Example \ref{lae212}, we see that if the only non-zero entry of a row in the row-reduced echelon matrix \((A’|B’)\) lies in the last column, then
the linear system \(AX=B\) has no solution. Suppose that the system \(AX=B\) consists of \(m\) linear equations in \(n\) unknowns. If the solution of the linear system \(AX=B\) exists, then a {\bf general solution} of \(AX=B\) has the following form
\begin{equation}{\label{laeq213}}\tag{32}
{\bf x}={\bf x}_{0}+t_{1}{\bf u}_{1}+t_{2}{\bf u}_{2}+\cdots +t_{n-r}{\bf u}_{n-r},
\end{equation}
where \(r\) is the number of non-zero rows in \((A’|B’)\) with \(r\leq m\). In other words, an arbitrary solution of \(AX=B\) can be expressed in terms of \(n-r\) parameters.
Determinants.
We first consider the determinants of order \(2\). Let
\[A=\left [\begin{array}{cc}
a & b \\ c & d
\end{array}\right ]\]
be a \(2\times 2\) matrix. Its determinant is denoted and defined by
\[\det (A)=\left |\begin{array}{cc}
a & b \\ c & d
\end{array}\right |=ad-bc.\]
For the \(3\times 3\) matrix
\[A=\left [\begin{array}{ccc}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{array}\right ].\]
We define its determinant according to the formulas called the expansion by row. Suppose that we consider the expansion by the first row. Then, the determinant \(\det (A)\) is defined by
\begin{align*} \det (A) & =\left |\begin{array}{ccc}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{array}\right |\\ & =a_{11}\cdot\left |\begin{array}{cc}
a_{22} & a_{23}\\
a_{32} & a_{33}
\end{array}\right |-a_{12}\cdot\left |\begin{array}{cc}
a_{21} & a_{23}\\
a_{31} & a_{33}
\end{array}\right |+a_{13}\cdot\left |\begin{array}{cc}
a_{21} & a_{22}\\
a_{31} & a_{32}
\end{array}\right |.\end{align*}
Let \(A^{(ij)}\) be the matrix obtained from \(A\) by deleting the \(i\)th row and the \(j\)th column. Then, the determinant can be expressed as
\[\det (A)=a_{11}\cdot\det (A^{(11)})-a_{12}\cdot\det (A^{(12)})+a_{13}\cdot\det (A^{(13)}).\]
We can also consider the expansion by the second row. In this case, the determinant is given by
\[\det (A)=-a_{21}\cdot\det (A^{(21)})+a_{22}\cdot\det (A^{(22)})-a_{23}\cdot\det (A^{(23)}).\]
We see that the sign is different. In general, the sign is determined according to the following pattern
\[A=\left [\begin{array}{ccc}
+ & – & +\\
– & + & -\\
+ & – & +
\end{array}\right ].\]
Therefore, we can similarly consider the expansion by the third row, which is given by
\[\det (A)=a_{31}\cdot\det (A^{(31)})-a_{32}\cdot\det (A^{(32)})+a_{33}\cdot\det (A^{(33)}).\]
On the other hand, we can also consider the expansion by columns. For example, if we consider the expansion by the first column, then we have
\[\det (A)=a_{11}\cdot\det (A^{(11)})-a_{21}\cdot\det (A^{(21)})+a_{31}\cdot\det (A^{(31)}).\]
In general, the determinant of order \(n\), i.e., the determinant of \(n\times n\) matrix is defined by
\begin{align*}
\det (A) & =(-1)^{i+1}\cdot a_{i1}\cdot\det (A^{(i1)})+(-1)^{i+2}\cdot a_{i2}\cdot\det (A^{(i2)})+\cdots+(-1)^{i+n}\cdot a_{in}\cdot\det (A^{(in)})\\
& =\sum_{j=1}^{n}(-1)^{i+j}\cdot a_{ij}\cdot\det (A^{(ij)}),
\end{align*}
which can be regarded as the expansion by the \(i\)th row. Also, the expansion by the \(j\)th column is given by
\begin{align*}
\det (A) & =(-1)^{j+1}\cdot a_{1j}\cdot\det (A^{(1j)})+(-1)^{j+2}\cdot a_{2j}\cdot\det (A^{(2j)})+\cdots+(-1)^{j+n}\cdot a_{nj}\cdot\det (A^{(nj)})\\
& =\sum_{i=1}^{n}(-1)^{i+j}\cdot a_{ij}\cdot\det (A^{(ij)}).
\end{align*}
Let \(A_{\cdot j}\) denote the \(j\)th column of \(A\). Then we also write
\[\det (A)=\det\left (A_{\cdot 1},A_{\cdot 2},\cdots ,A_{\cdot n}\right ).\]
Proposition. We have the following properties.
(i) Let \(A\) be the unit matrix, i.e., \(a_{ii}=1\) for \(i=1,\cdots ,n\) and \(a_{ij}=0\) for \(i\neq j\) and \(i,j=1,\cdots ,n\). Then \(\det (A)=1\).
(ii) If any two rows of a matrix are equal, then its determinant is zero. If any two columns of a matrix are equal, then its determinant is also zero.
(iii) If we interchange any two rows of a matrix, then the determinant changes by a sign. If we interchange any two columns of a matrix, then the determinant also changes by a sign.
(iv) If we add a scalar multiple of one column to another column, then the determinate does not change. If we add a scalar multiple of one row to another row, then the determinate does not also change.
(v) If \(B\) is a matrix obtained by multiplying a row of \(A\) by a non-zero scalar \(k\), then \(\det (B)=k\cdot\det (A)\). If \(B\) is a matrix obtained by multiplying a column of \(A\) by a non-zero scalar \(k\), then we also have \(\det (B)=k\cdot\det (A)\).
Proof. If \(A\) is a unit matrix, then \(\det (A)=(-1)^{i+1}a_{ii}\det (A^{(ii)})\), where \(A^{(ii)}\) is also an \((n-1)\times (n-1)\) unit matrix. Therefore, by induction, we can obtain \(\det (A)=1\), which proves part (i). \(\blacksquare\)
Example. We wish to compute the determinant
\[\left |\begin{array}{rrrr}
1 & 3 & 1 & 1\\ 2 & 1 & 5 & 2\\ 1 & -1 & 2 & 3\\ 4 & 1 & -3 & 7
\end{array}\right |.\]
We add the third row to the second row, and then add the third row to the fourth row. Then, we have
\[\left |\begin{array}{rrrr}
1 & 3 & 1 & 1\\ 3 & 0 & 7 & 5\\ 1 & -1 & 2 & 3\\ 4 & 1 & -3 & 7
\end{array}\right |=\left |\begin{array}{rrrr}
1 & 3 & 1 & 1\\ 3 & 0 & 7 & 5\\ 1 & -1 & 2 & 3\\ 5 & 0 & -1 & 10
\end{array}\right |.\]
Three times the third row and add it to the first row. We obtain
\[\left |\begin{array}{rrrr}
4 & 0 & 7 & 0\\ 3 & 0 & 7 & 5\\ 1 & -1 & 2 & 3\\ 5 & 0 & -1 & 10
\end{array}\right |.\]
Then, we expand it according to the second column to obtain
\[\left |\begin{array}{rrr}
4 & 7 & 0\\ 3 & 7 & 5\\ 5 & -1 & 10
\end{array}\right |.\]
Finally, we subtract twice the second row from the first row, and then from the third row, which obtains
\[\left |\begin{array}{rrr}
-2 & 7 & 0\\ 3 & 7 & 5\\ -1 & -15 & 0
\end{array}\right |.\]
We expand according to the third column to obtain \(-5\cdot (30-7)=-115\). \(\sharp\)
Example. If \(x_{1},\cdots ,x_{n}\) are real numbers. The following matrix
\[V_{n}=\left [\begin{array}{rrrr}
1 & x_{1} & \cdots & x_{1}^{n-1}\\
1 & x_{2} & \cdots & x_{2}^{n-1}\\
\vdots & \vdots & \cdots & \vdots\\
1 & x_{n} & \cdots & x_{n}^{n-1}\\
\end{array}\right ]\]
is called the Vandermonde matrix. Using the induction on \(n\), we can obtain
\[\det (V_{n})=\prod_{i<j}\left (x_{j}-x_{i}\right ).\]
To perform the induction, we can multiply each column by \(x_{1}\) and subtract it from the next column on the right, starting from the right-hand side. Then, we can obtain
\[V_{n}=\left (x_{n}-x_{1}\right )\cdot\left (x_{n-1}-x_{1}\right )\cdots\left (x_{2}-x_{1}\right )\cdot V_{n-1}.\]
Proposition. We have the following properties.
(i) If \(A\) is an \(n\times n\) matrix, then \(\det (A)=\det (A^{\top})\).
(ii) If \(A\) and \(B\) are \(n\times n\) matrices, then \[\det (AB)=\det (A)\cdot\det (B).\]
(iii) If the \(n\times n\) matrix \(A\) is invertible, then \[\det (A^{-1})=\frac{1}{\det (A)}.\]
Proof. Part (iii) can be obtained by the following equalities
\[1=\det (I)=\det (AA^{-1})=\det (A)\cdot\det (A^{-1}).\]
\begin{equation}{\label{lat36}}\tag{33}\mbox{}\end{equation}
Theorem \ref{lat36}. Let \(A\) be an \(n\times n\) matrix such that \(\det (A)\neq 0\). Then \(A\) is invertible. The \(ij\)th entry \(a_{ij}^{*}\) of \(A^{-1}\) is given by
\[a_{ij}^{*}=\frac{(-1)^{i+j}\cdot\det (A^{(ji)})}{\det (A)}.\]
\begin{equation}{\label{lat29}}\tag{34}\mbox{}\end{equation}
Theorem \ref{lat29}. We have the following properties.
(i) Let \(v,w\in\mathbb{R}^{2}\). The area of parallelogram spanned by \(v\) and \(w\) is \(|\det (v,w)|\).
(ii) Let \(u,v,w\in\mathbb{R}^{3}\). The volume of box spanned by \(u\), \(v\) and \(w\) is \(|\det (u,v,w)|\).
The properties of determinant can be used to solve the system of linear equations as shown below
\[\begin{array}{ccc}
a_{11}\cdot x_{1}+a_{12}\cdot x_{2}+\cdots +a_{1n}\cdot x_{n} & = & b_{1}\\
a_{21}\cdot x_{1}+a_{22}\cdot x_{2}+\cdots +a_{2n}\cdot x_{n} & = & b_{2}\\
\vdots & \vdots & \vdots\\
a_{n1}\cdot x_{1}+a_{n2}\cdot x_{2}+\cdots +a_{nn}\cdot x_{n} & = & b_{n}.
\end{array}\]
\begin{equation}{\label{lat28}}\tag{35}\mbox{}\end{equation}
Theorem \ref{lat28}. The coefficients of the system of linear equations can be written as a matrix
\[A=\left [\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & \cdots & a_{1n}\\
a_{21} & a_{22} & a_{23} & \cdots & a_{2n}\\
\vdots & \vdots & \vdots & \cdots & \vdots\\
a_{n1} & a_{n2} & a_{n3} & \cdots & a_{nn}
\end{array}\right ]\]
and let \(B\) be the column vector of the right-hand side of the system. If \(\det (A)\neq 0\), then
\[x_{j}=\frac{\det\left (A_{\cdot 1},\cdots ,B,\cdots ,A_{\cdot n}\right )}{\det (A)}\]
for \(j=1,\cdots ,n\), where \(B\) occurs in the \(j\)th column instead of \(A_{\cdot j}\) for \(j=1,\cdots ,n\).
The rule given in Theorem \ref{lat28} for obtaining the solution of system of linear equations by means of determinants is known as Cramer’s rule.
Example. Solve the following system of linear equations:
\begin{eqnarray*}&& 3x+2y+4z=1\\&& 2x-y+z=0\\&& x+2y+3z=1.\end{eqnarray*}
We have
\[x=\frac{\left |\begin{array}{rrr}
1 & 2 & 4\\ 0 & -1 & 1\\ 1 & 2 & 3
\end{array}\right |}{\left |\begin{array}{rrr}
3 & 2 & 4\\ 2 & -1 & 1\\ 1 & 2 & 3
\end{array}\right |}=-\frac{1}{5}, y=\frac{\left |\begin{array}{rrr}
3 & 1 & 4\\ 2 & 0 & 1\\ 1 & 1 & 3
\end{array}\right |}{\left |\begin{array}{rrr}
3 & 2 & 4\\ 2 & -1 & 1\\ 1 & 2 & 3
\end{array}\right |}=0\mbox{ and }z=\frac{\left |\begin{array}{rrr}
3 & 2 & 1\\ 2 & -1 & 0\\ 1 & 2 & 1
\end{array}\right |}{\left |\begin{array}{rrr}
3 & 2 & 4\\ 2 & -1 & 1\\ 1 & 2 & 3
\end{array}\right |}=-\frac{2}{5}.\]
