Section 5.8

The number \(1\) is the multiplicative identity since

\[ 1\cdot a=a\cdot1=a \]

For matrices we have the same identity.

\(I_{2}\) represents the \(2\times2\) identity matrix and is defined as $\( I_{2}=\left[\begin{matrix}1 & 0\\ 0 & 1 \end{matrix}\right] \)$

For example:

Let \(A=\left[\begin{matrix}-5 & 2\\ 3 & 4 \end{matrix}\right]\). Compute \(A\cdot I_{2}\) and \(I_{2}\cdot A\).

\[\begin{align*} \left[\begin{matrix}-5 & 2\\ 3 & 4 \end{matrix}\right]\cdot\left[\begin{matrix}1 & 0\\ 0 & 1 \end{matrix}\right] & =\left[\begin{matrix}(-5)(1)+(2)(0) & (-5)(0)+(2)(1)\\ (3)(1)+(4)(0) & (3)(0)+(4)(1) \end{matrix}\right]\\ & =\left[\begin{matrix}-5 & 2\\ 3 & 4 \end{matrix}\right] \end{align*}\]

and

\[\begin{align*} \left[\begin{matrix}1 & 0\\ 0 & 1 \end{matrix}\right]\cdot\left[\begin{matrix}-5 & 2\\ 3 & 4 \end{matrix}\right] & =\left[\begin{matrix}(1)(-5)+(0)(3) & (1)(2)+(0)(4)\\ (0)(-5)+(1)(3) & (0)(2)+(1)(4) \end{matrix}\right]\\ & =\left[\begin{matrix}-5 & 2\\ 3 & 4 \end{matrix}\right] \end{align*}\]

Definition 64 (\(n\times n\) Identity Matrix)

The \(n\times n\) identity matrix is \(I_{n}\) and is defined as

\[\begin{split} I_{n}=\left[\begin{matrix}1 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0\\ \vdots & \vdots & a_{ij} & \vdots\\ 0 & 0 & \cdots & 1 \end{matrix}\right]. \end{split}\]

The element \(a_{ij}=1\) when \(i=j\) (the diagonal elements), and \(a_{ij}=0\) otherwise.

We say \(\frac{1}{a}\) is the multiplicative inverse of \(a\) provided \(a\ne0\). That is,

\[ \frac{1}{a}\cdot a=a\cdot\frac{1}{a}=1. \]

Definition 65 (Multiplicative Inverse of Matrices)

We say \(A^{-1}\) is the multiplicative inverse matrix of matrix \(A\) if and only if

\[ AA^{-1}=A^{-1}A=I_{n} \]

Example 120

Find \(B^{-1}\) if

\[\begin{split} B=\left[\begin{matrix}-4 & 2 & 0\\ 1 & -1 & 2\\ 0 & 1 & 4 \end{matrix}\right] \end{split}\]

Solution:

First, we will construct the augmented matrix

\[\begin{split} \left[\begin{matrix}-4 & 2 & 0 & 1 & 0 & 0\\ 1 & -1 & 2 & 0 & 1 & 0\\ 0 & 1 & 4 & 0 & 0 & 1 \end{matrix}\right] \end{split}\]

then perform reduce row echelon form on this new matrix

\[\begin{align*} \left[\begin{matrix}-4 & 2 & 0 & 1 & 0 & 0\\ 1 & -1 & 2 & 0 & 1 & 0\\ 0 & 1 & 4 & 0 & 0 & 1 \end{matrix}\right] & \to_{-\frac{1}{4}\times R_{1}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\ 1 & -1 & 2 & 0 & 1 & 0\\ 0 & 1 & 4 & 0 & 0 & 1 \end{matrix}\right]\\ & \to_{(-1)\times R_{1}+R_{2}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\ 0 & -\frac{1}{2} & 2 & \frac{1}{4} & 1 & 0\\ 0 & 1 & 4 & 0 & 0 & 1 \end{matrix}\right]\\ & \to_{-2\times R_{2}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\ 0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\ 0 & 1 & 4 & 0 & 0 & 1 \end{matrix}\right]\\ & \to_{(-1)\times R_{2}+R_{3}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\ 0 & 1 & -6 & -\frac{1}{2} & -2 & 0\\ 0 & 0 & 8 & \frac{1}{2} & 2 & 1 \end{matrix}\right]\\ & \to_{\frac{1}{8}\times R_{3}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\ 0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\ 0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8} \end{matrix}\right]\\ & \to_{4\times R_{3}+R_{2}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\ 0 & 1 & 0 & -\frac{1}{4} & -1 & \frac{1}{2}\\ 0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8} \end{matrix}\right]\\ & \to_{\frac{1}{2}\times R_{2}+R_{1}}\left[\begin{matrix}1 & 0 & 0 & -\frac{3}{4} & -\frac{1}{2} & \frac{1}{4}\\ 0 & 1 & 0 & -\frac{1}{4} & -1 & \frac{1}{2}\\ 0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8} \end{matrix}\right] \end{align*}\]

Therefore, \(B^{-1}=\left[\begin{matrix}-\frac{3}{4} & -\frac{1}{2} & \frac{1}{4}\\ -\frac{1}{4} & -1 & \frac{1}{2}\\ \frac{1}{16} & \frac{1}{4} & \frac{1}{8} \end{matrix}\right]\).

Example 121

Find \(A^{-1}\), if possible, given that

\[\begin{split} A=\left[\begin{matrix}4 & -2 & 5\\ 0 & 1 & 0\\ -8 & 4 & -10 \end{matrix}\right]. \end{split}\]

Solution:

First, we construct the augmented matrix and the perform reduced row echelon form

\[\begin{align*} \left[\begin{matrix}4 & -2 & 5 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & 0\\ -8 & 4 & -10 & 0 & 0 & 1 \end{matrix}\right] & \to_{\frac{1}{4}\times R_{1}}\left[\begin{matrix}1 & -\frac{1}{2} & \frac{5}{4} & \frac{1}{4} & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & 0\\ -8 & 4 & -10 & 0 & 0 & 1 \end{matrix}\right]\\ & \to_{8\times R_{1}+R_{3}}\left[\begin{matrix}1 & -\frac{1}{2} & \frac{5}{4} & \frac{1}{4} & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 2 & 0 & 1 \end{matrix}\right] \end{align*}\]

Since \(a_{31}=a_{32}=a_{33}=0\) we will not be able to reach the desired transformation. Therefore, there does not exists an \(A^{-1}\).

Theorem 14 (Solving Matrix Equation)

Suppose \(A\) is an \(n\times n\) matrix with inverse \(A^{-1}\), \(X\) is an \(n\times1\) matrix of variables, and \(B\) is an \(n\times1\) matrix. The matrix equation

\[ AX=B \]

has the solution

\[ X=A^{-1}B. \]

Example 122

Solve the system

\[\begin{split} \begin{cases} 2x-3y & =4\\ x+5y & =2 \end{cases} \end{split}\]

Solution:

First, we will write the system in matrix equation form

\[\begin{split} \left[\begin{matrix}2 & -3\\ 1 & 5 \end{matrix}\right]\cdot\left[\begin{matrix}x\\ y \end{matrix}\right]=\left[\begin{matrix}4\\ 2 \end{matrix}\right] \end{split}\]

where \(A=\left[\begin{matrix}2 & -3\\ 1 & 5 \end{matrix}\right]\), \(X=\left[\begin{matrix}x\\ y \end{matrix}\right]\), and \(B=\left[\begin{matrix}4\\ 2 \end{matrix}\right]\).

Next, we want to find \(A^{-1}\)

\[\begin{align*} \left[\begin{matrix}2 & -3 & 1 & 0\\ 1 & 5 & 0 & 1 \end{matrix}\right] & \to_{\frac{1}{2}\times R_{1}}\left[\begin{matrix}1 & -\frac{3}{2} & \frac{1}{2} & 0\\ 1 & 5 & 0 & 1 \end{matrix}\right]\\ & \to_{(-1)\times R_{1}+R_{2}}\left[\begin{matrix}1 & -\frac{3}{2} & \frac{1}{2} & 0\\ 0 & \frac{13}{2} & -\frac{1}{2} & 1 \end{matrix}\right]\\ & \to_{\frac{2}{13}\times R_{2}}\left[\begin{matrix}1 & -\frac{3}{2} & \frac{1}{2} & 0\\ 0 & 1 & -\frac{1}{13} & \frac{2}{13} \end{matrix}\right]\\ & \to_{\frac{3}{2}\times R_{2}+R_{1}}\left[\begin{matrix}1 & 0 & \frac{5}{13} & \frac{3}{13}\\ 0 & 1 & -\frac{1}{13} & \frac{2}{13} \end{matrix}\right]\implies A^{-1}=\left[\begin{matrix}\frac{5}{13} & \frac{3}{13}\\ -\frac{1}{13} & \frac{2}{13} \end{matrix}\right]. \end{align*}\]

Using the solution of the matrix equation theorem we have

\[\begin{align*} X & =A^{-1}B\\ & =\left[\begin{matrix}\frac{5}{13} & \frac{3}{13}\\ -\frac{1}{13} & \frac{2}{13} \end{matrix}\right]\cdot\left[\begin{matrix}4\\ 2 \end{matrix}\right]\\ & =\left[\begin{matrix}\frac{5}{13}\cdot 4+\frac{3}{13}\cdot 2\\ -\frac{1}{13}\cdot 4+\frac{2}{13}\cdot 2 \end{matrix}\right]\\ & =\left[\begin{matrix}2\\ 0 \end{matrix}\right]=\left[\begin{matrix}x\\ y \end{matrix}\right]=X \end{align*}\]

which means the solution set of the system is \(\{(2,0)\}\).

Example 123

Solve the system

\[\begin{split} \begin{cases} -4x+2y & =12\\ x-y+2z & =7\\ y+4z & =20 \end{cases} \end{split}\]

Solution:

First, we write the system in matrix equation form

\[\begin{split} \left[\begin{matrix}-4 & 2 & 0\\ 1 & -1 & 2\\ 0 & 1 & 4 \end{matrix}\right]\cdot\left[\begin{matrix}x\\ y\\ z \end{matrix}\right]=\left[\begin{matrix}12\\ 7\\ 20 \end{matrix}\right] \end{split}\]

where \(A=\left[\begin{matrix}-4 & 2 & 0\\ 1 & -1 & 2\\ 0 & 1 & 4 \end{matrix}\right]\), \(X=\left[\begin{matrix}x\\ y\\ z \end{matrix}\right]\), and \(B=\left[\begin{matrix}12\\ 7\\ 20 \end{matrix}\right]\).

Next, we will find \(A^{-1}\)

\[\begin{align*} \left[\begin{matrix}-4 & 2 & 0 & 1 & 0 & 0\\ 1 & -1 & 2 & 0 & 1 & 0\\ 0 & 1 & 4 & 0 & 0 & 1 \end{matrix}\right] & \to_{-\frac{1}{4}\times R_{1}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\ 1 & -1 & 2 & 0 & 1 & 0\\ 0 & 1 & 4 & 0 & 0 & 1 \end{matrix}\right]\\ & \to_{(-1)\times R_{1}+R_{2}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\ 0 & -\frac{1}{2} & 2 & \frac{1}{4} & 1 & 0\\ 0 & 1 & 4 & 0 & 0 & 1 \end{matrix}\right]\\ & \to_{-2\times R_{2}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\ 0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\ 0 & 1 & 4 & 0 & 0 & 1 \end{matrix}\right]\\ & \to_{(-1)\times R_{2}+R_{3}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\ 0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\ 0 & 0 & 8 & \frac{1}{2} & 2 & 1 \end{matrix}\right]\\ & \to_{\frac{1}{8}\times R_{3}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\ 0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\ 0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8} \end{matrix}\right]\\ & \to_{\frac{1}{2}\times R_{2}+R_{1}}\left[\begin{matrix}1 & 0 & -2 & -\frac{1}{2} & -1 & 0\\ 0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\ 0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8} \end{matrix}\right]\\ & \to_{2\times R_{3}+R_{1}}\left[\begin{matrix}1 & 0 & 0 & -\frac{3}{8} & -\frac{1}{2} & \frac{1}{4}\\ 0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\ 0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8} \end{matrix}\right]\\ & \to_{4\times R_{3}+R_{2}}\left[\begin{matrix}1 & 0 & 0 & -\frac{3}{8} & -\frac{1}{2} & \frac{1}{4}\\ 0 & 1 & 0 & -\frac{1}{4} & -1 & \frac{1}{2}\\ 0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8} \end{matrix}\right]\implies A^{-1}=\left[\begin{matrix}-\frac{3}{8} & -\frac{1}{2} & \frac{1}{4}\\ -\frac{1}{4} & -1 & \frac{1}{2}\\ \frac{1}{16} & \frac{1}{4} & \frac{1}{8} \end{matrix}\right] \end{align*}\]

Using the solution of the matrix theorem we have

\[\begin{align*} X & =A^{-1}B\\ & =\left[\begin{matrix}-\frac{3}{8} & -\frac{1}{2} & \frac{1}{4}\\ -\frac{1}{4} & -1 & \frac{1}{2}\\ \frac{1}{16} & \frac{1}{4} & \frac{1}{8} \end{matrix}\right]\cdot\left[\begin{matrix}12\\ 7\\ 20 \end{matrix}\right]\\ & =\left[\begin{matrix}-\frac{3}{8}\cdot 12+(-\frac{1}{2})\cdot 7+\frac{1}{4}\cdot 20\\ -\frac{1}{4}\cdot 12+(-1)\cdot 7+\frac{1}{2}\cdot 20\\ \frac{1}{16}\cdot 12+\frac{1}{4}\cdot 7+\frac{1}{8}\cdot 20 \end{matrix}\right]\\ & =\left[\begin{matrix}-3\\ 0\\ 5 \end{matrix}\right]=\left[\begin{matrix}x\\ y\\ z \end{matrix}\right]=X \end{align*}\]

Therefore, the solution to the system is \(\{(-3,0,5)\}\).