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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 11: Q. 89 (page 758)

Consider the $2$ by $2$ squre matrix
$A = [\begin{matrix} a & b \\ c & d \end{matrix}]$
If $D = a d - b c 鈮� 0$ , show that A is nonsingular and that
$A^{鈭� 1} = \frac{1}{D} [\begin{array}{rr} d 鈥呪赌呪赌呪赌� 鈭� b \\ 鈭� c 鈥呪赌呪赌呪赌� a \end{array}]$

Short Answer

Expert verified

The matrix A is nonsingular.

Step by step solution

01

Step 1. Reduce the augmented matrix into its echelon form

We have the given matrix $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$

Let us reduce the augmented matrix $[A 鈭� I_{2}]$ into its echelon form.

$A = [\begin{array}{cccc} a & b & 1 & 0 \\ c & d & 0 & 1 \end{array}]$

Now perform the operation, $R_{1} = \frac{R_{1}}{2}$

$鈫� [\begin{array}{cccc} 1 & \frac{b}{a} & \frac{1}{a} & 0 \\ c & d & 0 & 1 \end{array}]$

$R_{2} = R_{2} - C R_{1}$

$鈫� [\begin{array}{cccc} 1 & \frac{b}{a} & \frac{1}{a} & 0 \\ 0 & \frac{a d 鈭� b c}{a} & 鈭� \frac{c}{a} & 1 \end{array}]$

$R_{2} = \frac{a R_{2}}{a d 鈭� b c}$

$鈫� [\begin{array}{cccc} 1 & \frac{b}{a} & \frac{1}{a} & 0 \\ 0 & 1 & \frac{c}{b c 鈭� a d} & \frac{a}{a d 鈭� b c} \end{array}]$

$R_{1} = R_{1} 鈭� \frac{b R_{2}}{a}$

$鈫� [\begin{array}{cccc} 1 & 0 & \frac{d}{a d 鈭� b c} & 鈭� \frac{b}{a d 鈭� b c} \\ 0 & 1 & c & \frac{a}{a d 鈭� b c} \end{array}]$

02

Step 2. Inverse of the matrix A

$A^{鈭� 1} = [\begin{array}{cc} \frac{d}{a d 鈭� b c} & 鈭� \frac{b}{a d 鈭� b c} \\ 鈭� \frac{c}{a d 鈭� b c} & \frac{a}{a d 鈭� b c} \end{array}] = \frac{1}{a d 鈭� b c} [\begin{array}{cc} d & 鈭� b \\ 鈭� c & a \end{array}]$

Here, let $D = a d 鈭� b c 鈮� 0$

Therefore, $A^{鈭� 1} = \frac{1}{D} [\begin{array}{cc} d & 鈭� b \\ 鈭� c & a \end{array}]$

Hence, we have proved that matrix A is a non-singular matrix.

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