Chapter 11: Q. 89 (page 758)

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

Short Answer

Expert verified

The matrix A is nonsingular.

Step by step solution

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 $[A ∣ I_{2}]$ into its echelon form.

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

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

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

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

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

$R_{2} = \frac{a R_{2}}{a d - b c}$

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

$R_{1} = R_{1} - \frac{b R_{2}}{a}$

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

Step 2. Inverse of the matrix A

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

Here, let $D = a d - b c \neq 0$

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

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

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

Short Answer

Step by step solution

Step 1. Reduce the augmented matrix into its echelon form

Step 2. Inverse of the matrix A

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Consider the 2by 2squre matrixA=abcdIf D=ad-bc≠0, show that A is nonsingular and thatA−1=1Dd −b−c a

Short Answer

Step by step solution

Step 1. Reduce the augmented matrix into its echelon form

Step 2. Inverse of the matrix A

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Applied Mathematics

Statistics

Discrete Mathematics

Calculus

Probability and Statistics

Study anywhere. Anytime. Across all devices.

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