# Get Answers to all your Questions

### Answers (1)

Answer:

$A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 2 & 3 \\ -1 & -5 \end{array}\right]$

Hint:

Here, we use basic concept of determinant and inverse of matrix

$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$

Given:

$A=\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]$

Solution:

$A^{2}=\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]=\left[\begin{array}{ll} 25-3 & 15-6 \\ -5+2 & -3+4 \end{array}\right]=\left[\begin{array}{cc} 22 & 9 \\ -3 & 1 \end{array}\right]$

Now,

\begin{aligned} &A^{2}-3 A-7 I=0 \\\\ &{\left[\begin{array}{ll} 22 & 9 \\ -3 & 1 \end{array}\right]-3\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]-7\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]} \\\\ &{\left[\begin{array}{ll} 22-15-7 & 9-9-0 \\ -3+3-0 & 1+6-7 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]} \\\\ &A^{2}-3 A-7 I=0 \end{aligned}

Multiply $A^{-1}$  both side

\begin{aligned} &A \times A \times A^{-1}-3 A \times A^{-1}-7 L A^{-1}=0 \\\\ &A-3 I-7 A^{-1}=0 \\\\ &7 A^{-1}=A-3 I \end{aligned}

\begin{aligned} &=\frac{1}{7}\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]-3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\\\ &A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 2 & 3 \\ -1 & -5 \end{array}\right] \end{aligned}

View full answer

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support