# Get Answers to all your Questions

### Answers (1)

Answer:

$A^{-1}=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right]$

Hint:

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

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

Given:

$A=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right]$

Solution:

\begin{aligned} &|A|=0\left|\begin{array}{rr} -3 & 4 \\ -3 & 4 \end{array}\right|-1\left|\begin{array}{ll} 4 & 4 \\ 3 & 4 \end{array}\right|+(-1)\left|\begin{array}{rr} 4 & -3 \\ 3 & -3 \end{array}\right| \\ &=0-1(16-12)-1(-12+9) \\ &=-4+3=-1 \end{aligned}

Hence $A^{-1}$  exist

Cofactor of A are

\begin{aligned} &C_{11}=0, C_{12}=-4, C_{13}=-3 \\ &C_{21}=-1, C_{22}=3, C_{23}=-4 \\ &C_{31}=-3, C_{32}=3, C_{33}=-4 \\ &A d j(A)=C_{{ij}}^{T} \end{aligned}

\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 0 & -1 & 1 \\ -4 & 3 & -4 \\ -3 & 3 & -4 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \end{aligned}

\begin{aligned} &=-\frac{1}{1}\left[\begin{array}{ccc} 0 & -1 & 1 \\ -4 & 3 & -4 \\ -3 & 3 & -4 \end{array}\right] \\ &A^{-1}=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right] \end{aligned}

View full answer

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

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support