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### Answers (1)

Answer:

$A^{-1}= A^{3}$

Hint:

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

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

Given:

$A=\left[\begin{array}{rrr} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]$

Solution:

$A^{-1}= A^{3}$

RHS

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

Let’s find  $|A| \& \operatorname{Adj}(A)$

$|A|=3+6-8=1$

Cofactor of A

\begin{aligned} &C_{11}=1, C_{21}=-1, C_{31}=0 \\ &C_{12}=-2, C_{22}=3, C_{32}=-4 \\ &C_{13}=-2, C_{23}=3, C_{33}=-3 \end{aligned}

\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{1}\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right] \end{aligned}                         (1)

RHS

\begin{aligned} &A^{3}=A^{2} \times A \\ &A^{2}=\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 9-6+0 & -9+9-4 & 12-12+4 \\ 6-6+0 & -6+9-4 & 8-12+4 \\ 0-2+0 & 0+3-1 & 0-4+1 \end{array}\right] \\ &\end{aligned}

$=\left[\begin{array}{lll} 3 & -4 & 4 \\ 0 & -1 & 0 \\ 2 & 2 & 3 \end{array}\right]$

\begin{aligned} &A^{3}=A^{2} \times A=\left[\begin{array}{ccc} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{array}\right]\left[\begin{array}{ccc} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right] \\ & \end{aligned}

$=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right]$                                                       (2)

So, here equation (1) and (2)

LHS = RHS

$A^{-1}=A^{3}$

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