# Let and  If B is the inverse of matrix  $A$ , then $\dpi{100} \alpha$ is Option 1) 2 Option 2) -1 Option 3) -2 Option 4) 5

As we learnt in

Inverse of a matrix -

$A^{-1}=\frac{1}{\left | A \right |}\cdot adjA$

-

$A= \begin{bmatrix} 1 & -1 &1 \\ 2& 1 &-3 \\ 1&1 &1 \end{bmatrix}$

$10(B)= \begin{bmatrix} 4 & 2 &2 \\ -5& 0 &\alpha \\ 1&-2 &3 \end{bmatrix}$

and B is the inverse of A.

$\left | A \right |= 4+ 1\times 5 +1\times1= 10$

$B= \frac{adj (A)}{\left | A \right |}= \frac{adj (A)}{10}$

So $10 B= adj (A)$

Now $adj (A) = \begin{bmatrix} 4 &-5 &1 \\ 2& 0 &-2 \\ 2& 5 &3 \end{bmatrix}^{T}$$= \begin{bmatrix} 4 &2 &2 \\ -5& 0 &5 \\ 1& -2 &3 \end{bmatrix}$

$\therefore \alpha = 5$

Option 1)

2

Option 2)

-1

Option 3)

-2

Option 4)

5

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