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Mathematics Question on Matrices

Let A=(1−11\[0.3em]21−3\[0.3em]111)A=\begin{pmatrix} 1&-1 &1 \\\[0.3em] 2 &1 &-3 \\\[0.3em] 1 &1&1 \end{pmatrix} and  10 B=(422\[0.3em]−50α\[0.3em]1−23)\,10\,B=\begin{pmatrix} 4&2 &2 \\\[0.3em] -5 &0 & \alpha \\\[0.3em] 1 &-2&3 \end{pmatrix} . If B is the inverse of A , then α\alpha is

A

2

B

1

C

-2

D

5

Answer

5

Explanation

Solution

Since B = A−1^{-1} ∴\therefore BA = A−1^{-1} A = I ∴\therefore (10 B) A = 10 I = [1000 0100 0010]\begin{bmatrix}10&0&0\\\ 0&10&0\\\ 0&0&10\end{bmatrix} Also (10 B) A=[422\[0.3em]−50α\[0.3em]1−23][1−11\[0.3em]21−3\[0.3em]111] = \begin{bmatrix} 4&2 &2 \\\[0.3em] -5 &0 & \alpha \\\[0.3em] 1 &-2&3 \end{bmatrix} \begin{bmatrix} 1&-1 &1 \\\[0.3em] 2 &1 &-3 \\\[0.3em] 1 &1&1 \end{bmatrix} = [1000 −5+α5+α−5+α 0010]=\begin{bmatrix}10&0&0\\\ -5 + \alpha &5 + \alpha&-5 + \alpha\\\ 0&0&10\end{bmatrix} = 101 if α\alpha = 5 Hence α\alpha = 5