Let $A=\begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix}$ and $10B=\begin{bmatrix}4&2&2\\-5&0&α\\1&-2&3\end{bmatrix}$. If B is the inverse of A, then find the value of $α$.
Answer & explanation
Correct answer: option 3
Here,
$A=\begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix}$
$∴|A|=\begin{vmatrix}1&-1&1\\2&1&-3\\1&1&1\end{vmatrix}$
$= 1(1+3)+1(2+3)+1(2-1)$
$=4+5+1=10$
Now, $adj.A=\begin{bmatrix}4&-5&1\\2&0&-2\\2&5&3\end{bmatrix}^T=\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}$
$⇒B=A^{-1}=\frac{1}{10}\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}$
$⇒10B=\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}$
Hence, $α=5$.