Question

If the value of $x$ for which the matrix
$A=\left[\begin{array}{ccc}2& 0& 7\\ 0& 1& 0\\ 1& -2& 1\end{array}\right]$ is the inverse of
$B=\left[\begin{array}{ccc}-x& 14x& 7x\\ 0& 1& 0\\ x& -4x& -2x\end{array}\right]$ is $k,$ then $10k$ is

Answer 1

We have,

$B=\left[\begin{array}{ccc}-x& 14x& 7x\\ 0& 1& 0\\ x& -4x& -2x\end{array}\right]$

$|B|=-x(-2x)+x(-7x)=-5x^2$

$adjB={\left[\begin{array}{ccc}-2x& 0& -x\\ 0& -5x^2& 10x^2\\ -7x& 0& -x\end{array}\right]}^T$

$=\left[\begin{array}{ccc}-2x& 0& -7x\\ 0& -5x^2& 0\\ -x& 10x^2& -x\end{array}\right]$

$B^{-1}=\frac{adjB}{|B|}$

$B^{-1}=\frac{1}{-5x^2}\left[\begin{array}{ccc}-2x& 0& -7x\\ 0& -5x^2& 0\\ -x& 10x^2& -x\end{array}\right]$

$B^{-1}=\left[\begin{array}{ccc}2/5x& 0& 7/5x\\ 0& 1& 0\\ 1/5x& -2& 1/5x\end{array}\right]$

Now,

Since $A$ matrix is the inverse of $B$ matrix,

Therefore, $B^{-1}=A$

$\Rightarrow \left[\begin{array}{ccc}2/5x& 0& 7/5x\\ 0& 1& 0\\ 1/5x& -2& 1/5x\end{array}\right]=\left[\begin{array}{ccc}2& 0& 7\\ 0& 1& 0\\ 1& -2& 1\end{array}\right]$

By comparing we get
$x=\frac{1}{5}=k$
$\Rightarrow 10k=2$