Question

Let $A=\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right]$ and $10B=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]$. If B is the inverse of A, then find the value $\alpha$.

Answer 1

$GivenA=\left[\begin{array}{ccc}1& -1& 1\\ 2& -1& 3\\ 1& 1& 1\end{array}\right]and10B=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]andBisinverseofAi.e{A}^{-1}=B\Rightarrow A^{-1}A=BABA=I\Rightarrow 10BA=10I\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]\left[\begin{array}{ccc}1& -1& 1\\ 2& -1& 3\\ 1& 1& 1\end{array}\right]=\left[\begin{array}{ccc}10& 0& 0\\ 0& 10& 0\\ 0& 0& 10\end{array}\right]$

$\left[\begin{array}{ccc}10& -4& 12\\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0& 4& 1\end{array}\right]=\left[\begin{array}{ccc}10& 0& 0\\ 0& 10& 0\\ 0& 0& 10\end{array}\right]$

$comparing{a}_{21}onbothsides-5+\alpha =0Hence\alpha =5$