Question

If inverse of $A=\left[\begin{array}{ccc}1& 1& 1\\ 2& -1& -1\\ 1& -1& 1\end{array}\right]$ is $\frac{-1}{6}\left[\begin{array}{ccc}-2& -2& 0\\ -3& 0& \alpha \\ -1& 2& -3\end{array}\right]$ then $\alpha =$

• A. $0$
• B. $-3$
• C. $3$
• D. $2$

Answer 1

The correct option is C $3$

Given,

$A=\left[\begin{array}{ccc}1& 1& 1\\ 2& -1& -1\\ 1& -1& 1\end{array}\right]$

$A^{-1}={\left[\begin{array}{ccc}1& 1& 1\\ 2& -1& -1\\ 1& -1& 1\end{array}\right]}^{-1}$

$=\left[\begin{array}{ccccccc}1& 1& 1& \mid & 1& 0& 0\\ 2& -1& -1& \mid & 0& 1& 0\\ 1& -1& 1& \mid & 0& 0& 1\end{array}\right]$

$R_1\leftrightarrow R_2$

$=\left[\begin{array}{ccccccc}2& -1& -1& \mid & 0& 1& 0\\ 1& 1& 1& \mid & 1& 0& 0\\ 1& -1& 1& \mid & 0& 0& 1\end{array}\right]$

$R_2\leftarrow R_2-\frac{1}{2}\cdot R_1$

$R_3\leftarrow R_3-\frac{1}{2}\cdot R_1$

$=\left[\begin{array}{ccccccc}2& -1& -1& \mid & 0& 1& 0\\ 0& \frac{3}{2}& \frac{3}{2}& \mid & 1& -\frac{1}{2}& 0\\ 0& -\frac{1}{2}& \frac{3}{2}& \mid & 0& -\frac{1}{2}& 1\end{array}\right]$

$R_3\leftarrow R_3+\frac{1}{3}\cdot R_2$

$=\left[\begin{array}{ccccccc}2& -1& -1& \mid & 0& 1& 0\\ 0& \frac{3}{2}& \frac{3}{2}& \mid & 1& -\frac{1}{2}& 0\\ 0& 0& 2& \mid & \frac{1}{3}& -\frac{2}{3}& 1\end{array}\right]$

$R_3\leftarrow \frac{1}{2}\cdot R_3$

$R_2\leftarrow R_2-\frac{3}{2}\cdot R_3$

$=\left[\begin{array}{ccccccc}2& -1& -1& \mid & 0& 1& 0\\ 0& \frac{3}{2}& 0& \mid & \frac{3}{4}& 0& -\frac{3}{4}\\ 0& 0& 1& \mid & \frac{1}{6}& -\frac{1}{3}& \frac{1}{2}\end{array}\right]$

$R_1\leftarrow R_1+1\cdot R_3$

$R_2\leftarrow \frac{2}{3}\cdot R_2$

$=\left[\begin{array}{ccccccc}2& -1& 0& \mid & \frac{1}{6}& \frac{2}{3}& \frac{1}{2}\\ 0& 1& 0& \mid & \frac{1}{2}& 0& -\frac{1}{2}\\ 0& 0& 1& \mid & \frac{1}{6}& -\frac{1}{3}& \frac{1}{2}\end{array}\right]$

$R_1\leftarrow R_1+1\cdot R_2$

$=\left[\begin{array}{ccccccc}2& 0& 0& \mid & \frac{2}{3}& \frac{2}{3}& 0\\ 0& 1& 0& \mid & \frac{1}{2}& 0& -\frac{1}{2}\\ 0& 0& 1& \mid & \frac{1}{6}& -\frac{1}{3}& \frac{1}{2}\end{array}\right]$

$R_1\leftarrow \frac{1}{2}\cdot R_1$

$=\left[\begin{array}{ccccccc}1& 0& 0& \mid & \frac{1}{3}& \frac{1}{3}& 0\\ 0& 1& 0& \mid & \frac{1}{2}& 0& -\frac{1}{2}\\ 0& 0& 1& \mid & \frac{1}{6}& -\frac{1}{3}& \frac{1}{2}\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{1}{3}& \frac{1}{3}& 0\\ \frac{1}{2}& 0& -\frac{1}{2}\\ \frac{1}{6}& -\frac{1}{3}& \frac{1}{2}\end{array}\right]$

$=-\frac{1}{6}\left[\begin{array}{ccc}-2& -2& 0\\ -3& 0& 3\\ -1& 2& -3\end{array}\right]$

$\therefore \alpha =3$