Question

If $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 3& 4\end{array}\right],B=\left[\begin{array}{c}7\\ 16\\ 22\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and $AX=B$, then $z$ is equal to

• A. 1
• B. -1
• C. -3
• D. 3

Answer 1

Answer: (D)

We have , $AX=B$
$\Rightarrow \left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 3& 4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}7\\ 16\\ 22\end{array}\right]$
This have solution if $A^{-1}$ exists i.e.,$IAI\ne 0$
$IAI=1(8-9)-1(4-3)+1\cdot (3-2)=-1-1+1-1\ne 0$.
Hence, $A^{-1}$ exists.
Now, $X=A^{-1}B$
$\Rightarrow {\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 3& 4\end{array}\right]}^{-1}-\left[\begin{array}{c}7\\ 16\\ 22\end{array}\right]$
Cofactor matrix of $A=\left[\begin{array}{ccc}-1& -1& 1\\ -1& 3& -2\\ 1& -2& 1\end{array}\right]$
$\therefore adjA=\left[\begin{array}{ccc}-1& -1& 1\\ -1& 3& -2\\ 1& -2& 1\end{array}\right]$
Now, $A^{-1}=\frac{adjA}{|A|}=\left[\begin{array}{ccc}1& 1& -1\\ 1& -3& 2\\ -1& 2& -1\end{array}\right]$
$X=A^{-1}B=\left[\begin{array}{ccc}1& 1& -1\\ 1& -3& 2\\ -1& 2& -1\end{array}\right]\left[\begin{array}{c}7\\ 16\\ 22\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}7+16-22\\ 7-48+44\\ -7+32-22\end{array}\right]=\left[\begin{array}{c}1\\ 3\\ 3\end{array}\right]$
Hence , $z=3$