Question

If $A= \begin{bmatrix}1 &1&1\\ 1&2&3\\ 1&3&4\end{bmatrix} , B= \begin{bmatrix}7\\ 16\\ 22\end{bmatrix},X= \begin{bmatrix}x\\ y\\ z\end{bmatrix}$ and $AX = B$, then $z$ is equal to

• A. 1
• B. -1
• C. -3
• D. 3

Answer 1

Answer: (D)

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