Question

If $A=\left[\begin{array}{ccc}x& 1& -x\\ 0& 1& -1\\ x& 0& 7\end{array}\right]$ and $det\left(A\right)=\left|\begin{array}{ccc}3& 0& 1\\ 2& -1& 0\\ 0& 6& 7\end{array}\right|$ then the value of $x$ is

• A. $-3$
• B. $3$
• C. $2$
• D. $-8$
• E. $-2$

Answer 1

The correct option is A $-3$

Given, $A=\left[\begin{array}{ccc}x& 1& -x\\ 0& 1& -1\\ x& 0& 7\end{array}\right]$. Then

det $\left(A\right)=x7-1(0+x)-x(0-x)$

$=7x-x+x^2=x^2+6x$ $...\left(i\right)$
Also $det\left(A\right)=\left|\begin{array}{ccc}3& 0& 1\\ 2& -1& 0\\ 0& 6& 7\end{array}\right|$
$=3(-7-0)-0+1\left(12\right)=-9$ $...\left(ii\right)$
From Eqs(i) and (ii), we get
$x^2+6x=-9$
$\Rightarrow x^2+6x+9=0\Rightarrow {(x+3)}^2=0$
$x+3=0$
$x=-3$