Question

Suppose the vectors $X_1$, $X_2$ and $X_3$ are the solutions of the system of linear equations, $Ax=b$ when the vector $b$ on the right side is equal to $b_1$, $b_2$ and $b_3$ respectively. if
$X_1=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],X_2=\left[\begin{array}{c}0\\ 2\\ 1\end{array}\right],X_3=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right],b_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],b_2=\left[\begin{array}{c}0\\ 2\\ 0\end{array}\right]$ and $b_3=\left[\begin{array}{c}0\\ 0\\ 2\end{array}\right]$, then the determinant of $A$ is equal to

• A. $2$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $4$

Answer 1

The correct option is A $2$

Let $A=\left[\begin{array}{ccc}{a}_1& a_2& a_3\\ a_{{}_4}& a_5& a_6\\ a_7& a_8& a_9\end{array}\right]$
Now, $A{X}_1=b_1$
$\Rightarrow \left[\begin{array}{ccc}{a}_1& a_2& a_3\\ a_{{}_4}& a_5& a_6\\ a_7& a_8& a_9\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$
$a_1+a_2+a_3=1$
$a_4+a_5+a_6=0$
$a_7+a_8+a_9=1$
Similarly,
$2a_2+a_3=0$
$2a_5+a_6=2$
$2a_8+a_9=0$
and
$a_3=0$
$a_6=0$
$a_9=2$
From the above equations, we have
$a_1=1,a_2=0,a_3=0$
$a_4=-1,a_5=1,a_6=0$
$a_7=-1,a_8=-1,a_9=2$

Now, $|A|=\left|\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ -1& -1& 2\end{array}\right|$
$\therefore \left|A\right|=1(2-0)=2$