Question

If $M$ is a $3\times 3$ matrix satisfying $M\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right],M\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right],M\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 12\end{array}\right]$, then the sum of the diagonal entries of $M$ is

• A. $9$
• B. $8$
• C. $10$
• D. $11$

Answer 1

The correct option is A

$9$

The explanation for the correct option

Let us assume that, $M=\left[\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right]$.

It is given that, $M\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right],M\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right],M\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 12\end{array}\right]$.

Now, $\left[\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right]\times \left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right]$

$$\begin{gathered} \Rightarrow \left[\begin{array}{c}0+a_2+0\\ 0+b_2+0\\ 0+c_2+0\end{array}\right]=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}{a}_2\\ b_2\\ c_2\end{array}\right]=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right] \end{gathered}$$

Compare both sides of the equation.

Thus, $a_2=-1$, $b_2=2$ and $c_2=3$.

Now, $\left[\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right]\times \left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}{a}_1-a_2+0\\ b_1-b_2+0\\ c_1-c_2+0\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$

Put the values $a_2=-1$, $b_2=2$ and $c_2=3$.

$$\begin{gathered} \Rightarrow \left[\begin{array}{c}{a}_1-\left(-1\right)\\ b_1-2\\ c_1-3\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}{a}_1\\ b_1\\ c_1\end{array}\right]=\left[\begin{array}{c}1-1\\ 1+2\\ -1+3\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}{a}_1\\ b_1\\ c_1\end{array}\right]=\left[\begin{array}{c}0\\ 3\\ 2\end{array}\right] \end{gathered}$$

Compare both sides of the equation.

Thus, $a_1=0$, $b_1=3$ and $c_1=2$.

Now, $\left[\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right]\times \left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 12\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}{a}_1+a_2+a_3\\ b_1+b_2+b_3\\ c_1+c_2+c_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 12\end{array}\right]$

Put the values $a_1=0$, $b_1=3$, $c_1=2$, $a_2=-1$, $b_2=2$ and $c_2=3$.

$$\begin{gathered} \Rightarrow \left[\begin{array}{c}0-1+a_3\\ 3+2+b_3\\ 2+3+c_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 12\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}-1+a_3\\ 5+b_3\\ 5+c_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 12\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}{a}_3\\ b_3\\ c_3\end{array}\right]=\left[\begin{array}{c}0-\left(-1\right)\\ 0-5\\ 12-5\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}{a}_3\\ b_3\\ c_3\end{array}\right]=\left[\begin{array}{c}1\\ -5\\ 7\end{array}\right] \end{gathered}$$

Compare both sides of the equation.

Thus, $a_3=1$, $b_3=-5$ and $c_3=7$.

Therefore, the sum of the diagonal entries of $M$ can be given by, $a_1+b_2+c_3=0+2+7=9$.

Hence, (A) is the correct option.