Question

Let 'M be a $3\times 3$ matrix such that $\left[012\right]M=\left[100\right]$ and $\left[345\right]M=\left[010\right],$ then $\left[678\right]$ is equal to

• A. $[21-2]$
• B. $\left[001\right]$
• C. $[-120]$
• D. $\left[9108\right]$

Answer 1

The correct option is C $[-120]$

Let $M$ be a $3\times 3$ matrix

$\left[\begin{array}{ccc}0& 1& 2\end{array}\right]m=\left[\begin{array}{ccc}1& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0& 1& 2\end{array}\right]\left[\begin{array}{ccc}{x}_1& x_2& x_3\\ x_4& x_5& x_6\\ x_7& x_8& x_9\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\end{array}\right]$

$\Rightarrow x_4+2x_7=1$ $\therefore$ solving for $m$

$x_5+2x_8=0$ like $\left(1\right)$

$x_6+2x_9=0$ we get

$\left[\begin{array}{ccc}6& 7& 8\end{array}\right]m=\left[\begin{array}{ccc}-1& 2& 0\end{array}\right]$

$\left[\begin{array}{ccc}3& 4& 5\end{array}\right]m=\left[\begin{array}{ccc}0& 1& 0\end{array}\right]$

$3x_1+4x_4+5x_7=0$

$3x_2+4x_5+5x_8=1$

$3x_3+4x_6+5x_9=0$

$\left[\begin{array}{ccc}6& 7& 8\end{array}\right]m=\left[\begin{array}{ccc}6x_1+7x_4+8x_7& 6x_2+7x_5+8x_8& 6x_3+7x_6+8x_9\end{array}\right]$

$6x_1+7x_4+8x_7=2(3x_1+4x_4+5x_7)-(x_4+2x_2)$

$=2\left(0\right)-1=-1-----\left(1\right)$