Question

Let M be a $3\times 3$ matrix satisfying $M\left[ \begin{matrix}
   0 \\
   1 \\
   0 \\
\end{matrix} \right]=\left[ \begin{matrix}
   -1 \\
   2 \\
   3 \\
\end{matrix} \right]$, $M\left[ \begin{matrix}
   1 \\
   -1 \\
   0 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 \\
   1 \\
   -1 \\
\end{matrix} \right]$ and $M\left[ \begin{matrix}
   1 \\
   1 \\
   1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 \\
   0 \\
   12 \\
\end{matrix} \right]$ then, find the sum of the diagonal entries of M.

Answer 1

Hint: M be a $3\times 3$ matrix which satisfies the given matrix multiplication.
So, we try to assume the matrix M.
Let M be $\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]$. We need to find the sum of the diagonal entries of M which is $\left( a+e+i \right)$.

Complete step-by-step solution:
Now, we break down the matrix multiplication to get the equation of the unknowns.
We have $M\left[ \begin{matrix}
   0 \\
   1 \\
   0 \\
\end{matrix} \right]=\left[ \begin{matrix}
   -1 \\
   2 \\
   3 \\
\end{matrix} \right]$ which implies $\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]\left[ \begin{matrix}
   0 \\
   1 \\
   0 \\
\end{matrix} \right]=\left[ \begin{matrix}
   -1 \\
   2 \\
   3 \\
\end{matrix} \right]$.
Breaking this, we get $b\times 1=b=-1$, $e\times 1=e=2$, $h\times 1=h=3$. We got value of 3 unknowns. We also have $M\left[ \begin{matrix}
   1 \\
   -1 \\
   0 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 \\
   1 \\
   -1 \\
\end{matrix} \right]$ which implies $\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]\left[ \begin{matrix}
   1 \\
   -1 \\
   0 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 \\
   1 \\
   -1 \\
\end{matrix} \right]$.
Breaking this, we get $a\times 1+b\times \left( -1 \right)=a-b=1$, $d\times 1+e\times \left( -1 \right)=d-e=1$, $g\times 1+h\times \left( -1 \right)=g-h=-1$.
We put values of b, e, h to get values of a, d, h.
So, $a+1=1\Rightarrow a=0$, $d-2=1\Rightarrow d=3$, $g-3=-1\Rightarrow g=2$.
Lastly, we have $M\left[ \begin{matrix}
   1 \\
   1 \\
   1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 \\
   0 \\
   12 \\
\end{matrix} \right]$ which implies $\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]\left[ \begin{matrix}
   1 \\
   1 \\
   1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 \\
   0 \\
   12 \\
\end{matrix} \right]$.
Breaking this, we get $a\times 1+b\times 1+c\times 1=a+b+c=0$, $d\times 1+e\times 1+f\times 1=d+e+f=0$, $g\times 1+h\times 1+i\times 1=g+h+i=12$.
We put values of 6 values we know to get the others.
So, $0-1+c=0\Rightarrow c=1$, $3+2+f=0\Rightarrow f=-5$, $2+3+i=12\Rightarrow i=7$.
We got all the 9 unknowns to get the matrix $M=\left[ \begin{matrix}
   0 & -1 & 1 \\
   3 & 2 & -5 \\
   2 & 3 & 7 \\
\end{matrix} \right]$.
The sum of the diagonal entries of M is $\left( a+e+i \right)=0+2+7=9$.

Note: We need to remember that we can’t use the inverse form here as the given multipliers are not a square matrix. Only square matrices have the inverse. The multipliers are all $3\times 1$ matrices. We need to get 9 equations to get the 9 unknowns.