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Question: Let M be a \(3\times 3\) matrix satisfying \(M\left[ \begin{matrix} 0 \\\ 1 \\\ 0 \\\...

Let M be a 3×33\times 3 matrix satisfying M[0 1 0 ]=[−1 2 3 ]M\left[ \begin{matrix} 0 \\\ 1 \\\ 0 \\\ \end{matrix} \right]=\left[ \begin{matrix} -1 \\\ 2 \\\ 3 \\\ \end{matrix} \right], M[1 −1 0 ]=[1 1 −1 ]M\left[ \begin{matrix} 1 \\\ -1 \\\ 0 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ 1 \\\ -1 \\\ \end{matrix} \right] and M[1 1 1 ]=[0 0 12 ]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.

Explanation

Solution

M be a 3×33\times 3 matrix which satisfies the given matrix multiplication.
So, we try to assume the matrix M.
Let M be [abc def ghi ]\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 (a+e+i)\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[0 1 0 ]=[−1 2 3 ]M\left[ \begin{matrix} 0 \\\ 1 \\\ 0 \\\ \end{matrix} \right]=\left[ \begin{matrix} -1 \\\ 2 \\\ 3 \\\ \end{matrix} \right] which implies [abc def ghi ][0 1 0 ]=[−1 2 3 ]\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×1=b=−1b\times 1=b=-1, e×1=e=2e\times 1=e=2, h×1=h=3h\times 1=h=3. We got value of 3 unknowns. We also have M[1 −1 0 ]=[1 1 −1 ]M\left[ \begin{matrix} 1 \\\ -1 \\\ 0 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ 1 \\\ -1 \\\ \end{matrix} \right] which implies [abc def ghi ][1 −1 0 ]=[1 1 −1 ]\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×1+b×(−1)=a−b=1a\times 1+b\times \left( -1 \right)=a-b=1, d×1+e×(−1)=d−e=1d\times 1+e\times \left( -1 \right)=d-e=1, g×1+h×(−1)=g−h=−1g\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⇒a=0a+1=1\Rightarrow a=0, d−2=1⇒d=3d-2=1\Rightarrow d=3, g−3=−1⇒g=2g-3=-1\Rightarrow g=2.
Lastly, we have M[1 1 1 ]=[0 0 12 ]M\left[ \begin{matrix} 1 \\\ 1 \\\ 1 \\\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\\ 0 \\\ 12 \\\ \end{matrix} \right] which implies [abc def ghi ][1 1 1 ]=[0 0 12 ]\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×1+b×1+c×1=a+b+c=0a\times 1+b\times 1+c\times 1=a+b+c=0, d×1+e×1+f×1=d+e+f=0d\times 1+e\times 1+f\times 1=d+e+f=0, g×1+h×1+i×1=g+h+i=12g\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⇒c=10-1+c=0\Rightarrow c=1, 3+2+f=0⇒f=−53+2+f=0\Rightarrow f=-5, 2+3+i=12⇒i=72+3+i=12\Rightarrow i=7.
We got all the 9 unknowns to get the matrix M=[0−11 32−5 237 ]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 (a+e+i)=0+2+7=9\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 3×13\times 1 matrices. We need to get 9 equations to get the 9 unknowns.