Matrices and Determinants 4 Question 34
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35. Let $M$ be a $3 \times 3$ matrix satisfying $M 1=2$,
======= ####35. Let $M$ be a $3 \times 3$ matrix satisfying $M\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right]$, $M\left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{c}1 \\ 1 \\ -1\end{array}\right]$ and $M\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{c}0 \\ 0 \\ 12\end{array}\right]$,
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Then, the sum of the diagonal entries of $M$ is …
(2011)
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Answer:
Correct Answer: 35. (9)
Solution:
- $\text { Let } \quad M=\left[\begin{array}{lll} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right]$
$ \begin{aligned} & \therefore \quad M\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]=\left[\begin{array}{r} -1 \\ 2 \\ 3 \end{array}\right] \Rightarrow M\left[\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right]=\left[\begin{array}{r} 1 \\ 1 \\ -1 \end{array}\right] \\ & M \cdot\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]=\left[\begin{array}{r} 0 \\ 0 \\ 12 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{l} a_2 \\ b_2 \\ c_2 \end{array}\right]=\left[\begin{array}{r} -1 \\ 2 \\ 3 \end{array}\right],\left[\begin{array}{l} a_1-a_2 \\ b_1-b_2 \\ c_1-c_2 \end{array}\right]=\left[\begin{array}{r} 1 \\ 1 \\ -1 \end{array}\right] \text {, } \\ & {\left[\begin{array}{l} a_1+a_2+a_3 \\ b_1+b_2+b_3 \\ c_1+c_2+c_3 \end{array}\right]=\left[\begin{array}{r} 0 \\ 0 \\ 12 \end{array}\right]} \\ & \Rightarrow \quad a_2=-1, b_2=2, c_2=3, a_1-a_2=1 \text {, } \\ & b_1-b_2=1, c_1-c_2=-1 \\ & \Rightarrow a_1+a_2+a_3=0, b_1+b_2+b_3=0 \\ & c_1+c_2+c_3=12 \\ & \therefore \quad a_1=0, b_2=2, c_3=7 \\ & \end{aligned} $
$\Rightarrow \quad$ Sum of diagonal elements $=0+2+7=9$
