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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
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 $M\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{c}0 \\ 0 \\ 12\end{array}\right]$ Then the sum of the diagonal entries of $M$ is
  • ✓
    $9$
  • B
    $8$
  • C
    $7$
  • D
    $6$
✓

Answer

Correct option: A.
$9$
Let $M$ be $\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right]$.
Now from the first condition $\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right]\left[\begin{array}{l}0 \\ 1 \\ o \end{array}\right]=\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right]$
we get the values of $b, e, h$ as $-1,2,3$ respectively.
Newly formed matrix $M=\left[\begin{array}{ccc}a & -1 & c \\ d & 2 & f \\ g & 3 & i\end{array}\right]$
From the second condition
$\left[\begin{array}{ccc} a & -1 & c \\ d & 2 & f \\ g & 3 & i \end{array}\right]\left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{c}1 \\ 1 \\ -1\end{array}\right]$
Solving the matrix, we get three equations
$\Rightarrow a+1=1, d-2=1, g-3=-1$
$\Rightarrow a= 0, d=3, g=2$
Modified matrix $M =\left[\begin{array}{ccc} 0 & -1 & c \\ 3 & 2 & f \\ 2 & 3 & i \end{array}\right]$
Now the third and the last condition, $\left[\begin{array}{ccc}0 & -1 & c \\ 3 & 2 & f \\ 2 & 3 & i \end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{c}0 \\ o \\ 12\end{array}\right]$
$\Rightarrow 2+3+ i =12 $
$\Rightarrow i =7$
Sum of the diagonal entries of $M = a + e + i = o +2+7=9$

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