SATHEE: Matrices and Determinants 4 Question 34

Matrices and Determinants 4 Question 34

35.

Let $M$ be a $3 \times 3$ matrix satisfying $M [\begin{array}{l} 0 \\ 1 \\ 0 \end{array}] = [\begin{matrix} - 1 \\ 2 \\ 3 \end{matrix}]$ , $M [\begin{matrix} 1 \\ - 1 \\ 0 \end{matrix}] = [\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix}]$ and $M [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}] = [\begin{matrix} 0 \\ 0 \\ 12 \end{matrix}]$ ,

Then, the sum of the diagonal entries of $M$ is …

(2011)

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Answer:

Correct Answer: 35. (9)

Solution:

$Let M = [\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}]$

$\begin{aligned} ∴ M [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] = [\begin{array}{r} - 1 \\ 2 \\ 3 \end{array}] \Rightarrow M [\begin{array}{r} 1 \\ - 1 \\ 0 \end{array}] = [\begin{array}{r} 1 \\ 1 \\ - 1 \end{array}] \\ M \cdot [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = [\begin{array}{r} 0 \\ 0 \\ 12 \end{array}] \\ \Rightarrow [\begin{matrix} a_{2} \\ b_{2} \\ c_{2} \end{matrix}] = [\begin{array}{r} - 1 \\ 2 \\ 3 \end{array}], [\begin{matrix} a_{1} - a_{2} \\ b_{1} - b_{2} \\ c_{1} - c_{2} \end{matrix}] = [\begin{array}{r} 1 \\ 1 \\ - 1 \end{array}], \\ [\begin{matrix} a_{1} + a_{2} + a_{3} \\ b_{1} + b_{2} + b_{3} \\ c_{1} + c_{2} + c_{3} \end{matrix}] = [\begin{array}{r} 0 \\ 0 \\ 12 \end{array}] \\ \Rightarrow a_{2} = - 1, b_{2} = 2, c_{2} = 3, a_{1} - a_{2} = 1, \\ b_{1} - b_{2} = 1, c_{1} - c_{2} = - 1 \\ \Rightarrow a_{1} + a_{2} + a_{3} = 0, b_{1} + b_{2} + b_{3} = 0 \\ c_{1} + c_{2} + c_{3} = 12 \\ ∴ a_{1} = 0, b_{2} = 2, c_{3} = 7 \end{aligned}$

$\Rightarrow$ Sum of diagonal elements $= 0 + 2 + 7 = 9$

Matrices and Determinants 4 Question 34

35.

Answer:

Solution:

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Matrices and Determinants 4 Question 34

35.

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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