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Provide solution for RD Sharma maths class 12 chapter Determinants exercise 5.1 question 9

Answers (1)

Answer:

\begin{aligned} |3 A|=27|A| \\ \end{aligned}

Given:

\begin{aligned} \mathrm{A}=\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right] \\ \text { To prove: }|3 A|=27|A| \end{aligned}

Solution:

\begin{aligned} &\mathrm{A}=\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right], 3 \mathrm{~A}=\left[\begin{array}{llc} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{array}\right] \\ &|3 A|=\mathrm{a}_{11} \mathrm{C}_{11}+\mathrm{a}_{21} \mathrm{C}_{21}+\mathrm{a}_{31} \mathrm{C}_{31} \\ &\quad=(-1)^{1+1} 3(36-0)+(-1)^{1+2} 0(0-0)+(-1)^{1+3} 3(0-0) \\ &\quad=3(36)+0+0 \\ &\quad=108 \quad \quad \quad \quad \quad \quad.....(1) \end{aligned}

\begin{aligned} &|A|=(-1)^{1+1} 1(4-0)+(-1)^{1+2} 0(0-0)+(-1)^{1+3} 1(0-0) \\ &27|A|=27(4+0+0) \\ &\quad=108 \quad \quad \quad \quad.....(2)\end{aligned} \\ F\! rom\; (1)\; and\; (2),\\ |3 A|=27|A| is proved.

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