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Please Solve R. D. Sharma class 12 Chapter 5 determinants Exercise  5.4 Question 25  Maths textbook Solution.

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Answer:x=-3, y=-1 \text { and } z=7

Hint: Solving determinant gives zero.

Given:

\begin{aligned} &3 x-y+2 z=6 \\ &2 x-y-z=2 \\ &3 x+6 y+5 z=20 \end{aligned}

Solution:

By Cramer’s rule:                                                                

Solving determinant,

|\mathrm{A}|=\left|\begin{array}{ccc} 3 & -1 & 2 \\ 2 & -1 & 1 \\ 3 & 6 & 5 \end{array}\right|

Expanding along 1^{st} row,

\begin{aligned} &=3(-5-6)+1(10-3)+2(12+3) \\ &=-33+7+30 \\ &=4 \end{aligned}

\begin{aligned} &\mathrm{D}_{\bar{x}}=\left|\begin{array}{ccc} 6 & -1 & 2 \\ 2 & -1 & 1 \\ 20 & 6 & 5 \end{array}\right|=-12 \\ &D_{y}=\left|\begin{array}{ccc} 3 & 6 & 2 \\ 2 & 2 & 1 \\ 3 & 20 & 5 \end{array}\right|=-4 \end{aligned}

D_{z}=\left|\begin{array}{ccc} 3 & -1 & 6 \\ 2 & -1 & 2 \\ 3 & 6 & 20 \end{array}\right|=28

By Cramer’s rule,

\begin{aligned} &\Rightarrow x=\frac{D_{x}}{D}=\frac{-12}{4}=-3 \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{-4}{4}=-1 \\ &\Rightarrow z=\frac{D_{z}}{D}=\frac{28}{4}=7 \end{aligned}

Concept: Solving matrix of order 3x3 by solving linear equations

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