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  • Explain Solution R.D. Sharma Class 12 Chapter 5 deteminants Exercise 5.4 Question 20 maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 5 deteminants Exercise 5.4 Question 20 maths Textbook Solution.

Answers (1)

Answer:\mathrm{x}=-2, \mathrm{y}=3, \mathrm{z}=\frac{3}{2} \text { and } \mathrm{w}=\frac{-1}{2}

Hint: Use Cramer’s rule to solve a system of linear equations

Given:

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

Solution:

Solving determinant,

|\mathbf{A}|=\left|\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -2 & 2 & 2 \\ 2 & 1 & -2 & 2 \\ 3 & -1 & 3 & -3 \end{array}\right|

\begin{aligned} C_{2} & \rightarrow C_{2}-C_{1} \\ C_{3} & \rightarrow C_{3}-C_{1} \\ C_{4} & \rightarrow C_{4}-C_{1} \end{aligned}

=\left|\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 & -3 & 1 & 1 \\ 2 & -1 & -4 & 0 \\ 3 & -4 & 0 & -6 \end{array}\right|

Expanding along R_{1},

=1\left|\begin{array}{ccc} -3 & 1 & 1 \\ -1 & -4 & 0 \\ -4 & 0 & -6 \end{array}\right|

\begin{aligned} &C_{2} \rightarrow C_{2}-C_{3} \\ &C_{1} \rightarrow C_{1}+3 C_{3} \end{aligned}

\begin{aligned} &=\left|\begin{array}{ccc} 0 & 0 & 1 \\ -1 & -4 & 0 \\ -22 & 6 & -6 \end{array}\right| \\ &=1(-6-88) \end{aligned}

=-94

\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{cccc} 2 & 1 & 1 & 1 \\ -6 & -2 & 2 & 2 \\ -5 & 1 & -2 & 2 \\ -3 & -1 & 3 & -3 \end{array}\right|=188

\mathrm{D}_{\mathrm{y}}=\left|\begin{array}{cccc} 1 & 2 & 1 & 1 \\ 1 & -6 & 2 & 2 \\ 2 & -5 & -2 & 2 \\ 3 & -3 & 3 & -3 \end{array}\right|=-282

\mathrm{D}_{z}=\left|\begin{array}{cccc} 1 & 1 & 2 & 1 \\ 1 & -2 & -6 & 2 \\ 2 & 1 & -5 & 2 \\ 3 & -1 & -3 & -3 \end{array}\right|=-141

\mathrm{D}_{\mathrm{w}}=\left|\begin{array}{cccc} 1 & 1 & 1 & 2 \\ 1 & -2 & 2 & -6 \\ 2 & 1 & -2 & -5 \\ 3 & -1 & 3 & -3 \end{array}\right|=47

By Cramer’s rule,

\begin{aligned} &\Rightarrow x=\frac{D_{x}}{D}=\frac{188}{-94}=-2 \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{-282}{-94}=3 \\ &\Rightarrow z=\frac{D_{z}}{D}=\frac{-141}{-94}=\frac{3}{2} \\ &\Rightarrow w=\frac{D_{w}}{D}=\frac{47}{-94}=\frac{-1}{2} \end{aligned}

Concept: Solving matrix of order 4x4 (Elementary row and column operations)

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