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Explain Solution R.D. Sharma Class 12 Chapter 5 deteminants Exercise 5.4 Question 28 maths Textbook Solution.

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Answer:x=k, y=2 k \text { and } z=3k

Hint: Use Cramer’s rule for system of linear equations.

Given:

\begin{aligned} &x+y-z=0 \\ &x-2 y+z=0 \\ &3 x+6 y-5 z=0 \end{aligned}

Solution: 

Solving determinant,

\begin{aligned} |A| &=\left|\begin{array}{ccc} 1 & 1 & -1 \\ 1 & -2 & 1 \\ 3 & 6 & -5 \end{array}\right| \\ &=1(10-6)-1(-5-3)-1(6+6) \\ &=4+8-12 \\ &=0 \end{aligned}

Now for x, the x column is replaced with constant column i.e.

\Rightarrow \mathrm{D}_{x}=\left|\begin{array}{ccc} 0 & 1 & -1 \\ 0 & -2 & 1 \\ 0 & 6 & -5 \end{array}\right|=0

If we are solving for y, the y column is replaced with constant column i.e.

\Rightarrow \mathrm{D}_{y}=\left|\begin{array}{ccc} 1 & 0 & -1 \\ 1 & 0 & 1 \\ 3 & 0 & -5 \end{array}\right|=0

If we are solving for z, the z column is replaced with constant column i.e.

\begin{aligned} &\Rightarrow D_{z}=\left|\begin{array}{ccc} 1 & 1 & 0 \\ 1 & -2 & 0 \\ 3 & 6 & 0 \end{array}\right|=0 \\ &\text { So, } D=D_{x}=D_{y}=D_{z}=0 \end{aligned}

The given system has either infinite solutions or it is inconsistent.

\begin{aligned} &x+y=z \\ &x-2 y=-z \end{aligned}

Using Cramer’s rule,

\begin{aligned} &\mathrm{D}=\left|\begin{array}{cc} 1 & 1 \\ 1 & -2 \end{array}\right|=-2-1=-3 \\ &\mathrm{D}_{x}=\left|\begin{array}{cc} z & 1 \\ -z & -2 \end{array}\right|=-2 \mathrm{z}+z=-z \end{aligned}

\begin{aligned} &\mathrm{D}_{\mathrm{y}}=\left|\begin{array}{cc} 1 & z \\ 1 & -z \end{array}\right|=-\mathrm{z}-\mathrm{z}=-2 \mathrm{z} \\ &\Rightarrow x=\frac{D_{x}}{D}=\frac{-z}{-3}=\frac{z}{3} \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{-2 z}{-3}=\frac{2 z}{3} \end{aligned}

Let z = 3k, then x = k and y = 2k

Concept: Solving matrix of order 3x3 by Cramer’s rule.

Note: When D = 0, there is either no solution or infinite solutions.


 

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