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  • Explain Solution R.D. Sharma Class 12 Chapter 5 deteminants Exercise 5.4 Question 29 maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 5 deteminants Exercise 5.4 Question 29 maths Textbook Solution.

Answers (1)

Answer:x=\frac{3 k+6}{5}, y=\frac{8+4 k}{5} \text { and } z=k

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

Given:

\begin{aligned} &2 x+y-2 z=4 \\ &x-2 y+z=-2 \\ &5 x-5 y+z=-2 \end{aligned}

Solution: 

Solving determinant,

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

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

\Rightarrow \mathrm{D}_{x}=\left|\begin{array}{ccc} 4 & 1 & -2 \\ -2 & -2 & 1 \\ -2 & -5 & 1 \end{array}\right|

Taking 2 common from C_{1},

=-2\left|\begin{array}{ccc} -2 & 1 & -2 \\ 1 & -2 & 1 \\ 1 & -5 & 1 \end{array}\right|=0 \quad\left(\because C_{1}=C_{3}\right)

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

\begin{array}{r} \Rightarrow \mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc} 2 & 4 & -2 \\ 1 & -2 & 1 \\ 5 & -2 & 1 \end{array}\right| \\ \quad \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+2 C_{3} \end{array}

=\left|\begin{array}{ccc} 2 & 0 & -2 \\ 1 & 0 & 1 \\ 5 & 0 & 1 \end{array}\right|=0 \quad\left(\because \text { Expanding along } \mathrm{R}_{1}\right)

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

\begin{aligned} & \Rightarrow \mathrm{D}_{z}=\left|\begin{array}{ccc} 2 & 1 & 4 \\ 1 & -2 & -2 \\ 5 & -5 & -2 \end{array}\right| \quad \text { Expanding along } \mathrm{R}_{1}, \\ =& 2(4-10)-1(-2+10)+4(-5+10) \\ =&-12-8+20 \\ =& 0 \end{aligned}

So,\mathrm{D}=\mathrm{D}_{\mathrm{x}}=\mathrm{D}_{\mathrm{x}}=\mathrm{D}_{2}=0

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

\begin{aligned} &x-2 y=-2-z \\ &5 x-5 y=-2-z \end{aligned}

Using Cramer’s rule,

\begin{aligned} &\mathrm{D}=\left|\begin{array}{ll} 1 & -2 \\ 5 & -5 \end{array}\right|=-5+10=5\\ &\mathrm{D}_{x}=\left|\begin{array}{ll} -2-z & -2 \\ -2-z & -5 \end{array}\right|=10+5 z-2 z-4=3 z+6\\ &\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ll} 1 & -2-z \\ 5 & -2-z \end{array}\right|=-2-\mathrm{z}+10+5 \mathrm{z}=8+4 \mathrm{z}\\ &\Rightarrow x=\frac{D_{x}}{D}=\frac{3 z+6}{5}\\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{8+4 z}{5}\\ &\text { Let } z=k, \text { then } x=\frac{3 z+6}{5} \text { and } y=\frac{8+4 z}{5} \end{aligned}

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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