#### Provide Solution For  R. D. Sharma Maths Class 12 Chapter 5  determinants Exercise 5.4  Question 30 Maths Textbook Solution.

Answer: $x=\frac{7-3 z}{2}, y=\frac{3 z-5}{2} \text { and } z=k$

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

Given:

\begin{aligned} &x-y+3 z=6 \\ &x+3 y-3 z=-4 \\ &5 x+3 y+3 z=10 \end{aligned}

Solution:

Solving determinant,

\begin{aligned} |A| &=\left|\begin{array}{ccc} 1 & -1 & 3 \\ 1 & 3 & -3 \\ 5 & 3 & 3 \end{array}\right| \\ &=1(9+9)+1(3+15)+3(3-15) \\ &=18+18-36 \\ &=0 \end{aligned}

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

$\Rightarrow \mathrm{D}_{x}=\left|\begin{array}{ccc} 6 & -1 & 3 \\ -4 & 3 & -3 \\ 10 & 3 & 3 \end{array}\right|$

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

$=\left|\begin{array}{ccc} 2 & 2 & 0 \\ -4 & 3 & -3 \\ 6 & 6 & 0 \end{array}\right|=0 \quad\left(\because \text { Expanding along } R_{1}\right)$

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

$\Rightarrow \mathrm{D}_{\mathrm{y}}=\left|\begin{array}{ccc} 1 & 6 & 3 \\ 1 & -4 & -3 \\ 5 & 10 & 3 \end{array}\right|$

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

$=\left|\begin{array}{ccc} 2 & 2 & 0 \\ 1 & -4 & -3 \\ 6 & 6 & 0 \end{array}\right|=0 \quad\left(\because \text { Expanding along } R_{1}\right)$

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

$\Rightarrow \mathrm{D}_{z}=\left|\begin{array}{ccc} 1 & -1 & 6 \\ 1 & 3 & -4 \\ 5 & 3 & 10 \end{array}\right|$

\begin{aligned} &R_{2} \rightarrow R_{2}-R_{1} \\ &R_{3} \rightarrow R_{3}-5 R_{1} \end{aligned}

\begin{aligned} &=\left|\begin{array}{ccc} 1 & -1 & 6 \\ 0 & 4 & -10 \\ 0 & 8 & -20 \end{array}\right|\\ &=0 \quad\left(\because \text { Expanding along } \mathrm{R}_{1}\right) \end{aligned}

So, $\mathrm{D}=\mathrm{D}_{\mathrm{x}}=\mathrm{D}_{\mathrm{y}}=\mathrm{D}_{z}=0$

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

\begin{aligned} &x-y=6-3 z \\ &x+3 y=-4+3 z \end{aligned}

Using Cramer’s rule,

\begin{aligned} &D=\left|\begin{array}{cc} 1 & -1 \\ 1 & 3 \end{array}\right|=3+1=4 \\ &D_{x}=\left|\begin{array}{cc} 6-3 z & -1 \\ -4+3 z & 3 \end{array}\right|=18-9 z-4+3 z=14-6 z \\ &D_{x}=\left|\begin{array}{lc} 1 & 6-3 z \\ 1 & -4+3 z \end{array}\right|=-4+3 z-6+3 z=6 z-10 \\ &\Rightarrow x=\frac{D_{x}}{D}=\frac{14-6 z}{4}=\frac{7-3 z}{2} \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{6 z-10}{4}=\frac{3 z-5}{2} \\ &\text { Let } z=k, \text { then } x=\frac{7-3 z}{2} \text { and } y=\frac{3 z-5}{2} \end{aligned}

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

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