#### Need Solution for R.D.Sharma Maths Class 12 Chapter 5 determinants  Exercise 5.4 Question 32  Maths Textbook Solution.

Answer: $\mathrm{x}=2, \mathrm{y}=3 \text { and } z=4$

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

Given:

Solution:

To form linear equation, let the rates of commissions on items A, B and C be x, y and z respectively. This can be expressed as a system of linear equations.

\begin{aligned} &2 \mathrm{x}+3 \mathrm{y}+4 \mathrm{z}=29 \\ &\mathrm{x}+\mathrm{y}+2 \mathrm{z}=13 \\ &3 \mathrm{x}+2 \mathrm{y}+z=16 \end{aligned}

Where x, y and z are number of cars $C_{1}, C_{2}\: and \: C_{3}$ respectively.

By Cramer’s rule, solving determinant:

\begin{aligned} D=&\left|\begin{array}{lll} 2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1 \end{array}\right| \\ & R_{1} \rightarrow R_{1}-4 R_{3} \\ & R_{2} \rightarrow R_{2}-2 R_{3} \\ &=\left|\begin{array}{ccc} -10 & -5 & 0 \\ -5 & -3 & 0 \\ 3 & 2 & 1 \end{array}\right| \\ =& 30-25 \\ =& 5 \end{aligned}

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

\begin{aligned} \Rightarrow D_{\bar{x}}=&\left|\begin{array}{ccc} 29 & 3 & 4 \\ 13 & 1 & 2 \\ 16 & 2 & 1 \end{array}\right| \\ & \quad \begin{aligned} R_{1} & \rightarrow R_{1}-4 R_{3} \\ R_{2} & \rightarrow R_{2}-2 R_{3} \end{aligned} \\ =&\left|\begin{array}{lll} -35 & -5 & 0 \\ -19 & -3 & 0 \\ 16 & 2 & 1 \end{array}\right| \\ =& 190-175 \\ =& 15 \end{aligned}

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

$\begin{array}{r} \Rightarrow \mathrm{D}_{\mathrm{y}}=\left|\begin{array}{lll} 2 & 29 & 4 \\ 1 & 13 & 2 \\ 3 & 16 & 1 \end{array}\right| \\ \quad R_{1} \rightarrow R_{1}-4 R_{3} \\ R_{2} \rightarrow R_{2}-2 R_{3} \end{array}$

\begin{aligned} &=\left|\begin{array}{ccc} -10 & -35 & 0 \\ -5 & -19 & 0 \\ 3 & 16 & 1 \end{array}\right| \\ &=190-175 \end{aligned}

$=15$

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

$\begin{array}{r} \Rightarrow \mathrm{D}_{z}=\left|\begin{array}{lll} 2 & 3 & 29 \\ 1 & 1 & 13 \\ 3 & 2 & 16 \end{array}\right| \\ R_{1} \rightarrow R_{1}-\left(R_{2}+R_{3}\right) \end{array}$

$=\left|\begin{array}{ccc} -2 & 0 & 0 \\ 1 & 1 & 13 \\ 3 & 2 & 16 \end{array}\right|$

\begin{aligned} &=-2(16-26) \\ &=20 \end{aligned}

Using Cramer’s rule,

\begin{aligned} &\Rightarrow x=\frac{D_{x}}{D}=\frac{10}{5}=2 \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{15}{5}=3 \\ &\Rightarrow z=\frac{D_{z}}{D}=\frac{20}{5}=4 \end{aligned}

$\therefore$ The number of cars produced of type $C_{1}, C_{2}\: and \: C_{3}$ are 2, 3 and 4 respectively.

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