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  • Please Solve R.D. Sharma class 12 Chapter 5 determinants Exercise 5.4 Question 31 Maths textbook Solution.

Please Solve R.D. Sharma class 12 Chapter 5 determinants Exercise  5.4 Question 31  Maths textbook Solution.

Answers (1)

Answer:\mathrm{x}=2, \mathrm{y}=4 \text { and } \mathrm{z}=11

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

Given:

Months sale of unit     Total commission drawn
  A B C  
Jan 90 100 20 800
Feb 130 50 40 900
March 60 100 30 850

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} &90 \mathrm{x}+100 \mathrm{y}+20 \mathrm{z}=800 \\ &130 \mathrm{x}+50 \mathrm{y}+40 \mathrm{z}=900 \\ &60 \mathrm{x}+100 \mathrm{y}+30 \mathrm{z}=850 \end{aligned}

By Cramer’s rule, solving determinant:

D=\left|\begin{array}{ccc} 90 & 100 & 20 \\ 130 & 50 & 40 \\ 60 & 100 & 30 \end{array}\right|

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

=\left|\begin{array}{ccc} -170 & 0 & -60 \\ 130 & 50 & 40 \\ -200 & 0 & -50 \end{array}\right|

\begin{aligned} &=-170(-2500)-60(10000) \\ &=425000-600000 \\ &=-175000 \end{aligned}

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

\Rightarrow \mathrm{D}_{\bar{x}}=\left|\begin{array}{ccc} 800 & 100 & 20 \\ 900 & 50 & 40 \\ 850 & 100 & 30 \end{array}\right|

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

=\left|\begin{array}{ccc} -1000 & 0 & -60 \\ 900 & 50 & 40 \\ -950 & 0 & -50 \end{array}\right|

\begin{aligned} &=-1000(-2500)-60(47500) \\ &=2500000-2850000 \\ &=-350000 \end{aligned}

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} 90 & 800 & 20 \\ 130 & 900 & 40 \\ 60 & 850 & 30 \end{array}\right|

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

\begin{aligned} &=\left|\begin{array}{ccc} 90 & 800 & 20 \\ -50 & -700 & 0 \\ -75 & -350 & 0 \end{array}\right| \\ &=20(17500-52500) \\ &=-700000 \end{aligned}

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

\Rightarrow \mathrm{D}_{z}=\left|\begin{array}{ccc} 90 & 100 & 800 \\ 130 & 50 & 900 \\ 60 & 100 & 850 \end{array}\right|

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

=\left|\begin{array}{ccc} -170 & 0 & -1000 \\ 130 & 50 & 900 \\ -200 & 0 & -950 \end{array}\right|

\begin{aligned} &=-170(-47500)-1000(10000) \\ &=8075000-10000000 \\ &=-1925000 \end{aligned}

Using Cramer’s rule,

\begin{aligned} &\Rightarrow x=\frac{D_{x}}{D}=\frac{-350000}{-175000}=2 \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{-700000}{-175000}=4 \\ &\Rightarrow z=\frac{D_{z}}{D}=\frac{-1925000}{-175000}=11 \end{aligned}

\therefore The rates of commission of items A, B and C are 2%, 4% and 11% respectively.

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

 

Posted by

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