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  • Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 21 maths textbook solution

Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 21 maths textbook solution

Answers (1)

Answer:

V\! ariety \; A: R s 5, V\! ariety\; B: R s 8, V\! ariety\; C: R s 8

Given:

According to the question

A+B+C=21,\;4 A+3 B+2 C=60,\;6 A+2 B+3 C=70

Hint:

X=A-1B is used to solve this problem.

And the determinant and co-factor of matrix A, take it’s transpose that will be Adj A using Adj A calculate A-1.

Solution:

As there are 3 variables of pen A, B, C

Meenu purchased a pen of each variety which costs her Rs21

Therefore

A+B+C=21

Similarly

For Jeevan

4 A+3 B+2 C=60

For Shikha

6 A+2 B+3 C=70

\begin{aligned} &{\left[\begin{array}{lll} 1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3 \end{array}\right]\left[\begin{array}{l} A \\ B \\ C \end{array}\right]=\left[\begin{array}{l} 21 \\ 60 \\ 70 \end{array}\right]} \\ &\text { Where } P=\left[\begin{array}{lll} 1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3 \end{array}\right], Q=\left[\begin{array}{l} 21 \\ 60 \\ 70 \end{array}\right] \\ &|P|=1(9-4)-1(12-12)+1(8-18) \\ &=-5 \neq 0 \end{aligned}

\begin{aligned} &P^{-1} \text { exists }\\ &c_{11}=5 \quad c_{12}=0 \quad c_{13}=-10\\ &c_{21}=-1 \quad c_{22}=-3 \quad c_{23}=4\\ &c_{31}=-1 \quad c_{32}=2 \quad c_{33}=-1 \end{aligned}

\begin{aligned} &\operatorname{adjP}=\left[\begin{array}{ccc} 5 & 0 & -10 \\ -1 & -3 & 4 \\ -1 & 2 & -1 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1 \end{array}\right] \\ &P^{-1}=\frac{1}{-5}\left[\begin{array}{ccc} 5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1 \end{array}\right] \end{aligned}

\begin{aligned} &X=P^{-1} Q \\ &=\frac{1}{-5}\left[\begin{array}{ccc} 5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1 \end{array}\right]\left[\begin{array}{l} 21 \\ 60 \\ 70 \end{array}\right] \\ &=\frac{1}{-5}\left[\begin{array}{c} 105-60-70 \\ 0-180+140 \\ -210+240-70 \end{array}\right] \end{aligned}

\begin{gathered} =\frac{1}{-5}\left[\begin{array}{l} -25 \\ -40 \\ -40 \end{array}\right] \\ X=\left[\begin{array}{l} 5 \\ 8 \\ 8 \end{array}\right] \end{gathered}

Therefore

                   Cost of A variety of pens= Rs5

                   Cost of B variety of pens= Rs8

                   Cost of C variety of pens= Rs8

Posted by

Gurleen Kaur

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