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  • Need solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 11

Need solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 11

Answers (1)

Answer:

\begin{aligned} &x=11, y=15, z=19 \end{aligned}

Given:

Let x, y, z  be the production level of 1st,2nd,3rd product respectively.

Hint:

 X = A-1B  is used for the problem

And the determination & cofactor of matrix A.Take transpose that will adjA using adjA find A-1

Solution:

According to the Question,

\begin{aligned} &x+y+z=45\: \: \: \: ......(1)\\ &-x+z=8\: \: \: \: ......(2)\\ &x+z=2 y\\ &x-2 y+z=0\: \: \: \: ......(3)\\ \end{aligned}

\begin{aligned} &{\left[\begin{array}{ccc} 1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & -2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 45 \\ 8 \\ 0 \end{array}\right]} \\ &A X=B \end{aligned}

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

C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].

\begin{aligned} &c_{11}=(-1)^{1+1}\left|\begin{array}{cc} 0 & 1 \\ -2 & 1 \end{array}\right|=2 \\ &c_{12}=(-1)^{1+2}\left|\begin{array}{cc} -1 & 1 \\ 1 & 1 \end{array}\right|=2 \\ &c_{13}=(-1)^{1+3}\left|\begin{array}{cc} -1 & 0 \\ 1 & -2 \end{array}\right|=2 \\ &c_{21}=(-1)^{2+1}\left|\begin{array}{cc} 1 & 1 \\ -2 & 1 \end{array}\right|=-3 \end{aligned}

\begin{aligned} &c_{22}=(-1)^{2+2}\left|\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right|=0 \\ &c_{23}=(-1)^{2+13}\left|\begin{array}{cc} 1 & 1 \\ 1 & -2 \end{array}\right|=3 \\ &c_{31}=(-1)^{3+1}\left|\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right|=1 \\ &c_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array}\right|=-2 \\ &c_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & 1 \\ -1 & 0 \end{array}\right|=1 \end{aligned}

\begin{aligned} &\operatorname{adj} A=\left[\begin{array}{ccc} 2 & 2 & 2 \\ -3 & 0 & 3 \\ 1 & -2 & 1 \end{array}\right]^{T}=\left[\begin{array}{ccc} 2 & -3 & 1 \\ 2 & 0 & -2 \\ 2 & 3 & 1 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} a d j A \end{aligned}

\begin{aligned} &=\frac{1}{6}\left[\begin{array}{ccc} 2 & -3 & 1 \\ 2 & 0 & -2 \\ 2 & 3 & 1 \end{array}\right] \\ X &=A^{-1} B \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{6}\left[\begin{array}{c} 90-24+0 \\ 90+0+0 \\ 90+24+0 \end{array}\right]=\frac{1}{6}\left[\begin{array}{c} 66 \\ 90 \\ 114 \end{array}\right]=\left[\begin{array}{c} 11 \\ 15 \\ 19 \end{array}\right]} \\ &x=11, y=15, z=19 \end{aligned}

Posted by

Gurleen Kaur

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