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Provide solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 1 subquestion (ii)

Answers (1)

Answer:

x=\frac{9}{2},\; \; y=-\frac{7}{2}

Given:

\left[\begin{array}{ll} 5 & 7 \\ 4 & 6 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} -2 \\ -3 \end{array}\right]

Hint:

X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A .

Solution:

\left[\begin{array}{ll} 5 & 7 \\ 4 & 6 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} -2 \\ -3 \end{array}\right]

        A X=B\\ |A|=\left[\begin{array}{ll}5 & 7 \\ 4 & 6\end{array}\right]=30-28=2 \neq 0\\ This\; has\; a\; unique\; solution\; given\; by\; X=A^{-1} B.\\ C_{i i} \; be\; the\; co-\! factor\; o\! f\; the\; elements\; a_{i j}\; in\; A=\left[a_{i j}\right].\; Then,

        \begin{aligned} &C_{11}=(-1)^{1+1}(6)=2 \quad, \quad C_{12}=(-1)^{1+2}(4)=-4 \\ &C_{21}=(-1)^{2+1}(7)=-7 \quad, \quad C_{22}=(-1)^{2+2}(5)=5 \\ &A=\left[\begin{array}{cc} 6 & -4 \\ -7 & 5 \end{array}\right] \end{aligned}

        \begin{aligned} \operatorname{adj} A &=\left[\begin{array}{cc} 6 & -4 \\ -7 & 5 \end{array}\right]^{T} \\ &=\left[\begin{array}{cc} 6 & -7 \\ -4 & 5 \end{array}\right] \\ A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{2}\left[\begin{array}{cc} 6 & -7 \\ -4 & 5 \end{array}\right] \end{aligned}

        \begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{2}\left[\begin{array}{cc} 6 & -7 \\ -4 & 5 \end{array}\right] \\ X &=A^{-1} B \\ &=\frac{1}{2}\left[\begin{array}{cc} 6 & -7 \\ -4 & 5 \end{array}\right]\left[\begin{array}{l} -2 \\ -3 \end{array}\right] \\ &=\frac{1}{2}\left[\begin{array}{c} -12+21 \\ 8-15 \end{array}\right] \end{aligned}

        \begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{9}{2} \\ -\frac{7}{2} \end{array}\right]} \\ &\therefore x=\frac{9}{2} \quad, \quad \mathrm{y}=-\frac{7}{2} \end{aligned}

Posted by

Gurleen Kaur

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