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  • Explain solution RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (xiv) maths

Explain solution RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (xiv) maths

Answers (1)

Answer:

\begin{aligned} &x=2 \quad, \quad y=1 \quad, \quad z=3 \end{aligned}

Given:

\begin{aligned} &x-y+2 z=7 \\ &3 x+4 y-5 z=-5 \\ &2 x-y+3 z=12 \end{aligned}

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 using Adj A calculate A-1.

Solution:

        \begin{aligned} &A=\left[\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{array}\right] \\ \end{aligned}

        \begin{aligned} |A|=\left|\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{array}\right| &=1(12-5)+1(9+10)+2(-3-8) \\ &=7+19-22 \\ &=4 \end{aligned}

        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} 4 & -5 \\ -1 & 3 \end{array}\right|=7 & \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 3 & -5 \\ 2 & 3 \end{array}\right|=-19 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 3 & 4 \\ 2 & -1 \end{array}\right|=-11 & , \quad C_{21}=(-1)^{2+1}\left|\begin{array}{rr} -1 & 2 \\ -1 & 3 \end{array}\right|=1 \end{aligned}

        \begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right|=-1 \quad \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & -1 \\ 2 & -1 \end{array}\right|=-1 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} -1 & 2 \\ 4 & -5 \end{array}\right|=-3 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & 2 \\ 3 & -5 \end{array}\right|=11 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & -1 \\ 3 & 4 \end{array}\right|=7 \end{aligned}

        \begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} 7 & -19 & -11 \\ 1 & -1 & -1 \\ -3 & 11 & 7 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array}\right] \end{aligned}

        \begin{aligned} A^{-1} &=\frac{1}{|A|} \text { adjA } \\ &=\frac{1}{4}\left[\begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array}\right] \end{aligned}

        \begin{aligned} X &=A^{-1} B \\ &=\frac{1}{4}\left[\begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array}\right]\left[\begin{array}{c} 7 \\ -5 \\ 12 \end{array}\right] \\ &=\frac{1}{4}\left[\begin{array}{c} 49-5-36 \\ -133+5+132 \\ -77+5+84 \end{array}\right] \end{aligned}

        \begin{aligned} &=\frac{1}{4}\left[\begin{array}{c} 8 \\ 4 \\ 12 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 2 \\ 1 \\ 3 \end{array}\right]} \\ &x=2, \quad y=1 \quad, \quad z=3 \end{aligned}

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