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  • Explain solution RD Sharma class 12 Chapter 6 Adjoint and Inverse of a Matrix exercise 6.1 question 31

Explain solution RD Sharma class 12 Chapter 6 Adjoint and Inverse of a Matrix exercise 6.1 question 31

Answers (1)

Answer:

X=\left[\begin{array}{cc} -7 & -5 \\ 7 & 6 \end{array}\right]                                                      

Hint:

Here, we use basic concept of determinant and inverse of matrix

Given:

A=\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right], B=\left[\begin{array}{cc} 14 & 7 \\ 7 & 7 \end{array}\right]

Solution:

\begin{aligned} &A X=B \\ &X=A^{-1} B \\ &|A|=-7 \end{aligned}                                                                  

Cofactor of A

\begin{aligned} &C_{11}=-2, C_{12}=1 \\ &C_{21}=-3, C_{22}=5 \\ &\operatorname{Adj}(A)=\left[\begin{array}{ll} -2 & 1 \\ -3 & 5 \end{array}\right]^{T}=\left[\begin{array}{cc} -2 & -3 \\ 1 & 5 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \end{aligned}

\begin{aligned} &A^{-1}=\frac{1}{-7}\left[\begin{array}{cc} 2 & 3 \\ -1 & -5 \end{array}\right] \\ &X=-\frac{1}{7}\left[\begin{array}{cc} 2 & 3 \\ -1 & -5 \end{array}\right]\left[\begin{array}{cc} 14 & 7 \\ 7 & 7 \end{array}\right] \end{aligned}

\begin{aligned} &=-\frac{1}{7}\left[\begin{array}{cc} 28+21 & 14+21 \\ -14-35 & -7-35 \end{array}\right] \\ &X=\left[\begin{array}{cc} -7 & -5 \\ 7 & 6 \end{array}\right] \end{aligned}

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