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  • Explain solution RD Sharma class 12 Chapter 6 Adjoint and Inverse of a Matrix exercise 6.1 question 33

Explain solution RD Sharma class 12 Chapter 6 Adjoint and Inverse of a Matrix exercise 6.1 question 33

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Answer:

X=\left[\begin{array}{cc} 9 & -14 \\ -16 & 25 \end{array}\right]

Hint:

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

A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)

Given:

A=\left[\begin{array}{ll} 2 & 1 \\ 5 & 3 \end{array}\right], B=\left[\begin{array}{ll} 5 & 3 \\ 3 & 2 \end{array}\right], C=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]

Solution:

Then the given equation becomes

\begin{aligned} &A \times B=I \\ &X=A^{-1} B^{-1} \\ &|A|=6-5=1 \\ &|B|=10-9=1 \end{aligned}

A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{1}\left[\begin{array}{cc} 3 & -1 \\ -5 & 2 \end{array}\right]

\begin{aligned} &B^{-1}=\frac{1}{|B|} \times A d j(B)=\frac{1}{1}\left[\begin{array}{cc} 2 & -3 \\ -3 & 5 \end{array}\right] \\ & \end{aligned}

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

\begin{aligned} &X=\left[\begin{array}{cc} 6+3 & -14 \\ -16 & 25 \end{array}\right] \\\end{aligned}

X=\left[\begin{array}{cc} 9 & -14 \\ -16 & 25 \end{array}\right]

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