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  • Provide Solution for RD Sharma Class 12 Chapter 6 Adjoint and Inverse Matrix Exercise 6.1 Question 13

Provide Solution for RD Sharma Class 12 Chapter 6 Adjoint and Inverse Matrix Exercise 6.1 Question 13

Answers (1)

Answer:

Hence proved  A-3 I=2\left(I+3 A^{-1}\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} 4 & 5 \\ 2 & 1 \end{array}\right]     

Solution:

Let’s find  |A|, A d j(A) \& A^{-1}

\begin{aligned} &|A|=4-10=-6 \\ &\operatorname{Adj}(A)=\left[\begin{array}{cc} 1 & -5 \\ -2 & 4 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{-6}\left[\begin{array}{cc} 1 & -5 \\ -2 & 4 \end{array}\right] \end{aligned}  

To show  A-3 I=2\left(I+3 A^{-1}\right)

LHS

A-3 I=\left[\begin{array}{ll} 4 & 5 \\ 2 & 1 \end{array}\right]-3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 5 \\ 2 & -2 \end{array}\right]                        (1)

RHS

\begin{aligned} &2\left(I+3 A^{-1}\right)=2 I+6 A^{-1}=2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\frac{6}{6}\left[\begin{array}{cc} -1 & 5 \\ 2 & -4 \end{array}\right] \\ &=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]+\left[\begin{array}{cc} -1 & 5 \\ 2 & -4 \end{array}\right] \end{aligned}

= \left[\begin{array}{cc} 1 & 5 \\ 2 & -2 \end{array}\right]                               (2) 

Here from equation (1) and (2)

A-3 I=2\left(I+3 A^{-1}\right)

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