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  • Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 10 Suquestion (i) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 10 Suquestion (i) Maths Textbook Solution.

Answers (1)

Answer:

Proved   \left ( AB \right )^{-1}=B^{-1}A^{-1}

Hint:

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

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

Given:

A=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right], B=\left[\begin{array}{ll} 4 & 6 \\ 3 & 2 \end{array}\right]        

Solution:

Let’s find  \left | A \right |

|A|=\left|\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right|=15-14=1

\operatorname{Adj}(A)=\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right]

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

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

Then let’s find  \left | B \right |Adj\left ( B \right ) \& \: B^{-1}

|B|=\left|\begin{array}{cc} 4 & 6 \\ 3 & 2 \end{array}\right|=8-18=-10

\operatorname{Adj}(B)=\left[\begin{array}{cc} 2 & -6 \\ -3 & 4 \end{array}\right]

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

B^{-1}=\frac{1}{-10} \times\left[\begin{array}{cc} 2 & -6 \\ -3 & 4 \end{array}\right]

Then find AB

\begin{aligned} &A \times B=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right] \times\left[\begin{array}{ll} 4 & 6 \\ 3 & 2 \end{array}\right] \\ &\end{aligned}

=\left[\begin{array}{cc} 12+6 & 18+4 \\ 28+15 & 42+10 \end{array}\right]=\left[\begin{array}{ll} 18 & 22 \\ 43 & 52 \end{array}\right]

Then let’s find  \left | AB \right |,Adj\left ( AB \right )  and inverse of  AB

|A B|=\left|\begin{array}{cc} 18 & 22 \\ 43 & 52 \end{array}\right|=936-946=-10

\operatorname{Adj}(A B)=\left[\begin{array}{cc} 52 & -22 \\ -43 & 18 \end{array}\right]

(A B)^{-1}=\frac{1}{-10} \times\left[\begin{array}{cc} 52 & -22 \\ -43 & 18 \end{array}\right]=\frac{1}{10}\left[\begin{array}{cc} -52 & 22 \\ 43 & -18 \end{array}\right]                 (1)

Now B^{-1}A^{-1}

=\frac{1}{10}\left[\begin{array}{cc} 2 & -6 \\ 3 & 4 \end{array}\right]\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right]         

=\frac{1}{10}\left[\begin{array}{cc} 10+42 & -4-18 \\ -15-28 & 6+12 \end{array}\right]                                

=\frac{1}{10}\left[\begin{array}{cc} -52 & 22 \\ 43 & -18 \end{array}\right]                           (2)

From equation (1) and (2)

\left ( AB \right )^{-1}=B^{-1}A^{-1}

Hence proved 

Posted by

infoexpert27

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