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  • Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 20

Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 20

Answers (1)

Answer:

      X = 9 \: and\: y =14  ,   A^{-1}=\frac{1}{14}\left[\begin{array}{cc} 5 & -3 \\ -2 & 4 \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} 4 & 3 \\ 2 & 5 \end{array}\right]

Solution:

\begin{aligned} &A^{2}=\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]=\left[\begin{array}{cc} 16+6 & 12+15 \\ 8+10 & 6+25 \end{array}\right]=\left[\begin{array}{cc} 22 & 27 \\ 18 & 31 \end{array}\right] \\\\ &A^{2}-x A+y I=\left[\begin{array}{ll} 22 & 37 \\ 18 & 31 \end{array}\right]-x\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 0 & y \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \end{aligned}

\begin{aligned} &22-4 x+y=0 \\ &4 x-y=22 \\ &x=9 \end{aligned}

So,

\begin{aligned} &4 x-y=22 \\ &4 \times 9-y=22 \\ &36-22=y \\ &y=14 \end{aligned}

So  A^{2}-9 A+14 I=0

Multiply by A^{-1}  both sides,

\begin{aligned} &A \times A \times A^{-1}-9 A \times A^{-1}+14 I \times A^{-1}=0 \\\\ &A-9 I+14 A^{-1}=0 \\\\ &A^{-1}=\frac{1}{14}[9 I-A] \end{aligned}

\begin{aligned} &A^{-1}=\frac{1}{14}\left[\begin{array}{ll} 9 & 0 \\ 0 & 9 \end{array}\right]-\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right] \\ \\&A^{-1}=\frac{1}{14}\left[\begin{array}{cc} 5 & -3 \\ -2 & 4 \end{array}\right] \end{aligned}                             

 

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