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  • Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 25 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 25 Maths Textbook Solution.

Answers (1)

Answer:

A^{-1}=\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]

Hint:

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

A^{-1}=\frac{1}{|A|} \times A d j(A)

Given:

A=\left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right]                                                        

Solution:

A^{3}-A^{2}-3 A-I_{3}=0

So,

A^{3}=A^{2} \times A

\begin{aligned} & \\ &A^{2}=A \times A=\left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1+0-6 & 0+0-8 & -2+0-2 \\ -2+2+6 & 0+1+8 & 4-2+2 \\ 3-8+3 & 0-4+4 & -6+8+1 \end{array}\right] \\\\ & \end{aligned}

=\left[\begin{array}{ccc} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{array}\right]

\begin{aligned} &A^{2} \times A=\left[\begin{array}{ccc} -5 & -8 & -4 \\ 6 & 9 & 2 \\ 3 & 4 & 1 \end{array}\right] \times\left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right]=\left[\begin{array}{ccc} -5+16-12 & 0-8+16 & 10-16-4 \\ 6-18+12 & 0-9+16 & -12+18+4 \\ -2-0+9 & 0-0-12 & 4+0+3 \end{array}\right] \\\\ & \end{aligned}

=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 0 & 7 & 10 \\ 7 & 12 & 7 \end{array}\right]

Now,

A^{3}-A^{2}-3 A-I

\begin{aligned} & \\ &{\left[\begin{array}{ccc} -1 & -8 & -10 \\ 0 & 7 & 10 \\ 7 & 12 & 7 \end{array}\right]-\left[\begin{array}{ccc} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{array}\right]-3\left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right]-\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]} \end{aligned}

\begin{aligned} &=\left[\begin{array}{ccc} -1+5 & -8+8 & -10+4 \\ 0-6 & 7-9 & 10-4 \\ 7+2 & 12-0 & 7-3 \end{array}\right]+\left[\begin{array}{ccc} -3-1 & 0 & 6-0 \\ 6-0 & 3-1 & -6-0 \\ -9-0 & -12+0 & -3-1 \end{array}\right] \\\\ &=\left[\begin{array}{ccc} 4 & 0 & -6 \\ -6 & -2 & 6 \\ 9 & 12 & 4 \end{array}\right]+\left[\begin{array}{ccc} -4 & 0 & 6 \\ 6 & 2 & -6 \\ -9 & -12 & -4 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \end{aligned}

Thus,

\begin{aligned} &A^{3}-A^{2}-3 A-I\\ &\text { now }\\ &(A A A) \times A^{-11}-A A A^{-1}-3 A A^{-1}-L A^{-1}=0\\ &A^{2}-A-3 A-I=0\\ &A^{-1}=\left(A^{2}-A-3 I\right) \end{aligned}

\begin{aligned} &=\left[\begin{array}{ccc} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{array}\right]-\left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right]-\left[\begin{array}{ccc} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right] \\ \end{aligned}

\begin{bmatrix} -5-1-3 &-8+0 & -4+2\\ 6+2-0 &9+1-3 &4-2 \\ -2-3-0 &-4 & 3-1-3 \end{bmatrix}=\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]

A^{-1}=\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]

 

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