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  • Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 24 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 24 Maths Textbook Solution.

Answers (1)

Answer:

A^{-1}=\frac{-1}{11}\left[\begin{array}{ccc} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \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}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]

Solution:

\begin{aligned} A^{3} &=A^{2} \times A \\\\ A^{2} &=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 1+1+2 & 1+2-1 & 1-3+3 \\ 1+2-6 & 1+4+3 & 1-6+9 \\ 2-1+6 & 2-2-3 & 2+3+9 \end{array}\right] \\\\ A^{2} &=\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right] \end{aligned}

\begin{aligned} &A^{2} \times A=\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 4+2+2 & 4+4-1 & 4-6+3 \\ -3+8-28 & -3+14+14 & -3-24-42 \\ 7-3+28 & 7-6-14 & 7+9+42 \end{array}\right] \\ \\&=\left[\begin{array}{ccc} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{array}\right] \end{aligned}

Now,

\begin{aligned} &A^{3}-6 A^{2}+5 A+11 I \\\\ &{\left[\begin{array}{ccc} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{array}\right]-6\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]+5\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]+11\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]} \end{aligned}

\left[\begin{array}{ccc} 8-24 & 7-12 & 1-6 \\ -23+18 & 27-48 & -69+84 \\ 32-42 & -13+18 & 58-84 \end{array}\right]+\left[\begin{array}{ccc} 5+11 & 5 & 5 \\ 5 & 10+11 & -15 \\ 10 & -5 & 26 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]

A^{3}-6 A^{2}+5 A+11 I=0

Now,

\begin{aligned} &(A \times A \times A) A^{-1}-6(A \times A) A^{-1}+5 A A^{-1}+11 I A^{-1}=0 \\\\ &A A\left(A^{-1} A\right)-6 A\left(A^{-1} A\right)+5 A^{-1} A=-11\left(A^{-1} I\right) \\\\ &A^{-1}=\frac{-1}{11}\left(A^{2}-6 A+5 I\right) \end{aligned}

Now,

A^{2}-6 A+5 I

\begin{aligned} & \\ &{\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]-6\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]+5\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]} \end{aligned}

\begin{aligned} &=\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]-\left[\begin{array}{ccc} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{array}\right]+\left[\begin{array}{ccc} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right] \\\\ &=\left[\begin{array}{ccc} -3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{array}\right] \end{aligned}

Hence,

A^{-1}=\frac{-1}{11}\left[\begin{array}{ccc} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{array}\right]

 

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