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  • Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 8 Subquestion (i).

Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 8 Subquestion (i).

Answers (1)

Answer:

            A^{-1}=\frac{1}{18} \times\left[\begin{array}{ccc} -5 & 1 & 7 \\ 1 & 7 & -5 \\ 7 & -5 & 1 \end{array}\right]

Hint:

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

Given:

A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{array}\right]           

Solution:

\begin{aligned} &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ &|A|=1\left|\begin{array}{ll} 3 & 1 \\ 1 & 2 \end{array}\right|-2\left|\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right|+3\left|\begin{array}{ll} 2 & 3 \\ 3 & 1 \end{array}\right| \\ &=1(6-1)-2(4-3)+3(2-9) \\ &|A|=5-2-21=-18 \end{aligned}

Let’s find  Adj \left ( A \right )

For that let’s find cofactor

\begin{aligned} &C_{11}=+(6-1)=5 \\ &C_{12}=-(4-3)=-1 \\ &C_{13}=+(2-9)=-7 \\ &C_{21}=-(4-3)=-1 \\ &C_{22}=+(2-9)=-7 \\ &C_{23}=-(1-6)=5 \\ &C_{31}=+(2-9)=-7 \\ &C_{32}=-(1-6)=5 \\ &C_{33}=+(3-4)=-1 \end{aligned}

\begin{aligned} &C_{i j}=\left[\begin{array}{ccc} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{array}\right] \\\\ &\operatorname{Adj}(A)=C_{\vec{j}}^{T} \end{aligned}

\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{array}\right] \\\\ &A^{-1}=\frac{1}{|A|} \times A d j(A) \end{aligned} 

\begin{aligned} A^{-1} &=-\frac{1}{18} \times\left[\begin{array}{ccc} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{array}\right]\\ \\ A^{-1} &=\frac{1}{18} \times\left[\begin{array}{ccc} -5 & 1 & 7 \\ 1 & 7 & -5 \\ 7 & -5 & 1 \end{array}\right] \end{aligned}

 

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