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  • Explain solution RD Sharma class 12 Chapter 6 Adjoint and Inverse of a Matrix exercise 6.1 question 2 Subquestion (i)

Explain solution RD Sharma class 12 Chapter 6 Adjoint and Inverse of a Matrix exercise 6.1 question 2 Subquestion (i)

Answers (1)

Answer:

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

Hint:

Here, we use basic concept of adjoint of matrix.

Given:

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

Solution:

Let’s find \left | A \right |

\begin{aligned} &|A|=1\left|\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right|-2\left|\begin{array}{ll} 2 & 2 \\ 2 & 1 \end{array}\right|+2\left|\begin{array}{ll} 2 & 1 \\ 2 & 2 \end{array}\right| \\\\ &=1(1-4)-2(2-4)+2(4-2) \\\\ &=(-3)+4+4 \\ \\&=5 \end{aligned}

Let’s find cofactors

\begin{aligned} &C_{11}=+(1-4)=-3 \\\\ &C_{12}=-(2-4)=2 \\\\ &C_{13}=+(4-2)=2 \\\\ &C_{21}=-(2-4)=2 \\\\ &C_{22}=+(1-4)=-3 \end{aligned}

\begin{aligned} &C_{23}=-(2-4)=2 \\\\ &C_{31}=+(4-2)=2 \\\\ &C_{32}=-(2-4)=2 \\\\ &C_{33}=+(1-4)=-3 \\\\ &C_{i j}=\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right] \end{aligned}

Let’s transpose it

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

Let’s verify below

\begin{aligned} \operatorname{Adj}(A) \times A &=|A| I=A \times \operatorname{Adj}(A) \\\\ \operatorname{Adj}(A) \times A &=\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right] \times\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right] \end{aligned}

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

=\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]                                                                              (1)

|A| I=5 \times\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]                                (2)

\begin{aligned} &A \times \operatorname{Adj}(A)=\left[\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right] \times\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right] \\\\ &=\left[\begin{array}{ccc} -3+4+4 & 0 & 0 \\ 0 & -3+4+4 & 0 \\ 0 & 0 & -3+4+4 \end{array}\right] \end{aligned}

=\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]                                                                                (3)

From equation (1), (2) and (3)

A \times A d j(A)=|A| I=\operatorname{Adj}(A) \times A

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