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

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

Answers (1)

Answer:

\operatorname{Adj}(A)=\left[\begin{array}{ccc} 2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & -1 \end{array}\right]

Hint:

Here, we use basic concept of adjoint of matrix.

Given:

A=\left[\begin{array}{ccc} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \end{array}\right]        

Solution:

Let’s find \left | A \right |

\begin{aligned} &|A|=1\left|\begin{array}{ll} 3 & 1 \\ 1 & 1 \end{array}\right|-2\left|\begin{array}{cc} 2 & 1 \\ -1 & 1 \end{array}\right|+5\left|\begin{array}{cc} 2 & 3 \\ -1 & 1 \end{array}\right| \\\\ &=1(3-1)-2(2+1)+5(2+3) \\\\ &=2-6+25=21 \end{aligned}

Let’s find cofactors

\begin{aligned} &C_{11}=+(3-1)=2 \\ &C_{12}=-(2+1)=-3 \\ &C_{13}=+(2+3)=5 \\ &C_{21}=-(2-5)=3 \\ &C_{22}=+(1+6)=6 \end{aligned}
\begin{aligned} &C_{23}=-(1+2)=-3 \\ &C_{31}=+(2-15)=-13 \\ &C_{32}=-(1-10)=9 \\ &C_{33}=+(3-4)=-1 \end{aligned}

C_{i j}=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 6 & -3 \\ -13 & 9 & -1 \end{array}\right]

Let’s take transpose of C_{i j}

\operatorname{Adj}(A)=\left[\begin{array}{ccc} 2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & -1 \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} 2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & -1 \end{array}\right] \times\left[\begin{array}{ccc} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21 \end{array}\right] \end{aligned}               (1)

|A| I=21 \times\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21 \end{array}\right]                                                                    (2)

A \times \operatorname{Adj}(A)=\left[\begin{array}{ccc} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \end{array}\right] \times\left[\begin{array}{ccc} 2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & 1 \end{array}\right]=\left[\begin{array}{ccc} 21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21 \end{array}\right]                   (3)

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

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

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