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  • Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverse of a Matrix Exercise 6.1 Question 1 Subquestion (i) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverse of a Matrix Exercise 6.1 Question 1 Subquestion (i) Maths Textbook Solution.

Answers (1)

Answer:

Adj \left ( A \right )=\left[\begin{array}{cc} 4 & -2 \\ -5 & -3 \end{array}\right]

Hint:

Here, we use basic concept of adjoint of matrix .

Given:

  Adj A=\left[\begin{array}{cc} -3 & 5 \\ 2 & 4 \end{array}\right]     

Solution:

A=\left[\begin{array}{cc} -3 & 5 \\ 2 & 4 \end{array}\right]

\begin{aligned} &|A|=-3 \times 4-5 \times 2 \\ &=-12-10=-22 \\ &C_{11}=4, C_{12}=-2, C_{21}=-5, C_{22}=-3 \end{aligned}                                                   

C_{i j}=\left[\begin{array}{cc} 4 & -2 \\ -5 & -3 \end{array}\right]                                                                                                

 Taking transpose,

\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{cc} 4 & -5 \\ -2 & -3 \end{array}\right]\\ \\ &\operatorname{Adj}(A) \times A=\left[\begin{array}{cc} 4 & -5 \\ -2 & -3 \end{array}\right]\left[\begin{array}{cc} -3 & 5 \\ 2 & 4 \end{array}\right] \\ \\&=\left[\begin{array}{cc} -12+(-10) & 20-20 \\ 6-6 & -10-12 \end{array}\right] \end{aligned}                                       

=\left[\begin{array}{cc} -22 & 0 \\ 0 & -22 \end{array}\right]                                                            (1)

|A| I=-22\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} -22 & 0 \\ 0 & -22 \end{array}\right]                   (2)

\begin{aligned} &A(\operatorname{Adj}(A))=\left[\begin{array}{cc} -3 & 5 \\ 2 & 4 \end{array}\right]\left[\begin{array}{cc} 4 & -5 \\ -2 & -3 \end{array}\right]\\ \end{aligned}

=\left[\begin{array}{cc} -22 & 0 \\ 0 & -22 \end{array}\right]                                                            (3)

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

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

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