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

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

Answers (1)

Answer:

\operatorname{Adj}(A)=\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]

Hint:

Here, we use basic concept of adjoint of matrix.

Given:

 A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]   

Solution:

Let

\begin{aligned} &A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \\\\ &|A|=\left|\begin{array}{ll} a & b \\ c & d \end{array}\right| \\\\ &|A|=a \times d-c \times b \\\\ &|A|=a d-c b \end{aligned}

Let’s find cofactor

\begin{aligned} &C_{11}=d, C_{12}=-c, C_{21}=-b, C_{22}=a \\ &C_{i j}=\left[\begin{array}{cc} d & -c \\ -b & a \end{array}\right] \end{aligned}

Let’s transpose  C_{ij}

\operatorname{Adj}(A)=\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]

Let’s prove this below

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

\begin{aligned} &\operatorname{Adj}(A) \times A=\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] \times\left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \\\\ &=\left[\begin{array}{cc} d a+(-b c) & d b-b d \\ -c a+a c & -c b+a d \end{array}\right] \\\\ &\operatorname{Adj}(A) \times A=\left[\begin{array}{cc} d a-b c & 0 \\ 0 & d a-b c \end{array}\right] \end{aligned}                                                (1)

MHS

\begin{aligned} &|A| I=a d-c b\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \\&=\left[\begin{array}{cc} a d-c b & 0 \\ 0 & a d-c b \end{array}\right] \\\\ &=\left[\begin{array}{cc} d a-b c & 0 \\ 0 & a d-c b \end{array}\right] \end{aligned}                                                 (2)

RHS

A \times A d j(A)=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \times\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]

=\left[\begin{array}{cc} a d-b c & 0 \\ 0 & d a-b c \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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