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Find the adjoint of the given matrix and verify in each case that A. (\operatorname{adj} A)=(\operatorname{adj} A)=m|A|.I.\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]
Adjoint and Inverse of a Matrix | CBSE | Class 12 | Maths | RS Aggarwal
Find the adjoint of the given matrix and verify in each case that A. (\operatorname{adj} A)=(\operatorname{adj} A)=m|A|.I.\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]

Solution:

Given matrix as A=\left(\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right).
Find: the adjoint of the matrix given.

Step: 1 Find the minor matrix of A.

M_{A}=\left(\begin{array}{ll} \left|\begin{array}{ll} 2 & 3 \\ 1 & 1 \end{array}\right| & \left|\begin{array}{ll} 1 & 3 \\ 3 & 1 \end{array}\right| & \left|\begin{array}{ll} 1 & 2 \\ 3 & 1 \end{array}\right| \\ \left|\begin{array}{ll} 1 & 2 \\ 1 & 1 \end{array}\right| & \left|\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right| & \left|\begin{array}{ll} 0 & 2 \\ 3 & 1 \end{array}\right| \\ \left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right| & \left|\begin{array}{ll} 0 & 2 \\ 1 & 3 \end{array}\right| & \left|\begin{array}{ll} 0 & 1 \\ 1 & 2 \end{array}\right| \end{array}\right)

=\left(\begin{array}{lll} -1 & -8 & -5 \\ -1 & -6 & -3 \\ -1 & -2 & -1 \end{array}\right)

Step: 2 Find the co-factor matrix of A.
\begin{array}{l} C_{A}=\left(\begin{array}{ccc} (-1)^{1+1}(-1) & (-1)^{1+2}(-8) & (-1)^{1+3}(-5) \\ (-1)^{2+1}(-1) & (-1)^{2+2}(-6) & (-1)^{2+3}(-3) \\ (-1)^{3+1}(-1) & (-1)^{3+2}(-2) & (-1)^{3+3}(-1) \end{array}\right) \\ =\left(\begin{array}{ccc} -1 & 8 & -5 \\ 1 & -6 & 3 \\ -1 & 2 & -1 \end{array}\right) . \end{array}

Step:3 By transpose of C_{A} we will have \operatorname{adj} A.
\operatorname{adj} A=C_{A}^{T}=\left(\begin{array}{ccc}-1 & 8 & -5 \\ 1 & -6 & 3 \\ -1 & 2 & -1\end{array}\right)^{T}=\left(\begin{array}{ccc}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right).

Let’s now calculate:
(i) A(\operatorname{adj} A)=\left(\begin{array}{ccc}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right)\left(\begin{array}{ccc}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right)
\begin{array}{l} =\left(\begin{array}{ccc} 0+8-10 & 0-6+6 & 0+2-2 \\ -1+16-15 & 1-12+9 & -1+4-3 \\ -3+8-5 & 3-6+3 & -3+2-1 \end{array}\right) \\ =\left(\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right) \\ =2 I_{3} \end{array}

(ii) (\operatorname{adj} A) A=\left(\begin{array}{ccc}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right)\left(\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right)
=\left(\begin{array}{lll} 0+1-3 & -1+2-1 & -2+3-1 \\ 0-6+6 & 8-12+2 & 16-18+2 \\ 0+3-3 & -5+6-1 & -10+9-1 \end{array}\right)
=\left(\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right)
=2 I_{3} .

(iii) |A|=\left|\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right|=0(2-3)-(1)(1-9)+2(1-6)=2
As a result from above cases it can be written as A(\operatorname{adj} A)=(\operatorname{adj} A)=|A| I_{2} and adjoint of given matrix is \left(\begin{array}{ccc}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right).

i

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