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Let A = $$\left[\begin{array}{ccc} {1} & {2} & {1} \\ {2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]$$. verify that $$[adj A] ^{–1} = adj (A ^{–1})$$
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QuestionMathsClass 12

Let A = $$\left[\begin{array}{ccc} {1} & {2} & {1} \\ {2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]$$. verify that $$[adj A] ^{–1} = adj (A ^{–1})$$

See the solution below.
4.6
4.6

Solution

Let $$A=\left[\begin{array}{ccc} {1} & {2} & {1} \\ {2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]$$
$$\therefore |A|=1(15-1) - 2(10-1) + 1(2-3) = 14-18-1= -5$$
Now, \begin{aligned} &A_{11}=14, A_{12}=-9, A_{13}=-1\\ &A_{21}=-9, A_{22}=4, A_{23}=1\\ &A_{31}=-1, A_{12}=1, A_{13}=-1 \end{aligned}
$$\therefore a d j A=\left[\begin{array}{ccc} {14} & {-9} & {-1} \\ {-9} & {4} & {1} \\ {-1} & {1} & {-1} \end{array}\right]$$
So, we have $$A^{-1}=\frac{1}{|A|}(a d j A)$$
= $$-\frac{1}{5}\left[\begin{array}{ccc} {14} & {-9} & {-1} \\ {-9} & {4} & {1} \\ {-1} & {1} & {-1} \end{array}\right]$$
Now,
$$|adj A| = 14(-4-1)+9(9+1)-1(-9+4)$$
$$= 14(-5)+9(10)-1(-5)$$
$$=-70+90+5=25$$
we have,
$$a d j A=\left[\begin{array}{ccc} {14} & {-9} & {-1} \\ {-9} & {4} & {1} \\ {-1} & {1} & {-1} \end{array}\right]$$
Now,
\begin{aligned} &A_{11}=5, A_{12}=-10, A_{13}=-5\\ &A_{21}=10, A_{22}=-15, A_{23}=5\\ &A_{31}=-5, A_{32}=-5, A_{33}=-25 \end{aligned}
So,
$$adj (adj A) = \left[\begin{array}{ccc} {5} & {-10} & {-5} \\ {10} & {-15} & {5} \\ {-5} & {-5} & {-25} \end{array}\right]$$
$$\therefore [\operatorname{adj} A]^{-1}=\frac{1}{|a d j A|}(\operatorname{adj}(\operatorname{adj} A))$$
$$= \frac{1}{25} \left[\begin{array}{ccc} {5} & {-10} & {-5} \\ {10} & {-15} & {5} \\ {-5} & {-5} & {-25} \end{array}\right]$$
$$= \frac{1}{5}\left[\begin{array}{ccc} {-1} & {2} & {-1} \\ {2} & {-3} & {-1} \\ {-1} & {-1} & {-5} \end{array}\right]$$
$$=\frac{-1}{5}\left[\begin{array}{lll} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right]$$$$\dots(i)$$
Now, $$A^{-1}=-\frac{1}{5}\left[\begin{array}{ccc} {14} & {-9} & {-1} \\ {-9} & {4} & {1} \\ {-1} & {1} & {-1} \end{array}\right]$$
$$\begin{array}{l} \text { Also, } \operatorname{adj}\left(A^{-1}\right)=\operatorname{adj}\left(\frac{-1}{5}\left[\begin{array}{rrr} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \end{array}\right]\right) \\ =\operatorname{adj}\left[\begin{array}{rrr} -\frac{14}{5} & \frac{9}{5} & \frac{1}{5} \\ \frac{9}{5} & -\frac{4}{5} & \frac{-1}{5} \\ \frac{1}{5} & \frac{-1}{5} & \frac{1}{5} \end{array}\right]=\left[\begin{array}{rrr} -\frac{5}{25} & \frac{-10}{25} & -\frac{5}{25} \\ \frac{-10}{25} & -\frac{15}{25} & -\frac{5}{25} \\ \frac{-5}{25} & -\frac{5}{25} & -\frac{25}{25} \end{array}\right]^{t} \\ \Rightarrow\left(\operatorname{adj} A^{-1}\right)=\frac{-1}{5}\left[\begin{array}{lll} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right]\dots(ii) \end{array}$$
Using equation $$(i)$$and, $$(ii)$$, we say that$$[adj A] ^{-1} = adj(A ^{-1})$$.