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Determinants — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDeterminants2 Marks
Question
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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Answer

Let $A=\left[\begin{array}{ccc} {1} & {-2} & {1} \\ {-2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]$
$\therefore |A|=1(15-1) + 2(-10-1) + 1(-2-3) = 14-22-5 = -13$
Now, $\begin{aligned} &A_{11}=14, A_{12}=11, A_{13}=-5\\ &A_{21}=11, A_{22}=4, A_{23}=-3\\ &A_{31}=-5, A_{12}=-3, A_{13}=-1 \end{aligned}$
$\therefore $ $a d j A=\left[\begin{array}{ccc} {14} & {11} & {-5} \\ {11} & {4} & {-3} \\ {-5} & {-3} & {-1} \end{array}\right]$
So, we have $A^{-1}=\frac{1}{|A|}(a d j A)$
= $-\frac{1}{13}\left[\begin{array}{ccc} {14} & {11} & {-5} \\ {11} & {4} & {-3} \\ {-5} & {-3} & {-1} \end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc} {-14} & {-11} & {5} \\ {-11} & {-4} & {3} \\ {5} & {3} & {1} \end{array}\right]$
Clearly, $a d j A=\left[\begin{array}{ccc} {-14} & {-11} & {5} \\ {-11} & {-4} & {3} \\ {5} & {3} & {1} \end{array}\right]$
$\therefore |adj A| = 14(-4-9)-11(-11-15)-5(-33+20)$
$= 14(-13)-11(-26)-5(-13)$
$= -182 + 286 + 65 = 169$
we have,
$adj (adj A) = \left[\begin{array}{ccc} {-13} & {26} & {-13} \\ {26} & {-39} & {-13} \\ {-13} & {-13} & {-65} \end{array}\right]$
$\therefore $ $[\operatorname{adj} A]^{\prime}=\frac{1}{|a d j A|}(\operatorname{adj}(\operatorname{adj} A))$
$= \frac{1}{169}\left[\begin{array}{ccc} {-13} & {26} & {-13} \\ {26} & {-39} & {-13} \\ {-13} & {-13} & {-65} \end{array}\right]$
$= \frac{1}{13}\left[\begin{array}{ccc} {-1} & {2} & {-1} \\ {2} & {-3} & {-1} \\ {-1} & {-1} & {-5} \end{array}\right]$
Now, $A^{-1}=\frac{1}{13}\left[\begin{array}{ccc} {-14} & {-11} & {5} \\ {-11} & {-4} & {3} \\ {5} & {3} & {1} \end{array}\right]$= $\left[\begin{array}{ccc} {-\frac{14}{13}} & {-\frac{11}{13}} & {\frac{5}{13}} \\ {-\frac{11}{13}} & {-\frac{4}{13}} & {\frac{3}{13}} \\ {\frac{5}{13}} & {\frac{3}{13}} & {\frac{1}{13}} \end{array}\right]$
$\therefore adj (A^{-1}) = \left[\begin{array}{ccc} {-\frac{4}{169}-\frac{9}{169}} & {-\left(-\frac{11}{169}-\frac{15}{169}\right)} & {-\frac{33}{169}+\frac{20}{169}} \\ {-\left(-\frac{11}{169}-\frac{15}{169}\right)} & {-\frac{14}{169}-\frac{25}{169}} & {-\left(-\frac{42}{169}+\frac{55}{169}\right)} \\ {-\frac{33}{169}+\frac{20}{169}} & {-\left(-\frac{42}{169}+\frac{55}{169}\right)} & {\frac{56}{169}-\frac{121}{169}} \end{array}\right]$
$= \frac{1}{169}\left[\begin{array}{ccc} {-13} & {26} & {-13} \\ {26} & {-39} & {-13} \\ {-13} & {-13} & {-65} \end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc} {-1} & {2} & {-1} \\ {2} & {-3} & {-1} \\ {-1} & {-1} & {-5} \end{array}\right]$
Hence$, [adj A]^{-1} = adj(A^{-1})$

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