Q : 8        Let A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}. Verify that,

                (i) \dpi{100} [adj A]^-^1 = adj (A^-^1)

Answers (1)
D Divya Prakash Singh

Given that A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix};

So, let us assume that A^{-1} = B matrix and adjA = C then;

|A| = 1(15-1) -2(10-1) +1(2-3) = 14-18-1 = -5 \neq 0

Hence its inverse exists;

A^{-1} = \frac{1}{|A|} adjA    or  B = \frac{1}{|A|}C;

so, we now calculate the value of adjA

Cofactors of A;

A_{11}= (-1)^{1+1}(15-1) = 14          A_{12}= (-1)^{1+2}(10-1) = -9

A_{13}= (-1)^{1+3}(2-3) = -1          A_{21}= (-1)^{2+1}(10-1) = -9

A_{22}= (-1)^{2+2}(5-1) = 4             A_{23}= (-1)^{2+3}(1-2) = 1

A_{31}= (-1)^{3+1}(2-3) = -1         A_{32}= (-1)^{3+2}(1-2) = 1???????

A_{33}= (-1)^{3+3}(3-4) = -1???????

\Rightarrow adjA =C= \begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}

A^{-1} =B= \frac{1}{|A|} adjA = \frac{1}{-5}\begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}

Finding the inverse of C;

|C| = 14(-4-1)+9(9+1)-1(-9+4) = -70+90+5 = 25 \neq 0

Hence its inverse exists;

C^{-1} = \frac{1}{|C|}adj C

Now, finding the adjC;

C_{11}= (-1)^{1+1}(-4-1) = -5                       C_{12}= (-1)^{1+2}(9+1) = -10

C_{13}= (-1)^{1+3}(-9+4) = -5                       C_{21}= (-1)^{2+1}(9+1) = -10

C_{22}= (-1)^{2+2}(-14-1) = -15                 C_{23}= (-1)^{2+3}(14-9) = -5

C_{31}= (-1)^{3+1}(-9+4) = -5                        C_{32}= (-1)^{3+2}(14-9) = -5

C_{33}= (-1)^{3+3}(56-81) = -25

adjC = \begin{bmatrix} -5 &-10 &-5 \\ -10& -15 & -5\\ -5& -5& -25 \end{bmatrix}

C^{-1} = \frac{1}{|C|}adjC = \frac{1}{25}\begin{bmatrix} -5 &-10 &-5 \\ -10& -15 & -5\\ -5& -5& -25 \end{bmatrix} = \begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}

or L.H.S. = C^{-1} = [adjA]^{-1} = \begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}

Now, finding the R.H.S.

adj (A^{-1}) = adj B

A^{-1} =B= \begin{bmatrix} \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{bmatrix}

Cofactors of B;

B_{11}= (-1)^{1+1}(\frac{-4}{25}-\frac{1}{25}) = \frac{-1}{5}          

B_{12}= (-1)^{1+2}(\frac{9}{25}+\frac{1}{25}) =- \frac{2}{5}

B_{13}= (-1)^{1+3}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}   

B_{21}= (-1)^{2+1}(\frac{9}{25}+\frac{1}{25}) = -\frac{2}{5}

B_{22}= (-1)^{2+2}(\frac{-14}{25}-\frac{1}{25}) = \frac{-3}{5}

 B_{23}= (-1)^{2+3}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}

B_{31}= (-1)^{3+1}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}

B_{32}= (-1)^{3+2}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}

B_{33}= (-1)^{3+3}(\frac{56}{25}-\frac{81}{25}) = -1

R.H.S. = adjB = adj(A^{-1}) =\begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}

 

Hence L.H.S. = R.H.S. proved.

 

 

 

 

 

 

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