Subjective Type

Let $$A=$$ $$\begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$$. Verify that (i)$${ [adj\quad A }]^{ -1 }=adj\quad { (A) }^{ -1 }$$ (ii)$${ (A }^{ -1 })^{ -1 }=A$$

Solution

Let $$A = \begin{bmatrix} 1&-1&1 \\-2&3 &1\\ 1&1&5 \end{bmatrix}$$

(i) $$[adj A]^{-1} = adj(A^{-1})$$

First we will calculate $$adj (A)$$ & $$A^{-1}$$

$$Adj\ A = \begin{bmatrix} A_{11}&A_{12}& A_{13} \\A_{21} & A_{22}& A_{23}\\ A_{31}& A_{32} & A_{33} \end{bmatrix}' = \begin{bmatrix} A_{11}&A_{21}& A_{31} \\A_{12} & A_{22}& A_{32}\\ A_{31}& A_{32} & A_{33} \end{bmatrix}$$

$$A = \begin{bmatrix}1 & -2&1 \\ -2& 3 &1\\ 1 & 1 &5\end{bmatrix}$$

$$M_{11} = \begin{vmatrix} 3& 1 \\ 1& 5\end{vmatrix} = 15 - 1 = 14$$

$$M_{12} = \begin{vmatrix}-2&1 \\1 &5\end{vmatrix} = -10 - 1 = -11$$

$$M_{13} = \begin{vmatrix}-2&3\\1 &1\end{vmatrix} = -2-3 = -5$$

$$M_{21} = \begin{vmatrix}-2&1\\1 &5\end{vmatrix} = -10-4 = -11$$

$$M_{22} = \begin{vmatrix}1&1\\1 &5\end{vmatrix} = 5-1 = 4$$

$$M_{23} = \begin{vmatrix}1&-2\\1 &1\end{vmatrix} = 1+2= 3$$

$$M_{31} = \begin{vmatrix}-2&1\\3&1\end{vmatrix} = -2-3 = -5$$

$$M_{32} = \begin{vmatrix}1&1\\-2 &1\end{vmatrix} = 1+2= 3$$

$$M_{33} = \begin{vmatrix}1&-2\\-2 &3\end{vmatrix} = 3-4= -5$$

$$A_{11} = (-1)^{1+1}M_{11} = (-1)^2 -14 = 14$$

$$A_{12} = (-1)^{1+2} M_{12} = (-1)^3 (-11) = 11$$

$$A_{13} = (-1)^{1+3}M_{13} = (-1)^4(-5) = -5$$

$$A_{21} = (-1)^{2+1}M_{21} = (-1)^3. (-11) = 11$$

$$A_{22} = (-1)^{2+2}M_{22}=(-1)^4. 4 = 4$$

$$A_{23} = (-1)^{2+3}M_{23}=(-1)^5(3)=-3$$

$$A_{31} = (-1)^{3+1}M_{31}=(-1)^4.(-5)=-5$$

$$A_{32} = (-1)^{3+2}M_{32} = (-1)^5.(3) = -3$$

$$A_{33} = (-1)^{3+3}M_{33} = (-1)^6.(-5) = -1$$

Thus, $$adj\ (A) = \begin{bmatrix} A_{11}&A_{21}& A_{31} \\A_{12} & A_{22}& A_{33}\\ A_{31}& A_{32} & A_{33} \end{bmatrix} = \begin{bmatrix} 14& 11&-5 \\11&4&-3\\ -5& -3& -1 \end{bmatrix} $$

Now,
$$A^{-1} = \dfrac{1}{|A|} adj\ (B)$$

Finding $$|A|$$
$$|A| = \begin{vmatrix} -1& -2& 1\\-2& 3 & 1\\ 1 & 1 &5\end{vmatrix}$$
$$=1(15 - 1) + 2 (-10 -1) + (-2 -3) = 14 - 22 - 5 = -13$$

therefore
$$A^{-1} = \dfrac{1}{|A|} adj \ (A) = \dfrac{1}{13} \begin{bmatrix} 14& 11& -5\\ 11& 4& -3\\-5 & -3&-1\end{bmatrix} $$

We need to verify
$$[adj\ A]^{-1} = adj \ (A^{-1})$$

Taking L.H.S
$$(adj\ A)^{-1}$$

Let $$B = adj\ (A)$$
$$B = \begin{bmatrix} 14& 11 &-5 \\11&4&-3 \\ -5& -3& -1\end{bmatrix} $$

Now,
$$B^{-1} = \dfrac{1}{|B|} adj \ (B)$$
exists if $$|B| \neq 0$$

$$|B| = \begin{vmatrix} 14&11&-5 \\11&4&-3\\-5 & -3&-1\end{vmatrix}$$
$$= 14(-4-9) + 1 (-11 - 15) - 5 (-33 + 20)$$
$$= 14(-13)-11(-26)-5(-13) = -182 + 286+65 = 169$$

Thus $$|B| =169 \neq 0$$
$$\therefore B^{-1}$$ exist

Now, calculating $$adj \ B$$
$$adj\ B = \begin{bmatrix} A_{11}&A_{12}& A_{13} \\A_{21} & A_{22}& A_{23}\\ A_{31}& A_{32} & A_{33} \end{bmatrix}' = \begin{bmatrix} A_{11}&A_{21}& A_{31} \\A_{12} & A_{22}& A_{32}\\ A_{31}& A_{32} & A_{33} \end{bmatrix}$$

Here $$A_{ij} $$ are the cofactors of matrix $$B$$
$$B = \begin{bmatrix}14 & 11&-5 \\ 11& 4 &-3\\ -5&-3 &-1\end{bmatrix}$$

$$M_{11} = \begin{vmatrix} 4&-3 \\-3&-1\end{vmatrix} = -4 -9 = -13$$

$$M_{12} = \begin{vmatrix} 11&-3 \\-5&-1\end{vmatrix} = -11 -15 = -26$$

$$M_{13} = \begin{vmatrix} 11&4 \\-5&-3\end{vmatrix} = -33 +20 = -13$$

$$M_{21} = \begin{vmatrix} 11&-5 \\-3&-1\end{vmatrix} = -11 -15 = -26$$

$$M_{22} = \begin{vmatrix} 14&-5 \\-5&-1\end{vmatrix} = -14 -25 = -39$$

$$M_{23} = \begin{vmatrix} 14&11 \\-5&-3\end{vmatrix} =(-42+55) = +13$$

$$M_{31} = \begin{vmatrix} 11&-5 \\4&-3\end{vmatrix} = -33-20 = -13$$

$$M_{32} = \begin{vmatrix} 14&5 \\-11&3\end{vmatrix} = -42+55 = 13$$

$$M_{33} = \begin{vmatrix} 14&11 \\-11&4\end{vmatrix} = 56-121= -65$$

Now, $$B^{-1} = \dfrac{1}{|B|} (adj\ B) = \dfrac{1}{169}\begin{bmatrix} -13& 26& 13 \\26&-39 &-13\\ -13 & -13 & -65 \end{bmatrix} $$

Taking $$13$$ common from all elements of the matrix
$$= \dfrac{13}{169} \begin{bmatrix} -1&2& -1\\ 2 & -3 & -1\\ -1& -1 & -5 \end{bmatrix} $$
$$= \dfrac{1}{13} \begin{bmatrix} -1&2& -1\\ 2 & -3 & -1\\ -1& -1 & -5 \end{bmatrix} $$
Thus, $$[adj\ A]^{-1} = B^{-1} = \dfrac{1}{13} \begin{bmatrix} -1&2& -1\\ 2 & -3 & -1\\ -1& -1 & -5 \end{bmatrix} $$
Taking R.H.S
$$adj\ (A^{-1})$$
$$A^{-1} = \dfrac{1}{13} \begin{bmatrix} -14& 11 & 5\\ -11 & -4&3\\ 5& 3 & 1 \end{bmatrix}$$
Let $$C = A^{-1}$$
$$C = \dfrac{1}{13} \begin{bmatrix} -14& 11 & 5\\ -11 & -4&3\\ 5& 3 & 1 \end{bmatrix}$$

$$= \begin{bmatrix}\dfrac{-14}{13} & \dfrac{-11}{13} & \dfrac{5}{13} \\ \dfrac{-11}{13} & \dfrac{-4}{13} & \dfrac{3}{13}\\ \dfrac{5}{13} & \dfrac{3}{13} & \dfrac{1}{13} \end{bmatrix}$$ [since 13 is divided to the matrix 13 gets divided to all elements of the matrix]

$$adj \ C = \begin{bmatrix} A_{11}&A_{12}& A_{13} \\A_{21} & A_{22}& A_{23}\\ A_{31}& A_{32} & A_{33} \end{bmatrix}' = \begin{bmatrix} A_{11}&A_{21}& A_{31} \\A_{12} & A_{22}& A_{32}\\ A_{31}& A_{32} & A_{33} \end{bmatrix}$$

Here $$A_{ij}$$ are the cofactors of matrix $$C$$

$$C = \begin{bmatrix}\dfrac{-14}{13} & \dfrac{-11}{13} & \dfrac{5}{13} \\ \dfrac{-11}{13} & \dfrac{-4}{13} & \dfrac{3}{13}\\ \dfrac{5}{13} & \dfrac{3}{13} & \dfrac{1}{13} \end{bmatrix}$$

$$M_{11} = \begin{vmatrix} \dfrac{-4}{13}&\dfrac{-3}{13} \\\dfrac{3}{13}&\dfrac{1}{13}\end{vmatrix} = \dfrac{-4}{169} - \dfrac{9}{169} =\dfrac{-13}{169} = \dfrac{-1}{13}$$

$$M_{12} = \begin{vmatrix} \dfrac{-11}{13}&\dfrac{3}{13} \\\dfrac{5}{13}&\dfrac{3}{13}\end{vmatrix} = \dfrac{-11}{169} - \dfrac{15}{169} =\dfrac{-26}{169} = \dfrac{-2}{13}$$

$$M_{13} = \begin{vmatrix} \dfrac{-11}{13}&\dfrac{-4}{13} \\\dfrac{5}{13}&\dfrac{3}{13}\end{vmatrix} = \dfrac{-33}{169} - \dfrac{20}{169} =\dfrac{-13}{169} = \dfrac{-1}{13}$$

$$M_{21} = \begin{vmatrix} \dfrac{-11}{13}&\dfrac{5}{13} \\\dfrac{3}{13}&\dfrac{1}{13}\end{vmatrix} = \dfrac{-11}{169} - \dfrac{55}{169} =\dfrac{-26}{169} = \dfrac{-2}{13}$$

$$M_{22} = \begin{vmatrix} \dfrac{-14}{13}&\dfrac{5}{13} \\\dfrac{5}{13}&\dfrac{1}{13}\end{vmatrix} = \dfrac{-14}{169} - \dfrac{25}{169} =\dfrac{-39}{169} = \dfrac{-3}{13}$$

$$M_{23} = \begin{vmatrix} \dfrac{-14}{13}&\dfrac{-11}{13} \\\dfrac{5}{13}&\dfrac{3}{13}\end{vmatrix} = \left(\dfrac{-42}{169} - \dfrac{55}{169}\right) =\dfrac{13}{169} = \dfrac{1}{13}$$

$$M_{31} = \begin{vmatrix} \dfrac{-11}{13}&\dfrac{5}{13} \\\dfrac{-4}{13}&\dfrac{3}{13}\end{vmatrix} = \dfrac{-33}{169} - \dfrac{20}{169} =\dfrac{-13}{169} = \dfrac{-1}{13}$$

$$M_{32} = \begin{vmatrix} \dfrac{-14}{13}&\dfrac{5}{13} \\\dfrac{-11}{13}&\dfrac{3}{13}\end{vmatrix} = \dfrac{-42}{169} - \dfrac{55}{169} =\dfrac{13}{169} = \dfrac{1}{13}$$

$$M_{33} = \begin{vmatrix} \dfrac{-14}{13}&\dfrac{-11}{13} \\\dfrac{-11}{13}&\dfrac{-4}{13}\end{vmatrix} = \dfrac{56}{169} - \dfrac{-121}{169} =\dfrac{-65}{169} = \dfrac{-5}{13}$$

$$A_{11} = (-1)^{1+1} M_{11} = (-1)^{2} \left(\dfrac{-1}{13}\right) = \dfrac{-1}{13}$$

$$A_{12} = (-1)^{1+2} M_{12} = (-1)^3\left(\dfrac{-2}{13} \right) = \dfrac{2}{13}$$

$$A_{13} = (-1)^{1+3} M_{13} = (-1)^4 . \dfrac{-1}{13} = \dfrac{-1}{13}$$

$$A_{21} = (-1)^{2+1} M_{21} = (-1)^3. \left(\dfrac{-2}{13}\right) = \dfrac{2}{13}$$

$$A_{22} = (-1)^{2+2}M_{22} = (-1)^4.\left(\dfrac{-3}{13}\right) = \dfrac{-3}{13}$$

$$A_{23} = (-1)^{2+3}M_{23} = (-1)^5. \left(\dfrac{1}{13}\right) = \dfrac{-1}{13}$$

$$A_{31} = (-1)^{3+1}M_{31} = (-1)^4. \left(\dfrac{-1}{13}\right) = \dfrac{-1}{13}$$

$$A_{32} = (-1)^{3+2} M_{32} = (-1)^5. \left(\dfrac{1}{13}\right) = \dfrac{-1}{13}$$

$$A_{33} = (-1)^{1+3}M_{33} = (-1)^6. \left(\dfrac{-5}{13}\right) = \dfrac{-5}{13}$$

Thus $$adj \ C = \begin{vmatrix} \dfrac{-1}{13}&\dfrac{2}{13}& \dfrac{-1}{13} \\ \dfrac{2}{13} & \dfrac{-3}{13}& \dfrac{-1}{13}\\ \dfrac{-1}{13}& \dfrac{-1}{13}& \dfrac{-5}{13} \end{vmatrix} = \dfrac{1}{13} \begin{bmatrix} -1 &2&-1 \\ 2& -3& -1\\ -1& -1& -5 \end{bmatrix} $$
$$\therefore adj \ (A^{-1}) = adj \ C = \dfrac{1}{13} \begin{bmatrix} -1 &2&-1 \\ 2& -3& -1\\ -1& -1& -5 \end{bmatrix} = R.H.S $$
Hence L.H.S = R.H.S
$$\therefore (adj \ A)^{-1} = adj\ (A^{-1})$$
(ii) $$(A^{-1})^{-1} = A$$
We have to find $$(A^{-1})^{-1}$$
So, $$(A^{-1})^{-1} = \dfrac{1}{|A^{-1}|}adj \ (A^{-1})$$
From First part,
$$A^{-1} = \dfrac{1}{13} \begin{bmatrix} -14& -11 & 5\\ -11&-4 & 3\\5&3&1\end{bmatrix} $$
Calculating $$|A^{-1}|$$
$$|A^{-1}| = \begin{vmatrix} \dfrac{1}{13}\begin{bmatrix} -14&-11 & 5 \\ -11&-4 & 3 \\ 5&3&1\end{bmatrix} \end{vmatrix}$$
Using $$|KA| = K^{n} |A|$$ Where $$n$$ is order of $$A$$
$$=\left(\dfrac{1}{13}\right)^3$$ $$\left(-14\begin{vmatrix} -4& 3 \\ 3 & 1 \end{vmatrix} - (-11) \begin{vmatrix} -11&3 \\ 5& 1 \end{vmatrix} +5 \begin{vmatrix} -11 & -4 \\ 5 & 3 \end{vmatrix}\right)$$
$$=\left(\dfrac{1}{13}\right)^3 (-14(-4-9)+11(-11-15)+5(-33+20))$$
$$= \left(\dfrac{1}{13}\right)^3(-14(-13)+11(-26)+5(-13))$$
$$=\left(\dfrac{1}{13}\right)^3 (182-286-65)=\left(\dfrac{1}{13}\right)^3(-169) = \dfrac{-169}{13\times 13\times 13} = \dfrac{1}{13}$$
Now, $$(A^{-1})^{-1} = \dfrac{1}{|A^{-1}|} (adj\ A^{-1})$$
Putting values
$$=\dfrac{1}{\frac{-1}{13}} \times \dfrac{1}{13} \begin{bmatrix} -1&2&-1 \\ 2&-3 & -1\\-1&-1&-5 \end{bmatrix} = -13\times \dfrac{1}{13} \begin{bmatrix} -1&2&-1 \\ 2&-3 & -1\\-1&-1&-5 \end{bmatrix} $$
$$= - \begin{bmatrix} -1&2&-1 \\ 2&-3 & -1\\-1&-1&-5 \end{bmatrix} = \begin{bmatrix} -1&2&-1 \\ 2&-3 & -1\\-1&-1&-5 \end{bmatrix} $$
$$=A$$
Thus, $$(A^{-1})^{-1} = A$$
Hence proved

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