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Mathematics Question on Determinants

Let A=[1−21 −231 115]A=\begin{bmatrix}1&-2&1\\\ -2&3&1\\\ 1&1&5\end{bmatrix}verify that -
(i)[adjA]−1=adj(A−1)(i)[adjA]^{-1}=adj(A^{-1})
(ii)(A−1)−1=A(ii)(A^{-1})^{-1}=A

Answer

A=[1−21 −231 115]A=\begin{bmatrix}1&-2&1\\\ -2&3&1\\\ 1&1&5\end{bmatrix}
∴∣A∣=1(15−1)+2(−10−1)+1(−2−3)=14−22−5=−13∴|A|=1(15-1)+2(-10-1)+1(-2-3)=14-22-5=-13
Now,A11=14,A12=11,A13=−5A_{11}=14,A_{12}=11,A_{13}=-5
A21=11,A22=4,A23=−3A_{21}=11,A_{22}=4,A_{23}=-3
A31=−5,A32=−3,A33=−1A_{31}=-5,A_{32}=-3,A_{33}=-1
∴adjA=[1411−5 114−3 −5−3−1]∴adjA=\begin{bmatrix}14& 11& -5\\\ 11& 4& -3\\\ -5& -3& -1\end{bmatrix}
∴A−1=1∣A∣(adjA)∴A^{-1}=\frac{1}{|A|}(adjA)
=−113[1411−5 114−3 −5−3−1]=\frac{-1}{13}\begin{bmatrix}14& 11& -5\\\ 11& 4& -3\\\ -5& -3& -1\end{bmatrix}
=113[−14−115 −11−43 531]=\frac{1}{13}\begin{bmatrix}-14& -11& 5\\\ -11& -4& 3\\\ 5& 3& 1\end{bmatrix}
(i)∣adjA∣=14(−4−9)−11(−11−15)−5(−33+20)(i)|adjA|=14(-4-9)-11(-11-15)-5(-33+20)
=14(−13)−11(−26)−5(−13)=14(-13)-11(-26)-5(-13)
=−182+286+65=169=-182+286+65=169
we have,
adj(adjA)=[−1326−13 26−39−13 −13−1365]adj(adjA)=\begin{bmatrix}-13& 26& -13\\\ 26& -39& -13\\\ -13& -13& 65\end{bmatrix}
∴[adjA]−1=1∣adjA∣(adj(adjA))∴[adjA]^{-1}=\frac{1}{|adjA|}(adj(adjA))
1169[−1326−13 26−39−13 −13−1365]\frac{1}{169}\begin{bmatrix}-13& 26& -13\\\ 26& -39& -13\\\ -13& -13& 65\end{bmatrix}
=113[−12−1 2−3−1 −1−1−5]=\frac{1}{13}\begin{bmatrix}-1&2&-1\\\ 2&-3&-1\\\ -1&-1&-5\end{bmatrix}
Now,A−1=113[−14−115 −11−43 531]=[−1413−1113513 −1113−413313 513313113]A^{-1}=\frac{1}{13}\begin{bmatrix}-14& -11& 5\\\ -11& -4& 3\\\ 5& 3& 1\end{bmatrix}=\begin{bmatrix}\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{bmatrix}
∴adj(A−1)=[−4/169−9/169−(−11/169−15/169)−33/169+20/169 −(−11/169−15/169)−14/169−25/169−(−42/169+55/169) −33/169+20/169−(−42/169+55/169)56/169−121/169]∴adj(A^{-1})=\begin{bmatrix}-4/169-9/169& -(-11/169-15/169)& -33/169+20/169\\\ -(-11/169-15/169)& -14/169-25/169& -(-42/169+55/169)\\\ -33/169+20/169& -(-42/169+55/169)& 56/169-121/169\end{bmatrix}
=1169[−1326−13 26−39−13 −13−1365]=\frac{1}{169}\begin{bmatrix}-13& 26& -13\\\ 26& -39& -13\\\ -13& -13& 65\end{bmatrix}
=113[−12−1 2−3−1 −1−1−5]=\frac{1}{13}\begin{bmatrix}-1&2&-1\\\ 2&-3&-1\\\ -1&-1&-5\end{bmatrix}
Hence,[adjA]−1=adj(A−1).Hence,[adjA]^{-1}=adj(A^{-1}).
(ii) we have shown that
A−1=113[−14−115 −11−43 531]A^{-1}=\frac{1}{13}\begin{bmatrix}-14& -11& 5\\\ -11& -4& 3\\\ 5& 3& 1\end{bmatrix}
And,adjA−1=113[−12−1 2−3−1 −1−1−5]adjA^{-1}=\frac{1}{13}\begin{bmatrix}-1&2&-1\\\ 2&-3&-1\\\ -1&-1&-5\end{bmatrix}Now,
∣A−1∣=(113)3[−14×(−13)+11×(−26)+5×(−13)]=(113)3×(−169)=−113|A^{-1}|=(\frac{1}{13})^3[-14\times(-13)+11\times(-26)+5\times(-13)]=(\frac{1}{13})^3\times(-169)=\frac{-1}{13}
∴(A−1)−1=adjA−1∣A−1∣=1(−113)×113[−12−1 2−3−1 −1−1−5]=[1−21 −231 115]=A∴(A^{-1})^{-1}=\frac{adjA^{-1}}{|A-1|}=\frac{1}{(\frac{-1}{13})}\times\frac{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}=A
∴(A−1)−1=A∴(A^{-1})^{-1}=A