Question: Given the matrices

$$\begin{gathered} \mathrm{A}=\left(1-11 \\ 40-1 \\ 4-20\right),\mathrm{B}=\left(101 \\ 211 \\ 212\right) \end{gathered}$$

1. Find${\mathbf{A}}^{\mathbf{-}\mathbf{1}}\mathbf{,}{\mathbf{B}}^{\mathbf{-}\mathbf{1}}\mathbf{,}{\mathbf{B}}^{\mathbf{-}\mathbf{1}}\mathbf{AB}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{B}}^{\mathbf{-}\mathbf{1}}{\mathbf{A}}^{\mathbf{-}\mathbf{1}}\mathbf{B}$
2. Show that the last two matrices are inverses, that is, that their product is the unit matrix.

In mathematics, matrix multiplication is a way that produces a matrix by multiplying two matrices.For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the alternate matrix. The inverse of a matrix is a matrix that when multiplied with the original matrix gives the identity matrix that is${\mathbf{A}}^{\mathbf{-}\mathbf{1}}\mathbf{A}\mathbf{=}\mathbf{I}$.

The given matrices are,

$$\begin{gathered} \mathrm{A}=\left(1-11 \\ 40-1 \\ 4-20\right),\mathrm{B}=\left(101 \\ 211 \\ 212\right) \end{gathered}$$

The values of ${\mathrm{A}}^{-1},{\mathrm{B}}^{-1},{\mathrm{B}}^{-1}\mathrm{AB}\mathrm{and}{\mathrm{B}}^{-1}{\mathrm{A}}^{-1}\mathrm{B}$is to be found and also it is to be proven that the last two matrices are inverses.

It is known that the inverse of matrix can be found out by formula,

${\mathrm{A}}^{-1}=\frac{{\mathrm{A}}^{\mathrm{T}}}{\left|\mathrm{A}\right|}$ ……(1)

$$\begin{gathered} \mathrm{Here}{\mathrm{A}}^{\mathrm{T}}=\left(\begin{array}{ccc}\left|\begin{array}{cc}{\mathrm{a}}_{22}& {\mathrm{a}}_{23}\\ {\mathrm{a}}_{32}& {\mathrm{a}}_{33}\end{array}\right|& \left|\begin{array}{cc}{\mathrm{a}}_{13}& {\mathrm{a}}_{12}\\ {\mathrm{a}}_{33}& {\mathrm{a}}_{32}\end{array}\right|& \left|\begin{array}{cc}{\mathrm{a}}_{12}& {\mathrm{a}}_{13}\\ {\mathrm{a}}_{22}& {\mathrm{a}}_{23}\end{array}\right|\\ \left|\begin{array}{cc}{\mathrm{a}}_{23}& {\mathrm{a}}_{21}\\ {\mathrm{a}}_{33}& {\mathrm{a}}_{31}\end{array}\right|& \left|\begin{array}{cc}{\mathrm{a}}_{11}& {\mathrm{a}}_{32}\\ {\mathrm{a}}_{31}& {\mathrm{a}}_{33}\end{array}\right|& \left|\begin{array}{cc}{\mathrm{a}}_{13}& {\mathrm{a}}_{11}\\ {\mathrm{a}}_{23}& {\mathrm{a}}_{21}\end{array}\right|\\ \left|\begin{array}{cc}{\mathrm{a}}_{21}& {\mathrm{a}}_{22}\\ {\mathrm{a}}_{31}& {\mathrm{a}}_{32}\end{array}\right|& \left|\begin{array}{cc}{\mathrm{a}}_{12}& {\mathrm{a}}_{11}\\ {\mathrm{a}}_{32}& {\mathrm{a}}_{31}\end{array}\right|& \left|\begin{array}{cc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}\\ {\mathrm{a}}_{32}& {\mathrm{a}}_{22}\end{array}\right|\end{array}\right)\mathrm{and}\left|\mathrm{A}\right|=\det \left(\mathrm{A}\right). \\ \mathrm{First}\mathrm{find}\mathrm{the}\mathrm{transpose}\mathrm{of}\mathrm{matrix}\mathrm{A}. \\ {\mathrm{A}}^{\mathrm{T}}\left(\begin{array}{ccc}\left|\begin{array}{cc}0& -1\\ -2& 0\end{array}\right|& \left|\begin{array}{cc}1& -1\\ 0& -2\end{array}\right|& \left|\begin{array}{cc}-1& 1\\ 0& -1\end{array}\right|\\ \left|\begin{array}{cc}-1& 4\\ 0& 4\end{array}\right|& \left|\begin{array}{cc}1& 1\\ 4& 0\end{array}\right|& \left|\begin{array}{cc}1& 1\\ -1& 4\end{array}\right|\\ \left|\begin{array}{cc}4& 0\\ 4& -2\end{array}\right|& \left|\begin{array}{cc}-1& 1\\ -2& 4\end{array}\right|& \left|\begin{array}{cc}1& -1\\ -2& 0\end{array}\right|\end{array}\right) \\ =\left(\begin{array}{ccc}-2& -2& 1\\ -4& -4& 5\\ -8& -2& 4\end{array}\right) \\ \mathrm{Find}\mathrm{determinant}\mathrm{of}\mathrm{matrix}\mathrm{A}. \\ \left|\mathrm{A}\right|=\left|\begin{array}{ccc}1& -1& 1\\ 4& 0& -1\\ 4& -2& 0\end{array}\right| \\ =1\left(0-\left(-2\right)\left(-1\right)\right)-\left(-1\right)\left(-1\right)\left(4\times 0-\left(4\right)\left(-1\right)\right)+\left(4\times \left(-2\right)-4\times 0\right) \\ =-2+4-8 \\ =-6 \end{gathered}$$

Put the above obtained values in equation (1).

$$\begin{gathered} {\mathrm{A}}^{-1}=\frac{1}{-6}\left(\begin{array}{ccc}-2& -2& 1\\ -4& -4& 5\\ -8& -2& 4\end{array}\right) \\ =\frac{1}{6}\left(\begin{array}{ccc}2& 2& -1\\ 4& 4& -5\\ 8& 2& -4\end{array}\right) \\ \mathrm{Similarly},\mathrm{find}\mathrm{the}\mathrm{transpose}\mathrm{of}\mathrm{matrix}\mathrm{B}. \end{gathered}$$

$$\begin{gathered} {\mathrm{B}}^{\mathrm{T}}=\left(\begin{array}{ccc}\left|\begin{array}{cc}1& 1\\ 1& 2\end{array}\right|& \left|\begin{array}{cc}1& 0\\ 2& 1\end{array}\right|& \left|\begin{array}{cc}0& 1\\ 1& 1\end{array}\right|\\ \left|\begin{array}{cc}1& 2\\ 2& 2\end{array}\right|& \left|\begin{array}{cc}1& 1\\ 2& 2\end{array}\right|& \left|\begin{array}{cc}1& 1\\ 1& 2\end{array}\right|\\ \left|\begin{array}{cc}2& 1\\ 2& 1\end{array}\right|& \left|\begin{array}{cc}0& 1\\ 1& 2\end{array}\right|& \left|\begin{array}{cc}1& 0\\ 1& 1\end{array}\right|\end{array}\right) \\ =\left(\begin{array}{ccc}1& 0& -1\\ -2& 0& 1\\ 0& 1& 1\end{array}\right) \\ \mathrm{Find}\mathrm{determinant}\mathrm{of}\mathrm{matrix}\mathrm{B}. \\ \left|\mathrm{B}\right|=\left|\begin{array}{ccc}1& 0& 1\\ 2& 1& 1\\ 2& 1& 2\end{array}\right| \\ =1\left(1\times 2-1\right)-\left(0\right)\left(2\times 2-2\times 1\right)+1\left(2\times 1-2\times 1\right) \\ =1-0+0 \\ =1 \\ \mathrm{Substitute}\mathrm{the}\mathrm{above}\mathrm{obtained}\mathrm{values}\mathrm{in}\mathrm{formula}\mathrm{of}\mathrm{finding}\mathrm{inverse}\mathrm{of}\mathrm{matrix}. \\ {\mathrm{B}}^{-1}=\frac{1}{1}\left(\begin{array}{ccc}1& 1& -1\\ -2& 0& 1\\ 0& -1& 1\end{array}\right) \\ =\left(\begin{array}{ccc}1& 1& -1\\ -2& 0& 1\\ 0& -1& 1\end{array}\right) \end{gathered}$$

Now, find the value of AB first, to find the value of ${\mathrm{B}}^{-1}\mathrm{AB}$

$$\begin{gathered} \mathrm{AB}=\left(\begin{array}{ccc}1& -1& 1\\ 4& 0& -1\\ 4& -2& 0\end{array}\right)\left(\begin{array}{ccc}1& 0& 1\\ 2& 1& 1\\ 2& 1& 2\end{array}\right) \\ =\left(\begin{array}{ccc}1& 0& 2\\ 2& -1& 2\\ 0& -2& 2\end{array}\right) \\ \mathrm{Now}\mathrm{multiplyabove}\mathrm{obtained}\mathrm{matrix}\mathrm{with}\mathrm{matrix}{\mathrm{B}}^{-1}. \\ {\mathrm{B}}^{-1}\mathrm{AB}=\left(\begin{array}{ccc}1& 1& -1\\ -2& 0& 1\\ 0& -1& 1\end{array}\right)\left(\begin{array}{ccc}1& 0& 2\\ 2& -1& 2\\ 0& -2& 2\end{array}\right) \\ =\left(\begin{array}{ccc}3& 1& 2\\ -2& -2& -2\\ -2& -1& 0\end{array}\right) \\ \mathrm{Similarly},\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{B}}^{-1}{\mathrm{A}}^{-1}\mathrm{B} \\ {\mathrm{B}}^{-1}{\mathrm{A}}^{-1}\mathrm{B}=\frac{1}{6}\left(\begin{array}{ccc}1& 1& -1\\ -2& 0& 1\\ 0& -1& 1\end{array}\right)\left(\begin{array}{ccc}2& 2& -1\\ 4& 4& -5\\ 8& 2& -4\end{array}\right)\left(\begin{array}{ccc}1& 0& 1\\ 2& 1& 1\\ 2& 1& 2\end{array}\right) \\ =\frac{1}{6}\left(\begin{array}{ccc}1& 1& -1\\ -2& 0& 1\\ 0& -1& 1\end{array}\right)\left(\begin{array}{ccc}4& 1& 2\\ 2& -1& -2\\ 4& -2& 2\end{array}\right) \\ =\frac{1}{6}\left(\begin{array}{ccc}2& 2& -2\\ -4& -4& -2\\ 2& -1& 4\end{array}\right) \end{gathered}$$

Substitute the obtained values of${\mathrm{B}}^{-1}\mathrm{AB}\mathrm{and}{\mathrm{B}}^{-1}{\mathrm{A}}^{-1}\mathrm{B}$ and in left-hand side.

$$\begin{gathered} \left({\mathrm{B}}^{-1}\mathrm{AB}\right)\left({\mathrm{B}}^{-1}{\mathrm{A}}^{-1}\mathrm{B}\right)=\frac{1}{6}\left(\begin{array}{ccc}3& 1& 2\\ -2& -2& -2\\ -2& -1& 0\end{array}\right)\left(\begin{array}{ccc}2& 2& -2\\ -4& -4& -2\\ 2& -1& 4\end{array}\right) \\ =\frac{1}{6}\left(\begin{array}{ccc}6& 0& 0\\ 0& 6& 0\\ 0& 0& 0\end{array}\right) \\ \frac{1}{6}\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right) \\ =I \end{gathered}$$

Which means that ${\mathrm{B}}^{-1}\mathrm{AB}={\left({\mathrm{B}}^{-1}{\mathrm{A}}^{-1}\mathrm{B}\right)}^{-1^{}}$

Therefore, the values of ${\mathrm{A}}^{-1},{\mathrm{B}}^{-1}\left(\mathrm{AB}\right)\mathrm{and}{\mathrm{B}}^{-1}{\mathrm{A}}^{-1}\mathrm{B}$are $\frac{1}{6}\left(\begin{array}{ccc}2& 2& -1\\ 4& 4& -5\\ 8& 2& -4\end{array}\right),\left(\begin{array}{ccc}1& 1& -1\\ -4& 0& 1\\ 0& -1& 1\end{array}\right),\left(\begin{array}{ccc}3& 1& 2\\ -2& -2& -2\\ 0& -1& 1\end{array}\right),\frac{1}{6}\left(\begin{array}{ccc}2& 2& -2\\ -4& -4& -2\\ 2& -1& 4\end{array}\right)$respectively and the last two matrices are proven to be inverses that is their product is an identity(unit) matrix.