Given a = ⎡ ⎢ ⎣ 5 0 4 2 3 2 1 2 1 ⎤ ⎥ ⎦ , B − 1 = ⎡ ⎢ ⎣ 1 3 3 1 4 3 1 3 4 ⎤ ⎥ ⎦ . Compute (Ab)−1. - Mathematics

Question

Given \[A = \begin{bmatrix}5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1\end{bmatrix}, B^{- 1} = \begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}\] . Compute (AB)⁻¹.

Solution

We have,
\[A = \begin{bmatrix}5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1\end{bmatrix}\]
\[ B^{- 1} = \begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}\]
\[\text{ We know }(AB )^{- 1} = B^{- 1} A^{- 1} \]
For matrix A,
\[ C_{11} = \begin{vmatrix}3 & 2 \\ 2 & 1\end{vmatrix} = - 1, C_{12} = - \begin{vmatrix}2 & 2 \\ 1 & 1\end{vmatrix} = 0\text{ and }C_{13} = \begin{vmatrix}2 & 3 \\ 1 & 2\end{vmatrix} = 1\]
\[ C_{21} = - \begin{vmatrix}0 & 4 \\ 2 & 1\end{vmatrix} = 8, C_{22} = \begin{vmatrix}5 & 4 \\ 1 & 1\end{vmatrix} = 1\text{ and }C_{23} = - \begin{vmatrix}5 & 0 \\ 1 & 2\end{vmatrix} = - 10\]
\[ C_{31} = \begin{vmatrix}0 & 4 \\ 3 & 2\end{vmatrix} = - 12, C_{32} = - \begin{vmatrix}5 & 4 \\ 2 & 2\end{vmatrix} = - 2\text{ and }C_{33} = \begin{vmatrix}5 & 0 \\ 2 & 3\end{vmatrix} = 15\]
Now,
\[adj (A) = \begin{bmatrix}- 1 & 0 & 1 \\ 8 & 1 & - 10 \\ - 12 & - 2 & 15\end{bmatrix}^T = \begin{bmatrix}- 1 & 8 & - 12 \\ 0 & 1 & - 2 \\ 1 & - 10 & 15\end{bmatrix}\]
\[\text{ and }\left| A \right| = - 1\]
\[ \therefore A^{- 1} = - \begin{bmatrix}- 1 & 8 & - 12 \\ 0 & 1 & - 2 \\ 1 & - 10 & 15\end{bmatrix} = \begin{bmatrix}1 & - 8 & 12 \\ 0 & - 1 & 2 \\ - 1 & 10 & - 15\end{bmatrix}\]
\[So, B^{- 1} A^{- 1} = \begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}\begin{bmatrix}1 & - 8 & 12 \\ 0 & - 1 & 2 \\ - 1 & 10 & - 15\end{bmatrix} = \begin{bmatrix}- 2 & 19 & - 27 \\ - 2 & 18 & - 25 \\ - 3 & 29 & - 42\end{bmatrix}\]

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Properties of Matrix Multiplication - Inverse of a Square Matrix by the Adjoint Method

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Q 15 Q 14 Q 16

Chapter 7: Adjoint and Inverse of a Matrix - Exercise 7.1 [Page 23]

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RD Sharma Mathematics [English] Class 12

Chapter 7 Adjoint and Inverse of a Matrix
Exercise 7.1 | Q 15 | Page 23

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