If a = ⎡ ⎢ ⎣ 1 − 2 3 0 − 1 4 − 2 2 1 ⎤ ⎥ ⎦ , Find ( a T ) − 1 . - Mathematics

Question

If \[A = \begin{bmatrix}1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1\end{bmatrix},\text{ find }\left( A^T \right)^{- 1} .\]

Solution

We know that (A^T)⁻¹ = (A⁻¹)^T.
\[A = \begin{bmatrix}1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1\end{bmatrix}\]
\[ A^{- 1} = \frac{1}{\left| A \right|}Adj . A\]
Now,
\[\left| A \right| = \begin{vmatrix}1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1\end{vmatrix}\]
\[ = 1\left( - 1 - 8 \right) - 2\left( - 8 + 3 \right)\]
\[ = - 9 + 10\]
\[ = 1\]
\[\text{ Now, to find Adj . A}\]
\[A_{11} = \left( - 1 \right)^{1 + 1} \left( - 9 \right) = - 9\]
\[ A_{12} = \left( - 1 \right)^{1 + 2} \left( 8 \right) = - 8\]
\[ A_{13} = \left( - 1 \right)^{1 + 3} \left( - 2 \right) = - 2\]
\[A_{21} = \left( - 1 \right)^{2 + 1} \left( - 8 \right) = 8\]
\[ A_{22} = \left( - 1 \right)^{2 + 2} \left( 7 \right) = 7 \]
\[ A_{23} = \left( - 1 \right)^{2 + 3} \left( - 2 \right) = 2\]
\[ A_{31} = \left( - 1 \right)^{3 + 1} \left( - 5 \right) = - 5\]
\[ A_{32} = \left( - 1 \right)^{3 + 2} \left( 4 \right) = - 4\]
\[ A_{33} = \left( - 1 \right)^{3 + 3} \left( - 1 \right) = - 1\]
Therefore,
\[Adj . A = \begin{bmatrix}- 9 & 8 & - 5 \\ - 8 & 7 & - 4 \\ - 2 & 2 & - 1\end{bmatrix}\]
Thus,
\[ A^{- 1} = \begin{bmatrix}- 9 & 8 & - 5 \\ - 8 & 7 & - 4 \\ - 2 & 2 & - 1\end{bmatrix} . \]
\[ \left( A^T \right)^{- 1} = \left( A^{- 1} \right)^T \]
\[ = \begin{bmatrix}- 9 & 8 & - 5 \\ - 8 & 7 & - 4 \\ - 2 & 2 & - 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 9 & - 8 & - 2 \\ 8 & 7 & 2 \\ - 5 & - 4 & - 1\end{bmatrix}\]
\[\text{ Hence, }\left( A^T \right)^{- 1} = \begin{bmatrix}- 9 & - 8 & - 2 \\ 8 & 7 & 2 \\ - 5 & - 4 & - 1\end{bmatrix} .\]

shaalaa.com

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

Is there an error in this question or solution?

Q 37 Q 36 Q 38

Chapter 7: Adjoint and Inverse of a Matrix - Exercise 7.1 [Page 25]

APPEARS IN

RD Sharma Mathematics [English] Class 12

Chapter 7 Adjoint and Inverse of a Matrix
Exercise 7.1 | Q 37 | Page 25

Video TutorialsVIEW ALL [3]

RELATED QUESTIONS

For the matrix A = `[(3,2),(1,1)]` find the numbers a and b such that A² + aA + bI = O.

If `A^(-1) =[(3,-1,1),(-15,6,-5),(5,-2,2)]` and `B = [(1,2,-2),(-1,3,0),(0,-2,1)]` find `(AB)^(-1)`

If x, y, z are nonzero real numbers, then the inverse of matrix A = `[(x,0,0),(0,y,0),(0,0,z)]` is ______.

Find the adjoint of the following matrix:
\[\begin{bmatrix}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix}\]

Verify that (adj A) A = |A| I = A (adj A) for the above matrix.

Find the inverse of the following matrix.

\[\begin{bmatrix}2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix}\]

Find the inverse of the following matrix and verify that \[A^{- 1} A = I_3\]

\[\begin{bmatrix}2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2\end{bmatrix}\]

For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]

\[A = \begin{bmatrix}2 & 1 \\ 5 & 3\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 5 \\ 3 & 4\end{bmatrix}\]

If \[A = \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\] , verify that \[A^2 - 4 A + I = O,\text{ where }I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\text{ and }O = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\] . Hence, find A⁻¹.

Show that

\[A = \begin{bmatrix}- 8 & 5 \\ 2 & 4\end{bmatrix}\] satisfies the equation \[A^2 + 4A - 42I = O\]. Hence, find A⁻¹.

If \[A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\], show that

\[A^2 - 5A + 7I = O\]. Hence, find A⁻¹.

If \[A = \begin{bmatrix}3 & - 2 \\ 4 & - 2\end{bmatrix}\], find the value of \[\lambda\] so that \[A^2 = \lambda A - 2I\]. Hence, find A⁻¹.

Show that \[A = \begin{bmatrix}5 & 3 \\ - 1 & - 2\end{bmatrix}\] satisfies the equation \[x^2 - 3x - 7 = 0\]. Thus, find A⁻¹.

Solve the matrix equation \[\begin{bmatrix}5 & 4 \\ 1 & 1\end{bmatrix}X = \begin{bmatrix}1 & - 2 \\ 1 & 3\end{bmatrix}\], where X is a 2 × 2 matrix.

\[\text{ If }A^{- 1} = \begin{bmatrix}3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1\end{bmatrix},\text{ find }\left( AB \right)^{- 1} .\]

Find the adjoint of the matrix \[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\] and hence show that \[A\left( adj A \right) = \left| A \right| I_3\].

\[\text{ If }A = \begin{bmatrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{bmatrix},\text{ find }A^{- 1}\text{ and show that }A^{- 1} = \frac{1}{2}\left( A^2 - 3I \right) .\]