Advertisements
Advertisements
प्रश्न
If \[A = \begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\] , show that \[A^{- 1} = A^3\]
उत्तर
\[\text{ We have, } A = \begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix}\]
\[ A^2 = \begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix}\begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix} = \begin{bmatrix} 9 - 6 + 0 & - 9 + 9 - 4 & 12 - 12 + 4\\6 - 6 + 0 & - 6 + 9 - 4 & 8 - 12 + 4\\0 - 2 + 0 & 0 + 3 - 1 & 0 - 4 + 1 \end{bmatrix} = \begin{bmatrix} 3 & - 4 & 4\\ 0 & - 1 & 0\\ - 2 & 2 & - 3 \end{bmatrix}\]
\[\text{ Now,} A^3 = A^2 \times A^{} = \begin{bmatrix} 3 & - 4 & 4\\ 0 & - 1 & 0\\ - 2 & 2 & - 3 \end{bmatrix}\begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix} = \begin{bmatrix} 9 - 8 & - 9 + 12 - 4 & 12 - 16 + 4\\ 0 - 2 + 0 & 0 + 3 + 0 & - 4\\ - 6 + 4 + 0 & 6 - 6 + 3 & - 8 + 8 - 3 \end{bmatrix} = \begin{bmatrix} 1 & - 1 & 0\\ - 2 & 3 & - 4\\ - 2 & 3 & - 3 \end{bmatrix}\]
\[\text{ Again, }A^3 \times A = \begin{bmatrix} 1 & - 1 & 0\\ - 2 & 3 & - 4\\ - 2 & 3 & - 3 \end{bmatrix}\begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 - 2 + 0 & - 3 + 3 + 0 & 4 - 4 + 0\\ - 6 + 6 & 6 - 9 + 4 & - 8 + 12 - 4\\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix} = I_3 [\text{ Identity matrix of order 3 }]\]
\[ \Rightarrow A^3 \times A = I_3 \]
\[ \Rightarrow A^3 = A^{- 1}\]
APPEARS IN
संबंधित प्रश्न
Find the inverse of the matrices (if it exists).
`[(2,-2),(4,3)]`
Find the adjoint of the following matrix:
\[\begin{bmatrix}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix}\]
Find the inverse of the following matrix:
Find the inverse of the following matrix.
Find the inverse of the following matrix.
Find the inverse of the following matrix and verify that \[A^{- 1} A = I_3\]
For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]
\[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 6 \\ 3 & 2\end{bmatrix}\]
For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]
\[A = \begin{bmatrix}2 & 1 \\ 5 & 3\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 5 \\ 3 & 4\end{bmatrix}\]
Find the inverse of the matrix \[A = \begin{bmatrix}a & b \\ c & \frac{1 + bc}{a}\end{bmatrix}\] and show that \[a A^{- 1} = \left( a^2 + bc + 1 \right) I - aA .\]
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.
Show that
If \[A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\], show that
Show that \[A = \begin{bmatrix}5 & 3 \\ - 1 & - 2\end{bmatrix}\] satisfies the equation \[x^2 - 3x - 7 = 0\]. Thus, find A−1.
Find the matrix X satisfying the matrix equation \[X\begin{bmatrix}5 & 3 \\ - 1 & - 2\end{bmatrix} = \begin{bmatrix}14 & 7 \\ 7 & 7\end{bmatrix}\]
Find the adjoint of the matrix \[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\] and hence show that \[A\left( adj A \right) = \left| A \right| I_3\].
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}5 & 2 \\ 2 & 1\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & 5 \\ 1 & 3\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}- 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix}\]
If \[A = \begin{bmatrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\text{ and }A \left( adj A = \right)\begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix}\], then find the value of k.
If A is an invertible matrix such that |A−1| = 2, find the value of |A|.
Find the inverse of the matrix \[\begin{bmatrix}3 & - 2 \\ - 7 & 5\end{bmatrix} .\]
Find the inverse of the matrix \[\begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\]
If \[A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}, B = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\] , find adj (AB).
If \[A = \begin{bmatrix}2 & 3 \\ 5 & - 2\end{bmatrix}\] , write \[A^{- 1}\] in terms of A.
If A is a singular matrix, then adj A is ______.
If A satisfies the equation \[x^3 - 5 x^2 + 4x + \lambda = 0\] then A-1 exists if _____________ .
If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is _____________ .
If \[A = \begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix}\], find A−1 and hence solve the system of linear equations 2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y + 2z = −3
Find A−1, if \[A = \begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix}\] . Hence solve the following system of linear equations:x + 2y + 5z = 10, x − y − z = −2, 2x + 3y − z = −11
`("aA")^-1 = 1/"a" "A"^-1`, where a is any real number and A is a square matrix.
If A is a square matrix of order 3 and |A| = 5, then |adj A| = ______.
If for a square matrix A, A2 – A + I = 0, then A–1 equals ______.
| To raise money for an orphanage, students of three schools A, B and C organised an exhibition in their residential colony, where they sold paper bags, scrap books and pastel sheets made by using recycled paper. Student of school A sold 30 paper bags, 20 scrap books and 10 pastel sheets and raised ₹ 410. Student of school B sold 20 paper bags, 10 scrap books and 20 pastel sheets and raised ₹ 290. Student of school C sold 20 paper bags, 20 scrap books and 20 pastel sheets and raised ₹ 440. |
Answer the following question:
- Translate the problem into a system of equations.
- Solve the system of equation by using matrix method.
- Hence, find the cost of one paper bag, one scrap book and one pastel sheet.
