Advertisements
Advertisements
प्रश्न
Verify that \[A^3 - 6 A^2 + 9A - 4I = O\] and hence find A−1.
Advertisements
उत्तर
\[A = \begin{bmatrix}2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2\end{bmatrix}\].
\[ \Rightarrow \left| A \right| = \begin{vmatrix} 2 & - 1 & 1\\ - 1 & 2 & - 1\\ 1 & - 1 & 2 \end{vmatrix} = 2 \times \left( 4 - 1 \right) + 1\left( - 2 + 1 \right) + 1\left( 1 - 2 \right) = 6 - 1 - 1 = 4 \]
\[\text{ Since, }\left| A \right| \neq 0 \]
\[\text{ Hence, }A^{- 1}\text{ exists }. \]
Now,
\[ A^2 = \begin{bmatrix} 2 & - 1 & 1\\ - 1 & 2 & - 1\\ 1 & - 1 & 2 \end{bmatrix}\begin{bmatrix} 2 & - 1 & 1\\ - 1 & 2 & - 1\\ 1 & - 1 & 2 \end{bmatrix} = \begin{bmatrix} 4 + 1 + 1 & - 2 - 2 - 1 & 2 + 1 + 2\\ - 2 - 2 - 1 & 1 + 4 + 1 & - 1 - 2 - 2\\ 2 + 1 + 2 & - 1 - 2 - 2 & 1 + 1 + 4 \end{bmatrix} = \begin{bmatrix} 6 & - 5 & 5\\ - 5 & 6 & - 5\\ 5 & - 5 & 6 \end{bmatrix} \]
\[ A^3 = A^2 . A = \begin{bmatrix} 6 & - 5 & 5\\ - 5 & 6 & - 5\\ 5 & - 5 & 6 \end{bmatrix}\begin{bmatrix} 2 & - 1 & 1\\ - 1 & 2 & - 1\\ 1 & - 1 & 2 \end{bmatrix} = \begin{bmatrix} 12 + 5 + 5 & - 6 - 10 - 5 & 6 + 5 + 10\\ - 10 - 6 - 5 & 5 + 12 + 5 & - 5 - 6 - 10\\ 10 + 5 + 6 & - 5 - 10 - 6 & 5 + 5 + 12 \end{bmatrix} = \begin{bmatrix} 22 & - 21 & 21\\ - 21 & 22 & - 21\\ 21 &- 21 & 22 \end{bmatrix} \]
\[\text{ Now, }A^3 - 6 A^2 + 9A - 4I = \begin{bmatrix} 22 & - 21 & 21\\ - 21 & 22 & - 21\\ 21 & - 21 & 22 \end{bmatrix} - 6\begin{bmatrix} 6 & - 5 & 5\\ - 5 & 6 & - 5\\ 5 & - 5 & 6 \end{bmatrix} + 9\begin{bmatrix} 2 & - 1 & 1\\ - 1 & 2 & - 1\\ 1 & - 1 & 2 \end{bmatrix} - 4\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \]
\[ = \begin{bmatrix} 22 - 36 + 18 - 4 & - 21 + 30 - 9 - 0 & 21 - 30 + 9 - 0\\ - 21 + 30 - 9 - 0 & 22 - 36 + 18 - 4 & - 21 + 30 - 9 - 0\\ 21 - 30 + 9 - 0 & - 21 + 30 - 9 - 0 & 22 - 36 + 18 - 4 \end{bmatrix} \]
\[ = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} = O [\text{ Null matrix }]\]
Hence proved .
\[\text{ Now,} A^3 - 6 A^2 + 9A - 4I = O\]
\[ \Rightarrow A^{- 1} \times \left( A^3 - 6 A^2 + 9A - 4I \right) = A^{- 1} \times O \]
\[ \Rightarrow A^2 - 6A + 9I - 4 A^{- 1} = O\]
\[ \Rightarrow 4 A^{- 1} = A^2 - 6A + 9I\]
\[ \Rightarrow 4 A^{- 1} = \begin{bmatrix} 6 & - 5 & 5\\ - 5 & 6 & - 5\\ 5 & - 5 & 6 \end{bmatrix} - 6\begin{bmatrix} 2 & - 1 & 1\\ - 1 & 2 & - 1\\ 1 & - 1 & 2 \end{bmatrix} + 9\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\]
\[ \Rightarrow 4 A^{- 1} = \begin{bmatrix} 6 - 12 + 9 & - 5 + 6 + 0 & 5 - 6 + 0\\ - 5 + 6 + 0 & 6 - 12 + 9 & - 5 + 6 + 0\\ 5 - 6 + 0 & - 5 + 6 + 0 & 6 - 12 + 9 \end{bmatrix} \]
\[ \Rightarrow 4 A^{- 1} = \begin{bmatrix} 3 & 1 & - 1\\ 1 & 3 & 1\\ - 1 & 1 & 3 \end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{4}\begin{bmatrix} 3 & 1 & - 1\\ 1 & 3 & 1\\ - 1 & 1 & 3 \end{bmatrix}\]
APPEARS IN
संबंधित प्रश्न
Find the inverse of the matrices (if it exists).
`[(-1,5),(-3,2)]`
Find the inverse of the matrices (if it exists).
`[(1,2,3),(0,2,4),(0,0,5)]`
Find the inverse of the matrices (if it exists).
`[(1,0,0),(3,3,0),(5,2,-1)]`
Find the inverse of the matrices (if it exists).
`[(1,-1,2),(0,2,-3),(3,-2,4)]`
If A is an invertible matrix of order 2, then det (A−1) is equal to ______.
Find the inverse of the following matrix:
Find the inverse of the following matrix.
For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]
\[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 6 \\ 3 & 2\end{bmatrix}\]
Show that
If \[A = \begin{bmatrix}4 & 3 \\ 2 & 5\end{bmatrix}\], find x and y such that
For the matrix \[A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3\end{bmatrix}\] . Show that
Show that the matrix, \[A = \begin{bmatrix}1 & 0 & - 2 \\ - 2 & - 1 & 2 \\ 3 & 4 & 1\end{bmatrix}\] satisfies the equation, \[A^3 - A^2 - 3A - I_3 = O\] . Hence, find A−1.
If \[A = \begin{bmatrix}- 1 & 2 & 0 \\ - 1 & 1 & 1 \\ 0 & 1 & 0\end{bmatrix}\] , show that \[A^2 = A^{- 1} .\]
If \[A = \begin{bmatrix}1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1\end{bmatrix},\text{ find }\left( A^T \right)^{- 1} .\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3\end{bmatrix}\]
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 adj \[A = \begin{bmatrix}2 & 3 \\ 4 & - 1\end{bmatrix}\text{ and adj }B = \begin{bmatrix}1 & - 2 \\ - 3 & 1\end{bmatrix}\]
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.
Find the inverse of the matrix \[\begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\]
If \[A = \begin{bmatrix}2 & 3 \\ 5 & - 2\end{bmatrix}\] , write \[A^{- 1}\] in terms of A.
If \[A = \begin{bmatrix}3 & 4 \\ 2 & 4\end{bmatrix}, B = \begin{bmatrix}- 2 & - 2 \\ 0 & - 1\end{bmatrix},\text{ then }\left( A + B \right)^{- 1} =\]
If A is a singular matrix, then adj A is ______.
If B is a non-singular matrix and A is a square matrix, then det (B−1 AB) is equal to ___________ .
For any 2 × 2 matrix, if \[A \left( adj A \right) = \begin{bmatrix}10 & 0 \\ 0 & 10\end{bmatrix}\] , then |A| is equal to ______ .
If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is _____________ .
If A and B are invertible matrices, which of the following statement is not correct.
(a) 3
(b) 0
(c) − 3
(d) 1
If the equation a(y + z) = x, b(z + x) = y, c(x + y) = z have non-trivial solutions then the value of `1/(1+"a") + 1/(1+"b") + 1/(1+"c")` is ____________.
If A = [aij] is a square matrix of order 2 such that aij = `{(1"," "when i" ≠ "j"),(0"," "when" "i" = "j"):},` then A2 is ______.
If A is a square matrix of order 3, |A′| = −3, then |AA′| = ______.
If A = `[(2, -3, 5),(3, 2, -4),(1, 1, -2)]`, find A–1. Use A–1 to solve the following system of equations 2x − 3y + 5z = 11, 3x + 2y – 4z = –5, x + y – 2z = –3
Given that A is a square matrix of order 3 and |A| = –2, then |adj(2A)| is equal to ______.
A furniture factory uses three types of wood namely, teakwood, rosewood and satinwood for manufacturing three types of furniture, that are, table, chair and cot.
The wood requirements (in tonnes) for each type of furniture are given below:
| Table | Chair | Cot | |
| Teakwood | 2 | 3 | 4 |
| Rosewood | 1 | 1 | 2 |
| Satinwood | 3 | 2 | 1 |
It is found that 29 tonnes of teakwood, 13 tonnes of rosewood and 16 tonnes of satinwood are available to make all three types of furniture.
Using the above information, answer the following questions:
- Express the data given in the table above in the form of a set of simultaneous equations.
- Solve the set of simultaneous equations formed in subpart (i) by matrix method.
- Hence, find the number of table(s), chair(s) and cot(s) produced.
| 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.
