Advertisements
Advertisements
प्रश्न
(A3)–1 = (A–1)3, where A is a square matrix and |A| ≠ 0.
विकल्प
True
False
Advertisements
उत्तर
This statement is True.
Explanation:
Since (AK)–1 = (A–1)K where K ∈ N
So, (A3)–1 = (A–1)3
APPEARS IN
संबंधित प्रश्न
The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?
Find the adjoint of the matrices.
`[(1,2),(3,4)]`
Find the inverse of the matrices (if it exists).
`[(2,1,3),(4,-1,0),(-7,2,1)]`
Find the inverse of the matrices (if it exists).
`[(1,0,0),(0, cos alpha, sin alpha),(0, sin alpha, -cos alpha)]`
For the matrix A = `[(3,2),(1,1)]` find the numbers a and b such that A2 + aA + bI = 0.
Let A = `[(1,2,1),(2,3,1),(1,1,5)]` verify that
- [adj A]–1 = adj(A–1)
- (A–1)–1 = A
Compute the adjoint of the following matrix:
Verify that (adj A) A = |A| I = A (adj A) for the above matrix.
Find the inverse of the following matrix:
Find the inverse of the following matrix.
Let \[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B = \begin{bmatrix}6 & 7 \\ 8 & 9\end{bmatrix} .\text{ Find }\left( AB \right)^{- 1}\]
If \[A = \begin{bmatrix}4 & 5 \\ 2 & 1\end{bmatrix}\] , then show that \[A - 3I = 2 \left( I + 3 A^{- 1} \right) .\]
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.
Let
\[F \left( \alpha \right) = \begin{bmatrix}\cos \alpha & - \sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\text{ and }G\left( \beta \right) = \begin{bmatrix}\cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ - \sin \beta & 0 & \cos \beta\end{bmatrix}\]
Show that
For the matrix \[A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3\end{bmatrix}\] . Show that
Verify that \[A^3 - 6 A^2 + 9A - 4I = O\] and hence find A−1.
If \[A = \begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\] , show that \[A^{- 1} = A^3\]
Find the matrix X for which
Find the matrix X satisfying the equation
If \[A = \begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}\] , find \[A^{- 1}\] and prove that \[A^2 - 4A - 5I = O\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & 5 \\ 1 & 3\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1\end{bmatrix}\]
Find the inverse of the matrix \[\begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\]
If A5 = O such that \[A^n \neq I\text{ for }1 \leq n \leq 4,\text{ then }\left( I - A \right)^{- 1}\] equals ________ .
If A satisfies the equation \[x^3 - 5 x^2 + 4x + \lambda = 0\] then A-1 exists if _____________ .
(a) 3
(b) 0
(c) − 3
(d) 1
If \[\begin{bmatrix}1 & - \tan \theta \\ \tan \theta & 1\end{bmatrix} \begin{bmatrix}1 & \tan \theta \\ - \tan \theta & 1\end{bmatrix} - 1 = \begin{bmatrix}a & - b \\ b & a\end{bmatrix}\], then _______________ .
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
Using matrix method, solve the following system of equations:
x – 2y = 10, 2x + y + 3z = 8 and -2y + z = 7
`("aA")^-1 = 1/"a" "A"^-1`, where a is any real number and A is a square matrix.
The value of `abs (("cos" (alpha + beta),-"sin" (alpha + beta),"cos" 2 beta),("sin" alpha, "cos" alpha, "sin" beta),(-"cos" alpha, "sin" alpha, "cos" beta))` is independent of ____________.
For matrix A = `[(2,5),(-11,7)]` (adj A)' is equal to:
For A = `[(3,1),(-1,2)]`, then 14A−1 is given by:
Read the following passage:
|
Gautam buys 5 pens, 3 bags and 1 instrument box and pays a sum of ₹160. From the same shop, Vikram buys 2 pens, 1 bag and 3 instrument boxes and pays a sum of ₹190. Also, Ankur buys 1 pen, 2 bags and 4 instrument boxes and pays a sum of ₹250. |
Based on the above information, answer the following questions:
- Convert the given above situation into a matrix equation of the form AX = B. (1)
- Find | A |. (1)
- Find A–1. (2)
OR
Determine P = A2 – 5A. (2)
| 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.
