Advertisements
Advertisements
प्रश्न
Find the inverse of the following matrix:
Advertisements
उत्तर
\[B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\]
\[\left| B \right| = 0 - 1 = - 1 \neq 0\]
B is a singular matrix; therefore, it is invertible .
\[\text{ Let }C_{ij}\text{ be a cofactor of }b_{ij}\text{ in B. }\]
Now,
\[ C_{11} = 0 \]
\[ C_{12} = - 1\]
\[ C_{21} = - 1\]
\[ C_{22} = 0\]
\[adjB = \begin{bmatrix}0 & - 1 \\ - 1 & 0\end{bmatrix}^T = \begin{bmatrix}0 & - 1 \\ - 1 & 0\end{bmatrix}\]
\[ \therefore B^{- 1} = \frac{1}{\left| B \right|}adjB = \frac{1}{- 1}\begin{bmatrix}0 & - 1 \\ - 1 & 0\end{bmatrix} = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\]
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?
Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. School A wants to award Rs x each, Rs y each and Rs z each for the three respective values to 3, 2 and 1 students, respectively with a total award money of Rs 1,600. School B wants to spend Rs 2,300 to award 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is Rs 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for an award.
Find the inverse of the matrices (if it exists).
`[(1,0,0),(0, cos alpha, sin alpha),(0, sin alpha, -cos alpha)]`
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.
Compute the adjoint of the following matrix:
\[\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}\]
Verify that (adj A) A = |A| I = A (adj A) for the above matrix.
If \[A = \begin{bmatrix}- 4 & - 3 & - 3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{bmatrix}\], show that adj A = A.
Find A (adj A) for the matrix \[A = \begin{bmatrix}1 & - 2 & 3 \\ 0 & 2 & - 1 \\ - 4 & 5 & 2\end{bmatrix} .\]
Find the inverse of the following matrix:
Find the inverse of the matrix \[A = \begin{bmatrix}a & b \\ c & \frac{1 + bc}{a}\end{bmatrix}\] and show that \[a A^{- 1} = \left( a^2 + bc + 1 \right) I - aA .\]
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 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.
prove that \[A^{- 1} = A^3\]
If \[A = \begin{bmatrix}- 1 & 2 & 0 \\ - 1 & 1 & 1 \\ 0 & 1 & 0\end{bmatrix}\] , show that \[A^2 = A^{- 1} .\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}7 & 1 \\ 4 & - 3\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}5 & 2 \\ 2 & 1\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix}\]
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:
\[\begin{bmatrix}1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & - 1 & 4 \\ 4 & 0 & 7 \\ 3 & - 2 & 7\end{bmatrix}\]
If A is symmetric matrix, write whether AT is symmetric or skew-symmetric.
If A is an invertible matrix such that |A−1| = 2, find the value of |A|.
Find the inverse of the matrix \[\begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\]
If A is an invertible matrix of order 3, then which of the following is not true ?
If \[A = \begin{bmatrix}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{bmatrix}\] , then the value of |adj A| is _____________ .
If A satisfies the equation \[x^3 - 5 x^2 + 4x + \lambda = 0\] then A-1 exists if _____________ .
If A = `[(0, 1, 3),(1, 2, x),(2, 3, 1)]`, A–1 = `[(1/2, -4, 5/2),(-1/2, 3, -3/2),(1/2, y, 1/2)]` then x = 1, y = –1.
(A3)–1 = (A–1)3, where A is a square matrix and |A| ≠ 0.
|A–1| ≠ |A|–1, where A is non-singular matrix.
|adj. A| = |A|2, where A is a square matrix of order two.
If A, B be two square matrices such that |AB| = O, then ____________.
A square matrix A is invertible if det A is equal to ____________.
Find the adjoint of the matrix A `= [(1,2),(3,4)].`
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 ____________.
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.
