Advertisements
Advertisements
Question
Find the inverse of the following matrix.
\[\begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{bmatrix}\]
Advertisements
Solution
\[A = \begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}3 & 1 \\ 1 & 2\end{vmatrix} = 5, C_{12} = - \begin{vmatrix}2 & 1 \\ 3 & 2\end{vmatrix} = - 1\text{ and }C_{13} = \begin{vmatrix}2 & 3 \\ 3 & 1\end{vmatrix} = - 7\]
\[ C_{21} = - \begin{vmatrix}2 & 3 \\ 1 & 2\end{vmatrix} = - 1, C_{22} = \begin{vmatrix}1 & 3 \\ 3 & 2\end{vmatrix} = - 7\text{ and }C_{23} = - \begin{vmatrix}1 & 2 \\ 3 & 1\end{vmatrix} = 5\]
\[ C_{31} = \begin{vmatrix}2 & 3 \\ 3 & 1\end{vmatrix} = - 7, C_{32} = - \begin{vmatrix}1 & 3 \\ 2 & 1\end{vmatrix} = 5\text{ and }C_{33} = \begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix} = - 1\]
\[adjA = \begin{bmatrix}5 & - 1 & - 7 \\ - 1 & - 7 & 5 \\ - 7 & 5 & - 1\end{bmatrix}^T = \begin{bmatrix}5 & - 1 & - 7 \\ - 1 & - 7 & 5 \\ - 7 & 5 & - 1\end{bmatrix}\]
\[\text{ and }\left| A \right| = - 18\]
\[ \therefore A^{- 1} = - \frac{1}{18}\begin{bmatrix}5 & - 1 & - 7 \\ - 1 & - 7 & 5 \\ - 7 & 5 & - 1\end{bmatrix}\]
APPEARS IN
RELATED QUESTIONS
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.
Verify A(adj A) = (adj A)A = |A|I.
`[(2,3),(-4,-6)]`
Let A = `[(3,7),(2,5)]` and B = `[(6,8),(7,9)]`. Verify that (AB)−1 = B−1A−1.
If A = `[(3,1),(-1,2)]` show that A2 – 5A + 7I = 0. Hence, find A–1.
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.
For the matrix
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 and verify that \[A^{- 1} A = I_3\]
If \[A = \begin{bmatrix}4 & 5 \\ 2 & 1\end{bmatrix}\] , then show that \[A - 3I = 2 \left( I + 3 A^{- 1} \right) .\]
Show that \[A = \begin{bmatrix}5 & 3 \\ - 1 & - 2\end{bmatrix}\] satisfies the equation \[x^2 - 3x - 7 = 0\]. Thus, 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.
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 satisfying the matrix equation \[X\begin{bmatrix}5 & 3 \\ - 1 & - 2\end{bmatrix} = \begin{bmatrix}14 & 7 \\ 7 & 7\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1\end{bmatrix}\]
If A is a square matrix, then write the matrix adj (AT) − (adj A)T.
If A is a non-singular symmetric matrix, write whether A−1 is symmetric or skew-symmetric.
If \[A = \begin{bmatrix}2 & 3 \\ 5 & - 2\end{bmatrix}\] be such that \[A^{- 1} = k A,\] then find the value of k.
If \[A = \begin{bmatrix}1 & - 3 \\ 2 & 0\end{bmatrix}\], write adj A.
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 \[A^2 - A + I = 0\], then the inverse of A is __________ .
If \[A = \begin{bmatrix}2 & 3 \\ 5 & - 2\end{bmatrix}\] be such that \[A^{- 1} = kA\], then k equals ___________ .
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
|adj. A| = |A|2, where A is a square matrix of order two.
Find the adjoint of the matrix A, where A `= [(1,2,3),(0,5,0),(2,4,3)]`
If A is a square matrix of order 3 and |A| = 5, then |adj A| = ______.
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
If A = `[(1/sqrt(5), 2/sqrt(5)),((-2)/sqrt(5), 1/sqrt(5))]`, B = `[(1, 0),(i, 1)]`, i = `sqrt(-1)` and Q = ATBA, then the inverse of the matrix A. Q2021 AT is equal to ______.
If A = `[(0, 1),(0, 0)]`, then A2023 is equal to ______.
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.
