Advertisements
Advertisements
प्रश्न
Find the inverse of the matrix \[A = \begin{bmatrix}a & b \\ c & \frac{1 + bc}{a}\end{bmatrix}\] and show that \[a A^{- 1} = \left( a^2 + bc + 1 \right) I - aA .\]
Advertisements
उत्तर
We have,
\[A = \begin{bmatrix}a & b \\ c & \frac{1 + bc}{a}\end{bmatrix}\]
\[\text{ So, }adj(A) = \begin{bmatrix}\frac{1 + bc}{a} & - b \\ - c & a\end{bmatrix}\]
\[\text{ and }\left| A \right| = 1\]
\[ \therefore A^{- 1} = \begin{bmatrix}\frac{1 + bc}{a} & - b \\ - c & a\end{bmatrix}\]
\[\text{ Now, }a A^{- 1} = \left( a^2 + bc + 1 \right)I - aA\]
\[\text{ LHS }= a A^{- 1} = a\begin{bmatrix}\frac{1 + bc}{a} & - b \\ - c & a\end{bmatrix} = \begin{bmatrix}1 + bc & - ba \\ - ca & a^2\end{bmatrix}\]
\[\text{ RHS }= \left( a^2 + bc + 1 \right)I - aA\]
\[ = a^2 + bc + 1\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} - \begin{bmatrix}a^2 & ba \\ ca & 1 + bc\end{bmatrix}\]
\[ = \begin{bmatrix}a^2 + bc + 1 & 0 \\ 0 & a^2 + bc + 1\end{bmatrix} - \begin{bmatrix}a^2 & ba \\ ca & 1 + bc\end{bmatrix}\]
\[ = \begin{bmatrix}1 + bc & - ba \\ - ca & a^2\end{bmatrix} =\text{ LHS }\]
Hence proved .
APPEARS IN
संबंधित प्रश्न
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 adjoint of the matrices.
`[(1,-1,2),(2,3,5),(-2,0,1)]`
Find the inverse of the matrices (if it exists).
`[(1,-1,2),(0,2,-3),(3,-2,4)]`
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.
Let A = `[(1,2,1),(2,3,1),(1,1,5)]` verify that
- [adj A]–1 = adj(A–1)
- (A–1)–1 = A
If \[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\] , show that adj A = 3AT.
Find the inverse of the following matrix.
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.
If \[A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\], show that
If \[A = \begin{bmatrix}4 & 3 \\ 2 & 5\end{bmatrix}\], find x and y such that
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:
\[\begin{bmatrix}1 & 6 \\ - 3 & 5\end{bmatrix}\]
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:
\[\begin{bmatrix}1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0\end{bmatrix}\]
If A is symmetric matrix, write whether AT is symmetric or skew-symmetric.
If A is a square matrix, then write the matrix adj (AT) − (adj A)T.
If \[A = \begin{bmatrix}2 & 3 \\ 5 & - 2\end{bmatrix}\] be such that \[A^{- 1} = k A,\] 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 \[A = \begin{bmatrix}3 & 1 \\ 2 & - 3\end{bmatrix}\], then find |adj A|.
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 for the matrix A, A3 = I, then A−1 = _____________ .
For non-singular square matrix A, B and C of the same order \[\left( A B^{- 1} C \right) =\] ______________ .
Let \[A = \begin{bmatrix}1 & 2 \\ 3 & - 5\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}\] and X be a matrix such that A = BX, then X is equal to _____________ .
If A is an invertible matrix, then det (A−1) is equal to ____________ .
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
If A = `[(x, 5, 2),(2, y, 3),(1, 1, z)]`, xyz = 80, 3x + 2y + 10z = 20, ten A adj. A = `[(81, 0, 0),(0, 81, 0),(0, 0, 81)]`
If A and B are invertible matrices, then which of the following is not correct?
A square matrix A is invertible if det A is equal to ____________.
For matrix A = `[(2,5),(-11,7)]` (adj A)' is equal to:
If `abs((2"x", -1),(4,2)) = abs ((3,0),(2,1))` then x is ____________.
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.
