Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
We have,
\[A = \begin{bmatrix}5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1\end{bmatrix}\]
\[ B^{- 1} = \begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}\]
\[\text{ We know }(AB )^{- 1} = B^{- 1} A^{- 1} \]
For matrix A,
\[ C_{11} = \begin{vmatrix}3 & 2 \\ 2 & 1\end{vmatrix} = - 1, C_{12} = - \begin{vmatrix}2 & 2 \\ 1 & 1\end{vmatrix} = 0\text{ and }C_{13} = \begin{vmatrix}2 & 3 \\ 1 & 2\end{vmatrix} = 1\]
\[ C_{21} = - \begin{vmatrix}0 & 4 \\ 2 & 1\end{vmatrix} = 8, C_{22} = \begin{vmatrix}5 & 4 \\ 1 & 1\end{vmatrix} = 1\text{ and }C_{23} = - \begin{vmatrix}5 & 0 \\ 1 & 2\end{vmatrix} = - 10\]
\[ C_{31} = \begin{vmatrix}0 & 4 \\ 3 & 2\end{vmatrix} = - 12, C_{32} = - \begin{vmatrix}5 & 4 \\ 2 & 2\end{vmatrix} = - 2\text{ and }C_{33} = \begin{vmatrix}5 & 0 \\ 2 & 3\end{vmatrix} = 15\]
Now,
\[adj (A) = \begin{bmatrix}- 1 & 0 & 1 \\ 8 & 1 & - 10 \\ - 12 & - 2 & 15\end{bmatrix}^T = \begin{bmatrix}- 1 & 8 & - 12 \\ 0 & 1 & - 2 \\ 1 & - 10 & 15\end{bmatrix}\]
\[\text{ and }\left| A \right| = - 1\]
\[ \therefore A^{- 1} = - \begin{bmatrix}- 1 & 8 & - 12 \\ 0 & 1 & - 2 \\ 1 & - 10 & 15\end{bmatrix} = \begin{bmatrix}1 & - 8 & 12 \\ 0 & - 1 & 2 \\ - 1 & 10 & - 15\end{bmatrix}\]
\[So, B^{- 1} A^{- 1} = \begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}\begin{bmatrix}1 & - 8 & 12 \\ 0 & - 1 & 2 \\ - 1 & 10 & - 15\end{bmatrix} = \begin{bmatrix}- 2 & 19 & - 27 \\ - 2 & 18 & - 25 \\ - 3 & 29 & - 42\end{bmatrix}\]
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?
Find the inverse of the matrices (if it exists).
`[(-1,5),(-3,2)]`
Find the inverse of the matrices (if it exists).
`[(1,0,0),(3,3,0),(5,2,-1)]`
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 = `[(2,-1,1),(-1,2,-1),(1,-1,2)]` verify that A3 − 6A2 + 9A − 4I = 0 and hence find A−1.
If A is an invertible matrix of order 2, then det (A−1) is equal to ______.
If x, y, z are nonzero real numbers, then the inverse of matrix A = `[(x,0,0),(0,y,0),(0,0,z)]` is ______.
Find the adjoint of the following matrix:
\[\begin{bmatrix}a & b \\ c & d\end{bmatrix}\]
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.
Given \[A = \begin{bmatrix}2 & - 3 \\ - 4 & 7\end{bmatrix}\], compute A−1 and show that \[2 A^{- 1} = 9I - A .\]
If \[A = \begin{bmatrix}4 & 3 \\ 2 & 5\end{bmatrix}\], find x and y such 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.
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:
\[\begin{bmatrix}5 & 2 \\ 2 & 1\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & 5 \\ 1 & 3\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}3 & 10 \\ 2 & 7\end{bmatrix}\]
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:
\[\begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\]
If A is a non-singular symmetric matrix, write whether A−1 is symmetric or skew-symmetric.
If A is an invertible matrix such that |A−1| = 2, find the value of |A|.
If \[A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}, B = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\] , find adj (AB).
If A is an invertible matrix, then which of the following is not true ?
If B is a non-singular matrix and A is a square matrix, then det (B−1 AB) is equal to ___________ .
If A and B are invertible matrices, which of the following statement is not correct.
(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 _______________ .
`("aA")^-1 = 1/"a" "A"^-1`, where a is any real number and A is a square matrix.
Find the adjoint of the matrix A, where A `= [(1,2,3),(0,5,0),(2,4,3)]`
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 ______.
A and B are invertible matrices of the same order such that |(AB)-1| = 8, If |A| = 2, then |B| is ____________.
| 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.
