Advertisements
Advertisements
Question
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3\end{bmatrix}\]
Solution
\[A = \begin{bmatrix}1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3\end{bmatrix}\]
We know
\[A = IA \]
\[ \Rightarrow \begin{bmatrix}1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix}1 & 2 & 0 \\ 0 & - 1 & - 1 \\ 0 & - 3 & 3\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ - 2 & 1 & 0 \\ - 1 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 - 2 R_1\text{ and }R_3 \to R_3 - R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & - 3 & 3\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 2 & - 1 & 0 \\ - 1 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to - R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - 2 \\ 0 & 1 & 1 \\ 0 & 0 & 6\end{bmatrix} = \begin{bmatrix}- 3 & 2 & 0 \\ 2 & - 1 & 0 \\ 5 & - 3 & 1\end{bmatrix} A \left[\text{ Applying }R_1 \to R_1 - 2 R_2\text{ and }R_3 \to R_3 + 3 R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}- 3 & 2 & 0 \\ 2 & - 1 & 0 \\ \frac{5}{6} & - \frac{1}{2} & \frac{1}{6}\end{bmatrix} A \left[\text{ Applying }R_3 \to \frac{1}{6} R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}- \frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & - \frac{1}{2} & \frac{- 1}{6} \\ \frac{5}{6} & - \frac{1}{2} & \frac{1}{6}\end{bmatrix}A \left[\text{ Applying }R_1 \to R_1 + 2 R_3\text{ and }R_2 \to R_2 - R_3 \right]\]
\[ \therefore A^{- 1} = \begin{bmatrix}- \frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & - \frac{1}{2} & \frac{- 1}{6} \\ \frac{5}{6} & - \frac{1}{2} & \frac{1}{6}\end{bmatrix}\]
APPEARS IN
RELATED QUESTIONS
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).
`[(1,2,3),(0,2,4),(0,0,5)]`
If A = `[(2,-1,1),(-1,2,-1),(1,-1,2)]` verify that A3 − 6A2 + 9A − 4I = O and hence find A−1
If x, y, z are nonzero real numbers, then the inverse of matrix A = `[(x,0,0),(0,y,0),(0,0,z)]` is ______.
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.
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:
Find the inverse of the following matrix.
Given \[A = \begin{bmatrix}2 & - 3 \\ - 4 & 7\end{bmatrix}\], compute A−1 and show that \[2 A^{- 1} = 9I - A .\]
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.
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 \[A = \begin{bmatrix}6 & 5 \\ 7 & 6\end{bmatrix}\] satisfies the equation \[x^2 - 12x + 1 = O\]. Thus, find A−1.
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 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:
\[\begin{bmatrix}- 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix}\]
If \[A = \begin{bmatrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\text{ and }A \left( adj A = \right)\begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix}\], then find the value of k.
If A is an invertible matrix such that |A−1| = 2, find the value of |A|.
Find the inverse of the matrix \[\begin{bmatrix}3 & - 2 \\ - 7 & 5\end{bmatrix} .\]
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 of order 3, then which of the following is not true ?
If \[A = \begin{bmatrix}3 & 4 \\ 2 & 4\end{bmatrix}, B = \begin{bmatrix}- 2 & - 2 \\ 0 & - 1\end{bmatrix},\text{ then }\left( A + B \right)^{- 1} =\]
For any 2 × 2 matrix, if \[A \left( adj A \right) = \begin{bmatrix}10 & 0 \\ 0 & 10\end{bmatrix}\] , then |A| is equal to ______ .
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 _____________ .
If for the matrix A, A3 = I, then A−1 = _____________ .
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 _____________ .
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
(A3)–1 = (A–1)3, where A is a square matrix and |A| ≠ 0.
Find the adjoint of the matrix A, where A `= [(1,2,3),(0,5,0),(2,4,3)]`
If `abs((2"x", -1),(4,2)) = abs ((3,0),(2,1))` then x is ____________.
If A is a square matrix of order 3 and |A| = 5, then |adj A| = ______.
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 ______.
If for a square matrix A, A2 – A + I = 0, then A–1 equals ______.
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.
