Advertisements
Advertisements
प्रश्न
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0\end{bmatrix}\]
Advertisements
उत्तर
\[A = \begin{bmatrix}1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0\end{bmatrix}\]
We know
\[A = IA \]
\[ \Rightarrow \begin{bmatrix}1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix}1 & 3 & - 2 \\ 0 & 9 & - 7 \\ 0 & - 5 & 4\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ - 2 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 + 3 R_1\text{ and }R_3 \to R_3 - 2 R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 3 & - 2 \\ 0 & 1 & - \frac{7}{9} \\ 0 & - 5 & 4\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ - 2 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to \frac{1}{9} R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{1}{3} \\ 0 & 1 & - \frac{7}{9} \\ 0 & 0 & \frac{1}{9}\end{bmatrix} = \begin{bmatrix}0 & - \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ - \frac{1}{3} & \frac{5}{9} & 1\end{bmatrix} A \left[\text{ Applying }R_1 \to R_1 - 3 R_2\text{ and }R_3 \to R_3 + 5 R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{1}{3} \\ 0 & 1 & - \frac{7}{9} \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}0 & - \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ - 3 & 5 & 9\end{bmatrix} A \left[\text{ Applying }R_3 \to 9 R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 3 & 5 & 9\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 + \frac{7}{9} R_3\text{ and }R_1 \to R_1 - \frac{1}{3} R_3 \right]\]
\[ \Rightarrow A^{- 1} = \begin{bmatrix}1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 3 & 5 & 9\end{bmatrix} \]
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,-1,2),(0,2,-3),(3,-2,4)]`
Let A = `[(1,2,1),(2,3,1),(1,1,5)]` verify that
- [adj A]–1 = adj(A–1)
- (A–1)–1 = A
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.
\[\begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{bmatrix}\]
Find the inverse of the following matrix and verify that \[A^{- 1} A = I_3\]
For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]
\[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 6 \\ 3 & 2\end{bmatrix}\]
Let
\[F \left( \alpha \right) = \begin{bmatrix}\cos \alpha & - \sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\text{ and }G\left( \beta \right) = \begin{bmatrix}\cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ - \sin \beta & 0 & \cos \beta\end{bmatrix}\]
Show that
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 \[A^{- 1} = A^3\]
Solve the matrix equation \[\begin{bmatrix}5 & 4 \\ 1 & 1\end{bmatrix}X = \begin{bmatrix}1 & - 2 \\ 1 & 3\end{bmatrix}\], where X is a 2 × 2 matrix.
Find the matrix X for which
Find the matrix X satisfying the equation
Find the adjoint of the matrix \[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\] and hence show that \[A\left( adj A \right) = \left| A \right| I_3\].
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & - 1 & 4 \\ 4 & 0 & 7 \\ 3 & - 2 & 7\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}- 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix}\]
If A is an invertible matrix of order 3, then which of the following is not true ?
If \[S = \begin{bmatrix}a & b \\ c & d\end{bmatrix}\], then adj A is ____________ .
If A is a singular matrix, then adj A is ______.
If A, B are two n × n non-singular matrices, then __________ .
If \[A = \begin{bmatrix}1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2\end{bmatrix},\text{ then aI + bA + 2 }A^2\] equals ____________ .
If a matrix A is such that \[3A^3 + 2 A^2 + 5 A + I = 0,\text{ then }A^{- 1}\] equal to _______________ .
An amount of Rs 10,000 is put into three investments at the rate of 10, 12 and 15% per annum. The combined income is Rs 1310 and the combined income of first and second investment is Rs 190 short of the income from the third. Find the investment in each using matrix method.
For what value of x, matrix `[(6-"x", 4),(3-"x", 1)]` is a singular matrix?
The value of `abs (("cos" (alpha + beta),-"sin" (alpha + beta),"cos" 2 beta),("sin" alpha, "cos" alpha, "sin" beta),(-"cos" alpha, "sin" alpha, "cos" beta))` is independent of ____________.
For A = `[(3,1),(-1,2)]`, then 14A−1 is given by:
Given that A is a square matrix of order 3 and |A| = –2, then |adj(2A)| is equal to ______.
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.
