Advertisements
Advertisements
प्रश्न
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}1 & 6 \\ - 3 & 5\end{bmatrix}\]
Advertisements
उत्तर
\[A = \begin{bmatrix} 1 & 6\\ - 3 & 5 \end{bmatrix}\]
We know
\[A = IA\]
\[ \Rightarrow \begin{bmatrix} 1 & 6\\ - 3 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}A\]
\[ \Rightarrow \begin{bmatrix} 1 & 6\\ - 3 + 3 & 5 + 18 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 + 3 & 1 + 0 \end{bmatrix}A [\text{ Applying }R_2 \to R_2 + 3 R_1 ]\]
\[ \Rightarrow \begin{bmatrix} 1 & 6\\ 0 & 23 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 3 & 1 \end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix} 1 & 6 - 6\\ 0 & 23 \end{bmatrix} = \begin{bmatrix} 1 - \frac{18}{23} & 0 - \frac{6}{23}\\ 3 & 1 \end{bmatrix}A [\text{ Applying }R_1 \to R_1 - \frac{6}{23} R_2 ]\]
\[ \Rightarrow \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{5}{23} & \frac{- 6}{23}\\ \frac{3}{23} & \frac{1}{23} \end{bmatrix}A [\text{ Applying }R_2 \to \frac{1}{23} R_2 ]\]
\[ \Rightarrow A^{- 1} = \begin{bmatrix} \frac{5}{23} & \frac{- 6}{23}\\ \frac{3}{23} & \frac{1}{23} \end{bmatrix} = \frac{1}{23}\begin{bmatrix} 5 & - 6\\ 3 & 1 \end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{23}\begin{bmatrix} 5 & - 6\\ 3 & 1 \end{bmatrix}\]
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 inverse of the matrices (if it exists).
`[(1,0,0),(3,3,0),(5,2,-1)]`
Find the inverse of the matrices (if it exists).
`[(2,1,3),(4,-1,0),(-7,2,1)]`
Let A = `[(3,7),(2,5)]` and B = `[(6,8),(7,9)]`. Verify that (AB)−1 = B−1A−1.
Find the adjoint of the following matrix:
\[\begin{bmatrix}- 3 & 5 \\ 2 & 4\end{bmatrix}\]
Find the adjoint of the following matrix:
Verify that (adj A) A = |A| I = A (adj A) for the above matrix.
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.
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.
\[\begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{bmatrix}\]
Find the inverse of the following matrix.
Find the inverse of the following matrix.
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 \[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B = \begin{bmatrix}6 & 7 \\ 8 & 9\end{bmatrix} .\text{ Find }\left( AB \right)^{- 1}\]
If \[A = \begin{bmatrix}4 & 5 \\ 2 & 1\end{bmatrix}\] , then show that \[A - 3I = 2 \left( I + 3 A^{- 1} \right) .\]
If \[A = \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\] , verify that \[A^2 - 4 A + I = O,\text{ where }I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\text{ and }O = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\] . Hence, find A−1.
If \[A = \begin{bmatrix}4 & 3 \\ 2 & 5\end{bmatrix}\], find x and y such that
For the matrix \[A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3\end{bmatrix}\] . Show 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.
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 matrix X for which
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}3 & 10 \\ 2 & 7\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2\end{bmatrix}\]
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:
\[\begin{bmatrix}1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0\end{bmatrix}\]
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 \[A = \begin{bmatrix}1 & 2 & - 1 \\ - 1 & 1 & 2 \\ 2 & - 1 & 1\end{bmatrix}\] , then ded (adj (adj A)) is __________ .
If A satisfies the equation \[x^3 - 5 x^2 + 4x + \lambda = 0\] then A-1 exists if _____________ .
If \[A = \begin{bmatrix}2 & 3 \\ 5 & - 2\end{bmatrix}\] be such that \[A^{- 1} = kA\], then k equals ___________ .
If A is an invertible matrix, then det (A−1) is equal to ____________ .
If \[A = \begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix}\], find A−1 and hence solve the system of linear equations 2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y + 2z = −3
If A = `[(0, 1, 3),(1, 2, x),(2, 3, 1)]`, A–1 = `[(1/2, -4, 5/2),(-1/2, 3, -3/2),(1/2, y, 1/2)]` then x = 1, y = –1.
If A, B be two square matrices such that |AB| = O, then ____________.
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 ____________.
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 ______.
