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
Find the adjoint of the matrices.
`[(1,2),(3,4)]`
Find the adjoint of the matrices.
`[(1,-1,2),(2,3,5),(-2,0,1)]`
Find the inverse of the matrices (if it exists).
`[(1,0,0),(3,3,0),(5,2,-1)]`
For the matrix A = `[(3,2),(1,1)]` find the numbers a and b such that A2 + aA + bI = 0.
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
Let A = `[(1, sin theta, 1),(-sin theta,1,sin theta),(-1, -sin theta, 1)]` where 0 ≤ θ ≤ 2π, then ______.
Compute the adjoint of the following matrix:
Verify that (adj A) A = |A| I = A (adj A) for the above matrix.
If \[A = \begin{bmatrix}- 4 & - 3 & - 3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{bmatrix}\], show that adj A = A.
Find the inverse of the following matrix.
Find the inverse of the following matrix.
For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]
\[A = \begin{bmatrix}2 & 1 \\ 5 & 3\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 5 \\ 3 & 4\end{bmatrix}\]
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
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.
If \[A = \begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\] , show that \[A^{- 1} = A^3\]
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:
\[\begin{bmatrix}1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1\end{bmatrix}\]
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 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 \[A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}, B = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\] , find adj (AB).
If A is an invertible matrix, then which of the following is not true ?
If \[A = \begin{bmatrix}3 & 4 \\ 2 & 4\end{bmatrix}, B = \begin{bmatrix}- 2 & - 2 \\ 0 & - 1\end{bmatrix},\text{ then }\left( A + B \right)^{- 1} =\]
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 _____________ .
|adj. A| = |A|2, where A is a square matrix of order two.
A square matrix A is invertible if det A is equal to ____________.
Find the adjoint of the matrix A, where A `= [(1,2,3),(0,5,0),(2,4,3)]`
Find x, if `[(1,2,"x"),(1,1,1),(2,1,-1)]` is singular
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 `abs((2"x", -1),(4,2)) = abs ((3,0),(2,1))` then x is ____________.
A and B are invertible matrices of the same order such that |(AB)-1| = 8, If |A| = 2, then |B| is ____________.
