Advertisements
Advertisements
Question
If A is an invertible matrix such that |A−1| = 2, find the value of |A|.
Solution
We know,
\[ \left| A \right|^{- 1} = \frac{1}{\left| A \right|}\]
\[ \Rightarrow 2 = \frac{1}{\left| A \right|} \left[ \because \left| A \right|^{- 1} = 2 \right]\]
\[ \Rightarrow \left| A \right| = \frac{1}{2}\]
APPEARS IN
RELATED QUESTIONS
Find the adjoint of the matrices.
`[(1,-1,2),(2,3,5),(-2,0,1)]`
Find the inverse of the matrices (if it exists).
`[(2,-2),(4,3)]`
Find the inverse of the matrices (if it exists).
`[(2,1,3),(4,-1,0),(-7,2,1)]`
Find the inverse of the matrices (if it exists).
`[(1,0,0),(0, cos alpha, sin alpha),(0, sin alpha, -cos alpha)]`
If A is an invertible matrix of order 2, then det (A−1) is equal to ______.
Let A = `[(1,-2,1),(-2,3,1),(1,1,5)]` verify that
- [adj A]–1 = adj (A–1)
- (A–1)–1 = A
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:
\[\begin{bmatrix}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix}\]
Find the inverse of the following matrix and verify that \[A^{- 1} A = I_3\]
For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]
\[A = \begin{bmatrix}2 & 1 \\ 5 & 3\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 5 \\ 3 & 4\end{bmatrix}\]
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.
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
prove that \[A^{- 1} = A^3\]
If \[A = \begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\] , show that \[A^{- 1} = A^3\]
If \[A = \begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}\] , find \[A^{- 1}\] and prove that \[A^2 - 4A - 5I = O\]
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:
\[\begin{bmatrix}3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0\end{bmatrix}\]
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, then which of the following is not true ?
If A is a singular matrix, 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 \[A^2 - A + I = 0\], then the inverse of A is __________ .
(a) 3
(b) 0
(c) − 3
(d) 1
If \[A = \begin{bmatrix}1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2\end{bmatrix},\text{ then aI + bA + 2 }A^2\] equals ____________ .
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
Using matrix method, solve the following system of equations:
x – 2y = 10, 2x + y + 3z = 8 and -2y + z = 7
`("aA")^-1 = 1/"a" "A"^-1`, where a is any real number and A is a square matrix.
|A–1| ≠ |A|–1, where A is non-singular matrix.
For what value of x, matrix `[(6-"x", 4),(3-"x", 1)]` is a singular matrix?
If A = [aij] is a square matrix of order 2 such that aij = `{(1"," "when i" ≠ "j"),(0"," "when" "i" = "j"):},` then A2 is ______.