Advertisements
Advertisements
प्रश्न
If x, y, z are nonzero real numbers, then the inverse of matrix A = `[(x,0,0),(0,y,0),(0,0,z)]` is ______.
पर्याय
`[(x^(-1),0,0),(0, y^(-1),0),(0,0,z^(-1))]`
`xyz[(x^(-1),0,0),(0,y^(-1),0),(0,0,z^(-1))]`
`1/(xyz)[(x,0,0),(0,y,0),(0,0,z)]`
`1/(xyz)[(1,0,0),(0,1,0),(0,0,1)]`
Advertisements
उत्तर
If x, y, z are nonzero real numbers, then the inverse of matrix A = `[(x,0,0),(0,y,0),(0,0,z)]` is `underlinebb([(x^(-1),0,0),(0, y^(-1),0),(0,0,z^(-1))])`.
Explanation:
Let, A = `[(x,0,0),(0,y,0),(0,0,z)]`
∴ |A| = x[yz − 0] = xyz
∴ A11 = `(-1)^(1 + 1)[(y,0),(0,z)]`
= (−1)2 [yz − 0]
= 1 × yz
= yz
A12 = `(-1)^(1 + 2)[(0,0),(0,z)]`
= (−1)3 [0 − 0]
= 0
A13 = `(-1)^(1 + 3)[(0,y),(0,0)]`
= (−1)4 [0 − 0]
= 0
A21 = `(-1)^(2 + 1)[(0,0),(0,z)]`
= (−1)3 [0 − 0]
= 0
A22 = `(-1)^(2 + 2)[(x,0),(0,z)]`
= (−1)4 [xz − 0]
= 1 × zx
= zx
A23 = `(-1)^(2 + 3)[(x,0),(0,0)]`
= (−1)5 [0 − 0]
= 0
A31 = `(-1)^(3 + 1)[(0,0),(0,z)]`
= (−1)4 [0 − 0]
= 0
A32 = `(-1)^(3 + 2)[(x,0),(0,0)]`
= (−1)5 [0 − 0]
= 0
A33 = `(-1)^(3 + 3)[(x,0),(0,y)]`
= (−1)6 [xy − 0]
= xy
∴ adj A = `[(yz,0,0),(0,zx,0),(0,0,xy)] = [(yz,0,0),(0,zx,0),(0,0,xy)]`
A−1 = `1/|A|` (adj A)
= `1/(xyz)[(yz,0,0),(0,zx,0),(0,0,xy)]`
= `[(1/x,0,0),(0,1/y,0),(0,0,1/z)]`
= `[(x^-1,0,0),(0,y^-1,0),(0,0,z^-1)]`
APPEARS IN
संबंधित प्रश्न
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)]`
If A is an invertible matrix of order 2, then det (A−1) is equal to ______.
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.
Find the adjoint of the following matrix:
\[\begin{bmatrix}- 3 & 5 \\ 2 & 4\end{bmatrix}\]
Find the adjoint of the following matrix:
\[\begin{bmatrix}a & b \\ c & d\end{bmatrix}\]
Compute 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:
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 and verify that \[A^{- 1} A = I_3\]
Let \[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B = \begin{bmatrix}6 & 7 \\ 8 & 9\end{bmatrix} .\text{ Find }\left( AB \right)^{- 1}\]
Given \[A = \begin{bmatrix}2 & - 3 \\ - 4 & 7\end{bmatrix}\], compute A−1 and show that \[2 A^{- 1} = 9I - A .\]
If \[A = \begin{bmatrix}4 & 5 \\ 2 & 1\end{bmatrix}\] , then show that \[A - 3I = 2 \left( I + 3 A^{- 1} \right) .\]
If \[A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\], show that
If \[A = \begin{bmatrix}4 & 3 \\ 2 & 5\end{bmatrix}\], find x and y such that
Show that \[A = \begin{bmatrix}5 & 3 \\ - 1 & - 2\end{bmatrix}\] satisfies the equation \[x^2 - 3x - 7 = 0\]. Thus, find A−1.
If \[A = \begin{bmatrix}- 1 & 2 & 0 \\ - 1 & 1 & 1 \\ 0 & 1 & 0\end{bmatrix}\] , show that \[A^2 = A^{- 1} .\]
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:
\[\begin{bmatrix}2 & 5 \\ 1 & 3\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix}\]
Find the inverse of the matrix \[\begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\]
If \[A = \begin{bmatrix}3 & 1 \\ 2 & - 3\end{bmatrix}\], then find |adj A|.
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 \[A = \begin{bmatrix}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{bmatrix}\] , then the value of |adj A| is _____________ .
The matrix \[\begin{bmatrix}5 & 10 & 3 \\ - 2 & - 4 & 6 \\ - 1 & - 2 & b\end{bmatrix}\] is a singular matrix, if the value of b is _____________ .
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
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.
|adj. A| = |A|2, where A is a square matrix of order two.
Find the adjoint of the matrix A, where A `= [(1,2,3),(0,5,0),(2,4,3)]`
Find the value of x for which the matrix A `= [(3 - "x", 2, 2),(2,4 - "x", 1),(-2,- 4,-1 - "x")]` 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 ____________.
If for a square matrix A, A2 – A + I = 0, then A–1 equals ______.
Read the following passage:
|
Gautam buys 5 pens, 3 bags and 1 instrument box and pays a sum of ₹160. From the same shop, Vikram buys 2 pens, 1 bag and 3 instrument boxes and pays a sum of ₹190. Also, Ankur buys 1 pen, 2 bags and 4 instrument boxes and pays a sum of ₹250. |
Based on the above information, answer the following questions:
- Convert the given above situation into a matrix equation of the form AX = B. (1)
- Find | A |. (1)
- Find A–1. (2)
OR
Determine P = A2 – 5A. (2)
