Advertisements
Advertisements
प्रश्न
If A = `[(2,-1,1),(-1,2,-1),(1,-1,2)]` verify that A3 − 6A2 + 9A − 4I = 0 and hence find A−1.
Advertisements
उत्तर
We have A = `[(2,-1,1),(-1,2,-1),(1,-1,2)]`
A2 = AA
= `[(2,-1,1),(-1,2,-1),(1,-1,2)] [(2,-1,1),(-1,2,-1),(1,-1,2)]`
= `[(4+1+1,-2-2-1,2+1+2),(-2-2-1,1+4+1,-1-2-2),(2+1+2,-1-2-2,1+1+4)]`
= `[(6,-5,5),(-5,6,-5),(5,-5,6)]`
A3 = A2A
= `[(6,-5,5),(-5,6,-5),(5,-5,6)] [(2,-1,1),(-1,2,-1),(1,-1,2)]`
= `[(12+5+5,-6-10-5,6+5+10),(-10-6-5,5+12+5,-5-6-10),(10+5+6,-5-10-6,5+5+12)]`
= `[(22,-21,21),(-21,22,-21),(21,-21,22)]`
Now, A3 − 6A2 + 9A − 4I
= `[(22,-21,21),(-21,22,-21),(21,-21,22)] - 6 [(6,-5,5),(-5,6,-5),(5,-5,6)] + 9 [(2,-1,1),(-1,2,-1),(1,-1,2)] - 4 [(1,0,0),(0,1,0),(0,0,1)]`
= `[(22,-21,21),(-21,22,-21),(21,-21,22)] - [(36,-30,30),(-30,36,-30),(30,-30,36)] + [(18,-9,9),(-9,18,-9),(9,-9,18)] - [(4,0,0),(0,4,0),(0,0,4)]`
= `[(22 - 36 + 18 - 4, -21 + 30 - 9 - 0, -21 - 30 + 9 - 0),(-21 + 30 - 9 - 0, 22 - 36 + 18 - 4, -21 - 30 + 9 - 0),(21 - 30 + 9 - 0, -21 + 30 - 9 - 0,22 - 36 + 18 - 4)]`
= `[(0,0,0),(0,0,0),(0,0,0)]`
= 0
Hence A3 − 6A2 + 9A − 4I = 0
Now, A3 − 6A2 + 9A − 4I = 0
⇒ 4I = A3 − 6A2 + 9A
Multiplying both sides by A−1, we get
⇒ `A^(−1) = 1/4 A^2 - 6/4 A + 9/4I`
= `1/4[(6,-5,5),(-5,6,-5),(5,-5,6)] - 6/4 [(2,-1,1),(-1,2,-1),(1,-1,2)] + 9/4 [(1,0,0),(0,1,0),(0,0,1)]`
= `1/4 [(6-12+9,-5+6+0,5-6+0),(-5+6+0,6-12+9,-5+6+0),(5-6+0,-5+6+0,6-12+9)]`
= `1/4 [(3,1,-1),(1,3,1),(-1,1,3)]`
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 adjoint of the matrices.
`[(1,-1,2),(2,3,5),(-2,0,1)]`
Verify A(adj A) = (adj A)A = |A|I.
`[(2,3),(-4,-6)]`
Find the inverse of the matrices (if it exists).
`[(1,-1,2),(0,2,-3),(3,-2,4)]`
Let A = `[(3,7),(2,5)]` and B = `[(6,8),(7,9)]`. Verify that (AB)−1 = B−1A−1.
For the matrix A = `[(3,2),(1,1)]` find the numbers a and b such that A2 + aA + bI = 0.
For the matrix A = `[(1,1,1),(1,2,-3),(2,-1,3)]` show that A3 − 6A2 + 5A + 11 I = 0. Hence, find A−1.
If x, y, z are nonzero real numbers, then the inverse of matrix A = `[(x,0,0),(0,y,0),(0,0,z)]` is ______.
Compute the adjoint of the following matrix:
Verify that (adj A) A = |A| I = A (adj A) for the above matrix.
If \[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\] , show that adj A = 3AT.
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.
Find the inverse of the following matrix.
For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]
\[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 6 \\ 3 & 2\end{bmatrix}\]
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.
For the matrix \[A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3\end{bmatrix}\] . Show that
Verify that \[A^3 - 6 A^2 + 9A - 4I = O\] and hence find A−1.
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 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:
\[\begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\]
If A is a non-singular symmetric matrix, write whether A−1 is symmetric or skew-symmetric.
If \[A = \begin{bmatrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\text{ and }A \left( adj A = \right)\begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix}\], then find the value of k.
If A is an invertible matrix such that |A−1| = 2, find the value of |A|.
If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is _____________ .
(a) 3
(b) 0
(c) − 3
(d) 1
If \[\begin{bmatrix}1 & - \tan \theta \\ \tan \theta & 1\end{bmatrix} \begin{bmatrix}1 & \tan \theta \\ - \tan \theta & 1\end{bmatrix} - 1 = \begin{bmatrix}a & - b \\ b & a\end{bmatrix}\], then _______________ .
If A is an invertible matrix, then det (A−1) is equal to ____________ .
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.
If A = `[(x, 5, 2),(2, y, 3),(1, 1, z)]`, xyz = 80, 3x + 2y + 10z = 20, ten A adj. A = `[(81, 0, 0),(0, 81, 0),(0, 0, 81)]`
(A3)–1 = (A–1)3, where A is a square matrix and |A| ≠ 0.
Find the adjoint of the matrix A, where A `= [(1,2,3),(0,5,0),(2,4,3)]`
For matrix A = `[(2,5),(-11,7)]` (adj A)' is equal to:
