Advertisements
Advertisements
प्रश्न
Verify A(adj A) = (adj A)A = |A|I.
`[(1,-1,2),(3,0,-2),(1,0,3)]`
Advertisements
उत्तर
Let A = `[(1,-1,2),(3,0,-2),(1,0,3)]`
|A| = 1[0 − 0] + 1[9 + 2] + 2[0 − 0]
= 1 × 11
= 11
A11 = `(-1)^(1 + 1) |(0,-2),(0,3)|`
= (−1)2 [0 − 0]
= 0
A12 = `(-1)^(1 + 2) |(3,-2),(1,3)|`
= (−1)3 [9 + 2]
= −11
A13 = `(-1)^(1 + 3) |(3,0),(1,0)|`
= (−1)4 [0 − 0]
= 0
A21 = `(-1)^(2 + 1) |(-1,2),(0,3)|`
= (−1)3 [−3 − 0]
= −1 × (−3)
= 3
A22 = `(- 1)^(2 + 2) |(1, 2),(1,3)|`
= (−1)4 [3 − 2]
= 1 × 1
= 1
A23 = `(-1)^(2+ 3) |(1,-1),(1,0)|`
= (−1)5 [0 + 1]
= −1
A31 = `(1)^(3 + 1) |(-1,2),(0,-2)|`
= (−1)4 [2 − 0]
= 1 × 2
= 2
A32 = `(-1)^(3 + 2) |(1,2),(3,-2)|`
= (−1)5 [−2 − 6]
= −1 × (−8)
= 8
A33 = `(-1)^(3 + 3) |(1,-1),(3,0)|`
= (−1)6 [0 + 3]
= 1 × 3
= 3
adj A = `[(0,-11,0),(3,1,-1),(2,8,3)] = [(0,3,2),(-11,1,8),(0,-1,3)]`
L.H.S. = A(adj A) = `[(1,-1,2),(3,0,-2),(1,0,3)] [(0,3,2),(-11,1,8),(0,-1,3)]`
= `[(1 xx 0 + (- 1) xx (- 11) + 2 xx 0, 1 xx 3 + (- 1) xx 1 + 2 xx (- 1), 1 xx 2 + (- 1) xx 8 + 2 xx 3),(3 xx 0 + 0 xx (- 11) + (- 2) xx 0, 3 xx 3 + 0 xx 1 + (- 2) xx (- 1), 3 xx 2 + 0 xx 8 + (- 2) xx 3),(1 xx 0 + 0 xx (- 11) + 3 xx 0, 1 xx 3 + 0 xx 1 + 3 xx (-1), 1 xx 2 + 0 xx 8 + 3 + 3)]`
= `[(0+11+0,3 - 1 - 2, 2 - 8 + 6),(0+0+0, 9 + 0 + 2, 6 + 0 - 6),(0 + 0 + 0, 3 + 0 - 3, 2 + 0 + 9)]`
= `[(11,0,0),(0,11,0),(0,0,11)]`
= `11[(1,0,0),(0,1,0),(0,0,1)]`
= 11 · I
= |A| · I
R.H.S. = (adj A)A `= [(0,3,2),(-11,1,8),(0,-1,3)][(1,-1,2),(3,0,-2),(1,0,3)]`
= `[(0xx3 + 3 xx 3 + 2 xx 1,0xx(-1) + 3 xx 0 + 2 xx 0,0 xx 2 + 3 xx (- 2) + 2 xx 3),(-11xx1 + 1 xx 3 + 8 xx 1, -11 xx (- 1) + 1 xx 0 + 8 xx 0,-11 xx 2 + 1 xx(- 2) + 8 xx3),(0 xx 1 + (- 1) xx 3 + 3 xx 1, 0xx(- 1) + (- 1) xx 0 + 3 xx 0, 0xx2 + (- 1)xx (- 2) + 3 xx 3)]`
`= [(0 + 9 + 2, 0 + 0 + 0, 0 - 6 + 6),(- 11 + 3 + 8, 11 + 0 + 0, - 22 - 2 + 24),(0 - 3 + 3, 0 + 0 + 0, 0 + 2 + 9)]`
= `[(11,0,0),(0,11,0),(0,0,11)]`
= `11 [(1,0,0),(0,1,0),(0,0,1)]`
= 11 · I
= |A| · I
Hence, A(adj A) = (adj A)A = |A|I
APPEARS IN
संबंधित प्रश्न
Verify A(adj A) = (adj A)A = |A|I.
`[(2,3),(-4,-6)]`
Find the inverse of the matrices (if it exists).
`[(-1,5),(-3,2)]`
Let A = `[(3,7),(2,5)]` and B = `[(6,8),(7,9)]`. Verify that (AB)−1 = B−1A−1.
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 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:
\[\begin{bmatrix}- 3 & 5 \\ 2 & 4\end{bmatrix}\]
Compute the adjoint of the following matrix:
Verify that (adj A) A = |A| I = A (adj A) for the above matrix.
Find the inverse of the following matrix.
Find the inverse of the following matrix.
Find the inverse of the following matrix.
Find the inverse of the following matrix.
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
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
If \[A = \begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\] , show that \[A^{- 1} = A^3\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}5 & 2 \\ 2 & 1\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1\end{bmatrix}\]
If \[A = \begin{bmatrix}1 & - 3 \\ 2 & 0\end{bmatrix}\], write adj A.
If \[A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}, B = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\] , find adj (AB).
If A5 = O such that \[A^n \neq I\text{ for }1 \leq n \leq 4,\text{ then }\left( I - A \right)^{- 1}\] equals ________ .
If \[A^2 - A + I = 0\], then the inverse of A is __________ .
If a matrix A is such that \[3A^3 + 2 A^2 + 5 A + I = 0,\text{ then }A^{- 1}\] 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 and B are invertible matrices, then which of the following is not correct?
(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)]`
Find x, if `[(1,2,"x"),(1,1,1),(2,1,-1)]` is singular
If the equation a(y + z) = x, b(z + x) = y, c(x + y) = z have non-trivial solutions then the value of `1/(1+"a") + 1/(1+"b") + 1/(1+"c")` is ____________.
If A is a square matrix of order 3, |A′| = −3, then |AA′| = ______.
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 ______.
