Advertisements
Advertisements
Question
Let A = `[(1,2,1),(2,3,1),(1,1,5)]` verify that
- [adj A]–1 = adj(A–1)
- (A–1)–1 = A
Advertisements
Solution
A = `[(1,2,1),(2,3,1),(1,1,5)]`
∴ |A| = 1(3 × 5 − 1 × 1) − 2(2 × 5 − 1 × 1) + 1(2 × 1 − 3 × 1)
= 1(15 − 1) −2(10 − 1) + 1(2 − 3)
= 14 − 18 − 1
= −5
Now, A11 = 14, A12 = −9, A13 = −1
A21 = −9, A22 = 4, A23 = 1
A31 = −1, A32 = 1, A33 = −1
∴ adj A = `[(14,-9,-1),(-9,4,1),(-1,1,-1)]`
∴ A−1 = `1/|A|` (adj A)
= `-1/5[(14,-9,-1),(-9,4,1),(-1,1,-1)]`
(i) L.H.S. = |adj A|
= 14(−4 − 1) + 9(9 + 1) − 1(−9 + 4)
= 14(−5) + 9(10) − 1(−5)
= 70 − 90 + 5
= 25
We have,
adj(adj A) = `[(-5,-10,-5),(-10,-15,-5),(-5,-5,-25)]`
Hence its inverse exists;
[adj A]−1 = `1/|adj A|` [adj(adj A)]
= `1/25[(-5,-10,-5),(-10,-15,-5),(-5,-5,-25)]`
L.H.S. = (adj A)−1 = `[(-1/5,-2/5,-1/5),(-2/5,-3/5,-1/5),(-1/5,-1/5,-1)]`
Now finding R.H.S.:
A−1 = `[((-14)/5,9/5,1/5),(9/5,(-4)/5,(-1)/5),(1/5,(-1)/5,1/5)]`
adj(A–1) = `[(-1/5,-2/5,-1/5),(-2/5,-3/5,-1/5),(-1/5,-1/5,-1)]`
Hence, [adj A]–1 = adj(A–1)
(ii) We have shown that:
A−1 = `-1/5[(14,-9,-1),(-9,4,1),(-1,1,-1)]`
A−1 = `1/|A| = 1/(-5)`
adj(A–1) = `[(-1/5,-2/5,-1/5),(-2/5,-3/5,-1/5),(-1/5,-1/5,-1)]`
Then,
(A−1)−1 = `1/|A^-1|` adj(A−1)
= `-1/(1/5) [(-1/5,-2/5,-1/5),(-2/5,-3/5,-1/5),(-1/5,-1/5,-1)]`
= `[(1,2,1),(2,3,1),(1,1,5)]`
Hence, (A–1)–1 = A
APPEARS IN
RELATED QUESTIONS
Find the adjoint of the matrices.
`[(1,2),(3,4)]`
Verify A(adj A) = (adj A)A = |A|I.
`[(1,-1,2),(3,0,-2),(1,0,3)]`
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)]`
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 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 and verify that \[A^{- 1} A = I_3\]
For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]
\[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 6 \\ 3 & 2\end{bmatrix}\]
Let \[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B = \begin{bmatrix}6 & 7 \\ 8 & 9\end{bmatrix} .\text{ Find }\left( AB \right)^{- 1}\]
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.
Show that \[A = \begin{bmatrix}5 & 3 \\ - 1 & - 2\end{bmatrix}\] satisfies the equation \[x^2 - 3x - 7 = 0\]. Thus, find A−1.
Show that \[A = \begin{bmatrix}6 & 5 \\ 7 & 6\end{bmatrix}\] satisfies the equation \[x^2 - 12x + 1 = O\]. Thus, find A−1.
For the matrix \[A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3\end{bmatrix}\] . Show that
If \[A = \begin{bmatrix}- 1 & 2 & 0 \\ - 1 & 1 & 1 \\ 0 & 1 & 0\end{bmatrix}\] , show that \[A^2 = A^{- 1} .\]
Find the matrix X satisfying the matrix equation \[X\begin{bmatrix}5 & 3 \\ - 1 & - 2\end{bmatrix} = \begin{bmatrix}14 & 7 \\ 7 & 7\end{bmatrix}\]
Find the matrix X for which
Find the matrix X satisfying the equation
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:
\[\begin{bmatrix}1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1\end{bmatrix}\]
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 non-singular symmetric matrix, write whether A−1 is symmetric or skew-symmetric.
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 satisfies the equation \[x^3 - 5 x^2 + 4x + \lambda = 0\] then A-1 exists if _____________ .
If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is _____________ .
Let \[A = \begin{bmatrix}1 & 2 \\ 3 & - 5\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}\] and X be a matrix such that A = BX, then X 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
Using matrix method, solve the following system of equations:
x – 2y = 10, 2x + y + 3z = 8 and -2y + z = 7
If A and B are invertible matrices, then which of the following is not correct?
If A = [aij] is a square matrix of order 2 such that aij = `{(1"," "when i" ≠ "j"),(0"," "when" "i" = "j"):},` then A2 is ______.
If for a square matrix A, A2 – A + I = 0, then A–1 equals ______.
A furniture factory uses three types of wood namely, teakwood, rosewood and satinwood for manufacturing three types of furniture, that are, table, chair and cot.
The wood requirements (in tonnes) for each type of furniture are given below:
| Table | Chair | Cot | |
| Teakwood | 2 | 3 | 4 |
| Rosewood | 1 | 1 | 2 |
| Satinwood | 3 | 2 | 1 |
It is found that 29 tonnes of teakwood, 13 tonnes of rosewood and 16 tonnes of satinwood are available to make all three types of furniture.
Using the above information, answer the following questions:
- Express the data given in the table above in the form of a set of simultaneous equations.
- Solve the set of simultaneous equations formed in subpart (i) by matrix method.
- Hence, find the number of table(s), chair(s) and cot(s) produced.
