Advertisements
Advertisements
Question
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1\end{bmatrix}\]
Solution
\[A = \begin{bmatrix}3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1\end{bmatrix}\]
We know
\[A = IA \]
\[ \Rightarrow \begin{bmatrix}3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - \frac{1}{3} \\ 2 & 3 & 0 \\ 0 & 4 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_1 \to \frac{1}{3} R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - \frac{1}{3} \\ 0 & 3 & \frac{2}{3} \\ 0 & 4 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{3} & 0 & 0 \\ - \frac{2}{3} & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 - 2 R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - \frac{1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 4 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{3} & 0 & 0 \\ - \frac{2}{9} & \frac{1}{3} & 0 \\ 0 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to \frac{1}{3} R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - \frac{1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & \frac{1}{9}\end{bmatrix} = \begin{bmatrix}\frac{1}{3} & 0 & 0 \\ - \frac{2}{9} & \frac{1}{3} & 0 \\ \frac{8}{9} & \frac{- 4}{3} & 1\end{bmatrix} A \left[\text{ Applying }R_3 \to R_3 - 4 R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - \frac{1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{3} & 0 & 0 \\ - \frac{2}{9} & \frac{1}{3} & 0 \\ 8 & - 12 & 9\end{bmatrix} A \left[\text{ Applying }R_3 \to 9 R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}3 & - 4 & 3 \\ - 2 & 3 & - 2 \\ 8 & - 12 & 9\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 - \frac{2}{9} R_3\text{ and }R_1 \to R_1 + \frac{1}{3} R_3 \right]\]
\[ \therefore A^{- 1} = \begin{bmatrix}3 & - 4 & 3 \\ - 2 & 3 & - 2 \\ 8 & - 12 & 9\end{bmatrix} \]
APPEARS IN
RELATED QUESTIONS
Find the adjoint of the matrices.
`[(1,2),(3,4)]`
Verify A (adj A) = (adj A) A = |A|I.
`[(2,3),(-4,-6)]`
Find the inverse of the matrices (if it exists).
`[(2,-2),(4,3)]`
If A = `[(3,1),(-1,2)]` show that A2 – 5A + 7I = O. Hence, find A–1.
For the matrix A = `[(3,2),(1,1)]` find the numbers a and b such that A2 + aA + bI = O.
Let A = `[(1, sin theta, 1),(-sin theta,1,sin 1),(-1, -sin theta, 1)]` where 0 ≤ θ≤ 2π, then ______.
If \[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\] , show that adj A = 3AT.
Find the inverse of the following matrix:
Find the inverse of the following matrix.
For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]
\[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 6 \\ 3 & 2\end{bmatrix}\]
Given \[A = \begin{bmatrix}2 & - 3 \\ - 4 & 7\end{bmatrix}\], compute A−1 and show that \[2 A^{- 1} = 9I - A .\]
If \[A = \begin{bmatrix}4 & 3 \\ 2 & 5\end{bmatrix}\], find x and y such that
Show that \[A = \begin{bmatrix}6 & 5 \\ 7 & 6\end{bmatrix}\] satisfies the equation \[x^2 - 12x + 1 = O\]. Thus, find A−1.
Show that the matrix, \[A = \begin{bmatrix}1 & 0 & - 2 \\ - 2 & - 1 & 2 \\ 3 & 4 & 1\end{bmatrix}\] satisfies the equation, \[A^3 - A^2 - 3A - I_3 = O\] . Hence, 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 matrix X for which
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:
\[\begin{bmatrix}5 & 2 \\ 2 & 1\end{bmatrix}\]
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:
\[\begin{bmatrix}2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1\end{bmatrix}\]
If A is a square matrix, then write the matrix adj (AT) − (adj A)T.
If A is a non-singular symmetric matrix, write whether A−1 is symmetric or skew-symmetric.
Find the inverse of the matrix \[\begin{bmatrix}3 & - 2 \\ - 7 & 5\end{bmatrix} .\]
Find the inverse of the matrix \[\begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\]
If A and B are invertible matrices, which of the following statement is not correct.
If \[A = \begin{bmatrix}2 & 3 \\ 5 & - 2\end{bmatrix}\] be such that \[A^{- 1} = kA\], then k equals ___________ .
(a) 3
(b) 0
(c) − 3
(d) 1
If a matrix A is such that \[3A^3 + 2 A^2 + 5 A + I = 0,\text{ then }A^{- 1}\] equal to _______________ .
If \[A = \begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix}\], find A−1 and hence solve the system of linear equations 2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y + 2z = −3
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
If A and B are invertible matrices, then which of the following is not correct?
`("aA")^-1 = 1/"a" "A"^-1`, where a is any real number and A is a square matrix.
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 A = [aij] is a square matrix of order 2 such that aij = `{(1"," "when i" ≠ "j"),(0"," "when" "i" = "j"):},` then A2 is ______.
For matrix A = `[(2,5),(-11,7)]` (adj A)' is equal to:
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.
