English

Find the Inverse of the Following Matrix. ⎡ ⎢ ⎣ 2 0 − 1 5 1 0 0 1 3 ⎤ ⎥ ⎦ - Mathematics

Advertisements
Advertisements

Question

Find the inverse of the following matrix.

\[\begin{bmatrix}2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix}\]
Advertisements

Solution

\[D = \begin{bmatrix}2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix}\]
Now, 
\[ C_{11} = \begin{vmatrix}1 & 0 \\ 1 & 3\end{vmatrix} = 3, C_{12} = - \begin{vmatrix}5 & 0 \\ 0 & 3\end{vmatrix} = - 15\text{ and }C_{13} = \begin{vmatrix}5 & 1 \\ 0 & 1\end{vmatrix} = 5\]
\[ C_{21} = - \begin{vmatrix}0 & - 1 \\ 1 & 3\end{vmatrix} = - 1, C_{22} = \begin{vmatrix}2 & - 1 \\ 0 & 3\end{vmatrix} = 6\text{ and }C_{23} = - \begin{vmatrix}2 & 0 \\ 0 & 1\end{vmatrix} = - 2\]
\[ C_{31} = \begin{vmatrix}0 & - 1 \\ 1 & 0\end{vmatrix} = 1, C_{32} = - \begin{vmatrix}2 & - 1 \\ 5 & 0\end{vmatrix} = - 5\text{ and }C_{33} = \begin{vmatrix}2 & 0 \\ 5 & 1\end{vmatrix} = 2\]
\[adjD = \begin{bmatrix}3 & - 15 & 5 \\ - 1 & 6 & - 2 \\ 1 & - 5 & 2\end{bmatrix}^T = \begin{bmatrix}3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2\end{bmatrix}\]
\[\text{ and }\left| D \right| = 1\]
\[ \therefore D^{- 1} = \begin{bmatrix}3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2\end{bmatrix}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Adjoint and Inverse of a Matrix - Exercise 7.1 [Page 23]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 7 Adjoint and Inverse of a Matrix
Exercise 7.1 | Q 8.4 | Page 23

RELATED QUESTIONS

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)]`


Find the inverse of the matrices (if it exists).

`[(1,0,0),(0, cos alpha, sin alpha),(0, sin alpha, -cos alpha)]`


For the matrix A = `[(3,2),(1,1)]` find the numbers a and b such that A2 + aA + bI = 0.


Compute the adjoint of the following matrix:

\[\begin{bmatrix}1 & 2 & 5 \\ 2 & 3 & 1 \\ - 1 & 1 & 1\end{bmatrix}\]

Verify that (adj A) A = |A| I = A (adj A) for the above matrix.


Compute the adjoint of the following matrix:

\[\begin{bmatrix}2 & - 1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & - 1\end{bmatrix}\]

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 the inverse of the following matrix.

\[\begin{bmatrix}2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2\end{bmatrix}\]

Find the inverse of the following matrix.

\[\begin{bmatrix}0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4\end{bmatrix}\]

Find the inverse of the following matrix.

\[\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & - \cos \alpha\end{bmatrix}\]

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.


Solve the matrix equation \[\begin{bmatrix}5 & 4 \\ 1 & 1\end{bmatrix}X = \begin{bmatrix}1 & - 2 \\ 1 & 3\end{bmatrix}\], where X is a 2 × 2 matrix.


Find the inverse by using elementary row transformations:

\[\begin{bmatrix}1 & 6 \\ - 3 & 5\end{bmatrix}\]


Find the inverse by using elementary row transformations:

\[\begin{bmatrix}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix}\]


Find the inverse by using elementary row transformations:

\[\begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\]


Find the inverse by using elementary row transformations:

\[\begin{bmatrix}2 & - 1 & 4 \\ 4 & 0 & 7 \\ 3 & - 2 & 7\end{bmatrix}\]


Find the inverse by using elementary row transformations:

\[\begin{bmatrix}3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1\end{bmatrix}\]    


If A is symmetric matrix, write whether AT is symmetric or skew-symmetric.


Find the inverse of the matrix \[\begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\]


If \[A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}, B = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\] , find adj (AB).


If \[A = \begin{bmatrix}3 & 1 \\ 2 & - 3\end{bmatrix}\], then find |adj A|.


If \[A = \begin{bmatrix}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{bmatrix}\] , then the value of |adj A| is _____________ .


If A satisfies the equation \[x^3 - 5 x^2 + 4x + \lambda = 0\] then A-1 exists if _____________ .


If for the matrix A, A3 = I, then A−1 = _____________ .


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 _____________ .


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)]`


If A = `[(0, 1, 3),(1, 2, x),(2, 3, 1)]`, A–1 = `[(1/2, -4, 5/2),(-1/2, 3, -3/2),(1/2, y, 1/2)]` then x = 1, y = –1.


(A3)–1 = (A–1)3, where A is a square matrix and |A| ≠ 0.


Find x, if `[(1,2,"x"),(1,1,1),(2,1,-1)]` is singular


Find the value of x for which the matrix A `= [(3 - "x", 2, 2),(2,4 - "x", 1),(-2,- 4,-1 - "x")]` is singular.


For what value of x, matrix `[(6-"x", 4),(3-"x", 1)]` is a singular matrix?


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 ____________.


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 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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×