English

If A, B Are Two N × N Non-singular Matrices, Then - Mathematics

Advertisements
Advertisements

Question

If A, B are two n × n non-singular matrices, then __________ .

Options

  • AB is non-singular

  • AB is singular

  • \[\left( AB \right)^{- 1} A^{- 1} B^{- 1}\]

  • (AB)−1 does not exist

MCQ

Solution

AB is non-singular

A and B are non - singular matrices of order n × n.
\[ \therefore \left| A \right| \neq 0\text{ and }\left| B \right| \neq 0 . . . \left( 1 \right)\]

A and B are of the same order, so AB is defined and is of the same order .

Thus, 
\[\left| AB \right| = \left| A \right|\left| B \right| \]
\[ \Rightarrow \left| AB \right| \neq 0 \left[\text{ Using }\left( 1 \right) \right] \]
Thus, AB is non - singular . 

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Adjoint and Inverse of a Matrix - Exercise 7.4 [Page 37]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 7 Adjoint and Inverse of a Matrix
Exercise 7.4 | Q 6 | Page 37

RELATED QUESTIONS

Find the adjoint of the matrices.

`[(1,-1,2),(2,3,5),(-2,0,1)]`


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

`[(2,-2),(4,3)]`


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

`[(1,2,3),(0,2,4),(0,0,5)]`


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)`


Let A = `[(1,-2,1),(-2,3,1),(1,1,5)]` verify that

  1. [adj A]–1 = adj (A–1)
  2. (A–1)–1 = A

Find the adjoint of the following matrix:

\[\begin{bmatrix}1 & \tan \alpha/2 \\ - \tan \alpha/2 & 1\end{bmatrix}\]
Verify that (adj A) A = |A| I = A (adj A) for the above matrix.

For the matrix 

\[A = \begin{bmatrix}1 & - 1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10\end{bmatrix}\] , show that A (adj A) = O.

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:

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

Find the inverse of the following matrix.

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

For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]

\[A = \begin{bmatrix}2 & 1 \\ 5 & 3\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 5 \\ 3 & 4\end{bmatrix}\]


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

(i) \[\left[ F \left( \alpha \right) \right]^{- 1} = F \left( - \alpha \right)\]
(ii) \[\left[ G \left( \beta \right) \right]^{- 1} = G \left( - \beta \right)\]
(iii) \[\left[ F \left( \alpha \right)G \left( \beta \right) \right]^{- 1} = G \left( - \beta \right)F \left( - \alpha \right)\]

If  \[A = \begin{bmatrix}4 & 3 \\ 2 & 5\end{bmatrix}\], find x and y such that 

\[A^2 = xA + yI = O\] . Hence, evaluate 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.


If \[A = \begin{bmatrix}2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2\end{bmatrix}\].
Verify that \[A^3 - 6 A^2 + 9A - 4I = O\]  and hence find A−1.

Find the adjoint of the matrix \[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\]  and hence show that \[A\left( adj A \right) = \left| A \right| I_3\]. 


Find the inverse by using elementary row transformations:

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


Find the inverse by using elementary row transformations:

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


Find the inverse by using elementary row transformations:

\[\begin{bmatrix}3 & 10 \\ 2 & 7\end{bmatrix}\]


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:

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


If A is symmetric matrix, write whether AT 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 \[A = \frac{1}{3}\begin{bmatrix}1 & 1 & 2 \\ 2 & 1 & - 2 \\ x & 2 & y\end{bmatrix}\] is orthogonal, then x + y =

(a) 3
(b) 0
(c) − 3
(d) 1


If \[A = \begin{bmatrix}1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2\end{bmatrix},\text{ then aI + bA + 2 }A^2\] equals ____________ .


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 (A1) is equal to ____________ .


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 `= [(1,2),(3,4)].`


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 A = `[(3,1),(-1,2)]`, then 14A−1 is given by:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×