English

Let a = ⎡ ⎢ ⎣ a 0 0 0 a 0 0 0 a ⎤ ⎥ ⎦ , Then an is Equal to - Mathematics

Advertisements
Advertisements

Question

Let A = \[\begin{bmatrix}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{bmatrix}\], then An is equal to

 

Options

  • \begin{bmatrix}a^n & 0 & 0 \\ 0 & a^n & 0 \\ 0 & 0 & a\end{bmatrix} 

  • \[\begin{bmatrix}a^n & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{bmatrix}\]

  • \[\begin{bmatrix}a^n & 0 & 0 \\ 0 & a^n & 0 \\ 0 & 0 & a^n\end{bmatrix}\]

  •  \[\begin{bmatrix}na & 0 & 0 \\ 0 & na & 0 \\ 0 & 0 & na\end{bmatrix}\]

MCQ

Solution

 \[\begin{bmatrix}a^n & 0 & 0 \\ 0 & a^n & 0 \\ 0 & 0 & a^n\end{bmatrix}\]

\[Here, \]

\[A = \begin{bmatrix}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{bmatrix}\]

\[ \Rightarrow A^2 = \begin{bmatrix}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{bmatrix}\begin{bmatrix}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{bmatrix} = \begin{bmatrix}a^2 & 0 & 0 \\ 0 & a^2 & 0 \\ 0 & 0 & a^2\end{bmatrix}\]

\[ \Rightarrow A^3 = \begin{bmatrix}a^2 & 0 & 0 \\ 0 & a^2 & 0 \\ 0 & 0 & a^2\end{bmatrix}\begin{bmatrix}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{bmatrix} = \begin{bmatrix}a^3 & 0 & 0 \\ 0 & a^3 & 0 \\ 0 & 0 & a^3\end{bmatrix}\]

\[\]

This pattern is applicable on all natural numbers .

\[ \therefore A^n = \begin{bmatrix}a^n & 0 & 0 \\ 0 & a^n & 0 \\ 0 & 0 & a^n\end{bmatrix}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Algebra of Matrices - Exercise 5.7 [Page 66]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 5 Algebra of Matrices
Exercise 5.7 | Q 8 | Page 66

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Compute the indicated product.

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


Compute the indicated products:

`[[a    b],[-b      a]][[a     -b],[b         a]]`


Compute the products AB and BA whichever exists in each of the following cases:

 [ab]`[[c],[d]]`+ [a, b, c, d] `[[a],[b],[c],[d]]`


Show that AB ≠ BA in each of the following cases:

`A = [[1,3,-1],[2,-1,-1],[3,0,-1]]` And `B= [[-2,3,-1],[-1,2,-1],[-6,9,-4]]`

 


For the following matrices verify the associativity of matrix multiplication i.e. (ABC = A(BC):

`A=[[4       2        3],[1       1          2],[3         0          1]]`=`B=[[1        -1          1],[0         1            2],[2           -1          1]]` and  `C= [[1       2       -1],[3       0         1],[0         0         1]]` 


If \[A = \begin{bmatrix}3 & - 5 \\ - 4 & 2\end{bmatrix}\] , find A2 − 5A − 14I.


If\[A = \begin{bmatrix}1 & 2 \\ 2 & 1\end{bmatrix}\] f (x) = x2 − 2x − 3, show that f (A) = 0


If A=then find λ, μ so that A2 = λA + μI

 

Find the value of x for which the matrix product`[[2       0           7],[0          1            0],[1       -2       1]]` `[[-x         14x          7x],[0         1            0],[x           -4x             -2x]]`equal an identity matrix.


Solve the matrix equations:

`[x1][[1,0],[-2,-3]][[x],[5]]=0`


If , then show that A is a root of the polynomial f (x) = x3 − 6x2 + 7x + 2.

 

`A=[[1,2,2],[2,1,2],[2,2,1]]`, then prove that A2 − 4A − 5I = 0


`A=[[3,2, 0],[1,4,0],[0,0,5]]` show that A2 − 7A + 10I3 = 0


Find the matrix A such that `=[[1,2,3],[4,5,6]]=[[-7,-8,-9],[2,4,6],[11,10,9]]`


`A=[[3,-5],[-4,2]]` then find A2 − 5A − 14I. Hence, obtain A3


If `P=[[x,0,0],[0,y,0],[0,0,z]]` and `Q=[[a,0,0],[0,b,0],[0,0,c]]` prove that `PQ=[[xa,0,0],[0,yb,0],[0,0,zc]]=QP`


`A=[[2,0,1],[2,1,3],[1,-1,0]]` , find A2 − 5A + 4I and hence find a matrix X such that A2 − 5A + 4I + = 0.

 

To promote making of toilets for women, an organisation tried to generate awarness through (i) house calls, (ii) letters, and (iii) announcements. The cost for each mode per attempt is given below:

(i) ₹50       (ii) ₹20       (iii) ₹40

The number of attempts made in three villages XY and Z are given below:

          (i)               (ii)              (iii)
X      400              300             100
Y      300              250               75
Z      500              400             150

Find the total cost incurred by the organisation for three villages separately, using matrices.

 

In a parliament election, a political party hired a public relations firm to promote its candidates in three ways − telephone, house calls and letters. The cost per contact (in paisa) is given in matrix A as
\[A = \begin{bmatrix}140 \\ 200 \\ 150\end{bmatrix}\begin{array} \text{Telephone}\\{\text{House calls }}\\ \text{Letters}\end{array}\]

The number of contacts of each type made in two cities X and Y is given in the matrix B as

\[\begin{array}"Telephone & House calls & Letters\end{array}\]

\[B = \begin{bmatrix}1000 & 500 & 5000 \\ 3000 & 1000 & 10000\end{bmatrix}\begin{array} \\City   X \\ City Y\end{array}\]

Find the total amount spent by the party in the two cities.

What should one consider before casting his/her vote − party's promotional activity of their social activities?

 

Let  `A =[[2,-3],[-7,5]]` And `B=[[1,0],[2,-4]]` verify that 

 (A + B)T = AT BT


If `A= [[3],[5],[2]]` And B=[1  0   4] , Verify that `(AB)^T=B^TA^T` 


Let `A= [[1,-1,0],[2,1,3],[1,2,1]]` And `B=[[1,2,3],[2,1,3],[0,1,1]]` Find `A^T,B^T` and verify that   (A + B)T = AT + BT


 For two matrices A and B,   \[A = \begin{bmatrix}2 & 1 & 3 \\ 4 & 1 & 0\end{bmatrix}, B = \begin{bmatrix}1 & - 1 \\ 0 & 2 \\ 5 & 0\end{bmatrix}\](AB)T = BT AT.

 


Express the matrix \[A = \begin{bmatrix}4 & 2 & - 1 \\ 3 & 5 & 7 \\ 1 & - 2 & 1\end{bmatrix}\] as the sum of a symmetric and a skew-symmetric matrix.

If A is an m × n matrix and B is n × p matrix does AB exist? If yes, write its order.

 

 If  \[A = \begin{bmatrix}2 & 1 & 4 \\ 4 & 1 & 5\end{bmatrix}and B = \begin{bmatrix}3 & - 1 \\ 2 & 2 \\ 1 & 3\end{bmatrix}\] . Write the orders of AB and BA.
 

 


If A = [aij] is a square matrix such that aij = i2 − j2, then write whether A is symmetric or skew-symmetric.


If `A=[[i,0],[0,i ]]` , n ∈ N, then A4n equals


If A and B are two matrices such that AB = A and BA = B, then B2 is equal to


If  \[A = \begin{bmatrix}1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} and B = \begin{bmatrix}1 & - 2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\] and AB = I3, then x + y equals 


If \[A = \begin{bmatrix}1 & - 1 \\ 2 & - 1\end{bmatrix}, B = \begin{bmatrix}a & 1 \\ b & - 1\end{bmatrix}\]and (A + B)2 = A2 + B2,   values of a and b are


If \[\begin{bmatrix}2x + y & 4x \\ 5x - 7 & 4x\end{bmatrix} = \begin{bmatrix}7 & 7y - 13 \\ y & x + 6\end{bmatrix}\] 


If  \[A = \begin{pmatrix}\cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{pmatrix},\] ,find adj·A and verify that A(adj·A) = (adj·A)A = |A| I3.


If X = `[(3, 1, -1),(5, -2, -3)]` and Y = `[(2, 1, -1),(7, 2, 4)]`, find X + Y


Show that if A and B are square matrices such that AB = BA, then (A + B)2 = A2 + 2AB + B2.


Three schools DPS, CVC, and KVS decided to organize a fair for collecting money for helping the flood victims. They sold handmade fans, mats, and plates from recycled material at a cost of Rs. 25, Rs.100, and Rs. 50 each respectively. The numbers of articles sold are given as

School/Article DPS CVC KVS
Handmade/fans 40 25 35
Mats 50 40 50
Plates 20 30 40

Based on the information given above, answer the following questions:

  • How many articles (in total) are sold by three schools?

If A = `[(a, b),(b, a)]` and A2 = `[(α, β),(β, α)]`, then ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×