Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Fill in the blanks in each of the If A and B are square matrices of the same order, then (i) (AB)’ = ________. (ii) (kA)’ = ________. (k is any scalar) (iii) [k (A - B)]’ = ________.

Fill in the blanks in each of the
If A and B are square matrices of the same order, then
(i) (AB)’ = ________.
(ii) (kA)’ = ________. (k is any scalar)
(iii) [k (A - B)]’ = ________.

Answers (1)

(i) (AB)’ = ________.

(AB)’ = B’A’

Let A be matrix of order m× n and B be of n× p.

A’ is of order n× m and B’ is of order p× n.

Hence, we get,  B’ A’ is of order p× m.

So, AB is of order m× p.

And (AB)’ is of order p× m.

We can see (AB)’ and B’ A’ are of same order p× m.

Hence proved,  (AB)’ = B’ A’

(ii) (kA)’ = ________. (k is any scalar)

If a scalar “k” is multiplied to any matrix the new matrix becomes

K times of the old matrix.

\begin{array}{l} \text { Eg: } A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \\ 2 A=2\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{ll} 2 & 4 \\ 6 & 8 \end{array}\right] \\ (2 A)=\left[\begin{array}{ll} 2 & 6 \\ 4 & 8 \end{array}\right] \\ A^{\prime}=\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right] \end{array}

Now 2A’ = 2\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{ll} 2 & 6 \\ 4 & 8 \end{array}\right]

Hence (2A)’ =2A’

Hence (kA)’ = k(A)’

(iii) [k (A - B)]’ = ________.

\begin{aligned} &A=\left[\begin{array}{ll} 5 & 7 \\ 7 & 6 \end{array}\right]\\ &A'=\left[\begin{array}{ll} 5 & 7 \\ 7 & 6 \end{array}\right]\\ &2A'=\left[\begin{array}{ll} 10 & 14 \\ 14 & 12 \end{array}\right]\\ &B=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\\ &B^{\prime}=\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\\ &2 B^{\prime}={2}\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\\ &=\left[\begin{array}{ll} 2 & 6 \\ 4 & 8 \end{array}\right]\\ &A-B=\left[\begin{array}{ll} 4 & 5 \\ 4 & 2 \end{array}\right] \end{aligned}

\begin{array}{l} \text { Now Let } k=2 \\ 2(A-B)=2\left[\begin{array}{ll} 4 & 5 \\ 4 & 2 \end{array}\right]=\left[\begin{array}{ll} 8 & 10 \\ 8 & 4 \end{array}\right] \\ {[2(A-B)]=\left[\begin{array}{cc} 8 & 8 \\ 10 & 4 \end{array}\right]} \\ 2 A^{\prime}-2 B'=\left[\begin{array}{ll} 10 & 14 \\ 14 & 12 \end{array}\right]-\left[\begin{array}{ll} 2 & 6 \\ 4 & 8 \end{array}\right]=\left[\begin{array}{cc} 8 & 8 \\ 10 & 4 \end{array}\right] \\ A^{\prime}-B'=\left[\begin{array}{ll} 5 & 7 \\ 7 & 6 \end{array}\right]-\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{ll} 4 & 4 \\ 5 & 3 \end{array}\right] \\ 2\left(A^{\prime}-B^{\prime}\right)=2\left[\begin{array}{ll} 4 & 4 \\ 5 & 3 \end{array}\right]=\left[\begin{array}{cc} 8 & 8 \\ 10 & 6 \end{array}\right] \\ \text { Hence we can see }[k(A-B)]^{\prime}=k(A)^{\prime}-k(B)^{\prime}=k\left(A^{\prime}-B^{\prime}\right) \end{array}

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question