Well-structured ISC S Chand Maths Class 12 Solutions Chapter 6 Matrices Ex 6(e) facilitate a deeper understanding of mathematical principles.
S Chand Class 12 ICSE Maths Solutions Chapter 6 Matrices Ex 6(e)
Question 1.
(i) If A = \(\left[\begin{array}{rrr}
1 & 4 & 2 \\
2 & 5 & 3 \\
3 & -1 & 0
\end{array}\right]\), find A + A’.
(ii) For matrix A = \(\left[\begin{array}{rrr}
-3 & 6 & 0 \\
4 & -5 & 8 \\
0 & -7 & -2
\end{array}\right]\), find \(\frac { 1 }{ 2 }\)(A – A’).
(iii) Find \(\frac { 1 }{ 2 }\)(A + A’) and \(\frac { 1 }{ 2 }\)(A – A’), where A = \(\left[\begin{array}{rrr}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]\).
Solution:
Question 2.
If B = \(\left[\begin{array}{rr}
1 & 3 \\
-2 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{rr}
-2 & 5 \\
3 & 4
\end{array}\right]\), find (BC)’.
Solution:
Now, BC = \(\left[\begin{array}{rr}
1 & 3 \\
-2 & 5
\end{array}\right]\left[\begin{array}{rr}
-2 & 5 \\
3 & 4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
-2+9 & 5+12 \\
4+15 & -10+20
\end{array}\right]=\left[\begin{array}{cc}
7 & 17 \\
19 & 10
\end{array}\right]\)
∴ (BC)’ = \(\left[\begin{array}{cc}
7 & 17 \\
19 & 10
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
7 & 19 \\
17 & 10
\end{array}\right]\)
Question 3.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 0 \\
3 & -1 & 4
\end{array}\right]\), find
(i) AA’
(ii) A’A
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & 2 & 0 \\
3 & -1 & 4
\end{array}\right]_{2 \times 3}\)
A’ = \(\left[\begin{array}{cc}
1 & 3 \\
2 & -1 \\
0 & 4
\end{array}\right]_{3 \times 2}\)
Since no. of columns in A = No. of rows in A’ = 3
∴ AA’ exists
Further No. of columns in A’ = No. of rows in A = 2
Thus, A’A exists.
Question 4.
If A = \(\left[\begin{array}{rr}
1 & -2 \\
1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
-1 & 2 \\
-1 & 1
\end{array}\right]\), Find
(i) A’B’ (ii) AB’
Solution:
Question 5.
Verify that (AB)’ = B’A’ if
(i) A = \(\left[\begin{array}{r}
3 \\
1 \\
-2
\end{array}\right]\), B = \(\left[\begin{array}{lll}
1 & -5 & 7
\end{array}\right]\)
(ii) A = \(\left[\begin{array}{ll}
2 & 3 \\
0 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ll}
3 & 4 \\
2 & 1
\end{array}\right]\)
(iii) A = \(\left[\begin{array}{lll}
-1 & 3 & 0 \\
-7 & 2 & 8
\end{array}\right]\), B = \(\left[\begin{array}{rr}
-5 & 0 \\
0 & 3 \\
1 & -8
\end{array}\right]\)
Solution:
Question 6.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\), Verify that (A + B)’ = A’ + B’.
Solution:
Now,
∴ From (1) & (2); we have
(A + B)’ = A’ + B’
Question 7.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 3 \\
-1 & 0 & 2 \\
1 & -3 & -1
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
4 & 5 & 6 \\
-1 & 0 & 1 \\
2 & 1 & 2
\end{array}\right]\), C = \(\left[\begin{array}{rrr}
-1 & -2 & 1 \\
-1 & 2 & 3 \\
-1 & -2 & 2
\end{array}\right]\),
find each of the following :
(i) 2A’ – B’
(ii) (A + B + C)’.
Is (A + B + C)’ = A’ + B’ + C’ ?
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & 2 & 3 \\
-1 & 0 & 2 \\
1 & -3 & -1
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
4 & 5 & 6 \\
-1 & 0 & 1 \\
2 & 1 & 2
\end{array}\right]\) & C = \(\left[\begin{array}{rrr}
-1 & -2 & 1 \\
-1 & 2 & 3 \\
-1 & -2 & 2
\end{array}\right]\)
from (1) & (2); we have
(A + B + C)’ = A’ + B’ + C’
Question 8.
If A = \(\left[\begin{array}{llll}
2 & 5 & 7 & 9
\end{array}\right]\) and B = \(\left[\begin{array}{l}
3 \\
0 \\
2 \\
4
\end{array}\right]\) then prove that (A + B’) = (A’ + B)’.
Solution:
Question 9.
Find x and y if the matrix
A = \(\frac{1}{3}\left[\begin{array}{rrr}
1 & 2 & 2 \\
2 & 1 & -2 \\
x & 2 & y
\end{array}\right]\) may satisfy the condition AA’ = A’A = I3.
Solution:
Given A = \(\frac{1}{3}\left[\begin{array}{rrr}
1 & 2 & 2 \\
2 & 1 & -2 \\
x & 2 & y
\end{array}\right]\)
∴ x + 4 + 2y = 0
⇒ x + 2y = – 4 … (1)
& 2x + 2 – 2y = 0
⇒ x – y = – 1 … (2)
on solving (1) & (2);
we have y = – 1; x = – 2
These values of x and y also satisfies the other conditions i.e. x² + y² + 4 = 9.
Now, A’A = I
⇒ \(\left[\begin{array}{ccc}
5+x^2 & 4+2 x & -2+x y \\
4+2 x & 9 & 2+2 y \\
-2+x y & 2+2 y & 8+y^2
\end{array}\right]=\left[\begin{array}{lll}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9
\end{array}\right]\)
5 + x² = 9 ⇒ x ± 2; 4 + 2x = 0 ⇒ x = – 2; – 2 + xy = 0
2 + 2y = 0 ⇒ y = – 1 ; 8 + y² = 9 ⇒ y = ± 1
Thus common values of x and y are ; x = – 2 & y = – 1
Question 10.
If A = \(\left[\begin{array}{rr}
2 & 3 \\
5 & -7
\end{array}\right]\), then verify that (A²)’ = (A’)².
Solution:
Given A = \(\left[\begin{array}{rr}
2 & 3 \\
5 & -7
\end{array}\right]\)
From (1) & (2) ; we have
(A²)’ = (A’)²
Question 11.
If A = \(\frac { 1 }{ 2 }\) , then verify that A’A = I.
Solution:
