OP Malhotra Class 12 Maths Solutions Chapter 6 Matrices Ex 6(a)

Regular engagement with ISC S Chand Maths Class 12 Solutions Chapter 6 Matrices Ex 6(a) can boost students confidence in the subject.

S Chand Class 12 ICSE Maths Solutions Chapter 6 Matrices Ex 6(a)

Question 1.
If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?
Solution:
We know that a matrix of order m x n has mn elements. Hence to find all possible order of matrix containing 8 elements, we will find ordered pair where product of whose components be equal to 8.
Thus, such ordered pairs are (2, 4), (4, 2), (1, 8), (8, 1). Hence required possible orders are 2 x 4, 4 x 2, 1 x 8 and 8 x 1. Now if a matrix containing 5 elements such possiblee ordered pairs are (1, 5) & (5, 1).
Hence, required possible ordered pairs are (1 x 5) & 5 x 1).

Question 2.
How many entries are there in (i) a 3 x 3 matrix, (ii) a 3 x 4 matrix, (iii) an m x n matrix, (iv) a square matrix of order n?
Solution:
Some we know that a matrix of order m x n containing mn elements.

1. Thus a matrix of order 3 x 3 containing 3 x 3 i.e. 9 entries.
2. A 3 x 4 matrix contains 3 x 4 i.e. 12 elements.
3. A m x n matrix contains mn elements.
4. A m x n matrix contains mn elements.
5. A square matrix of order n contains n x n i.e. n² elements.

Question 3.
Write out the matrix \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{23} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\)
given that a_ij = 4i – 3j
Solution:
Given a_ij = 4i – 3j
∴ a₁ = 4 – 3 = 1 ;
a₁₂ = 4 x 1 – 3 x 2 = – 2
a₁₃ = 4 – 3 x 3 = – 5
a₂₁ = 4 x 2 – 3 x 1 = 5
a₂₂ = 4 x 2 – 3 x 2 = 2
a₂₃ = 4 x 2 – 3 x 3 = – 1
a₃₁ = 4 x 3 – 3 x 1 = 9
a₃₂ = 4 x 3 – 3 x 2 = 6
a₃₃ = 4 x 3 – 3 x 3 = 3
Thus required matrix
= \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]=\left[\begin{array}{ccc}
1 & -2 & -5 \\
5 & 2 & -1 \\
9 & 6 & 3
\end{array}\right]\)

Question 4.
Construct a 2 x 2 matrix B – [b_ij] whose elements are given by
(i) b_ij = \(\frac{(i-2 j)^2}{2}\)
(ii) b_ij = \(\frac{1}{2}|-3 i+j|\)
Solution:
Given B = [b_ij]_{2 x 2} Here 1 ≤ i, j ≤ 2
(i) Here, b_ij = \(\frac{(i-2 j)^2}{2}\)
when i = 1, j = 1; b₁₁ = \(\frac{(1-2)^2}{2}=\frac{1}{2}\)
when i = 1, j = 2; b₁₂ = \(\frac{(1-4)^2}{2}=\frac{9}{2}\)
when i = 2, j = 1; b₂₁ = \(\frac{(2-2)^2}{2}\) = 0
when i = 2, j = 2; b₂₂ = \(\frac{(2-4)^2}{2}\) = 2
Thus, B = \(\left[\begin{array}{ll}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{array}\right]=\left[\begin{array}{cc}
1 / 2 & 9 / 2 \\
0 & 2
\end{array}\right]\)

(ii) Given b_ij = \(\frac{1}{2}\)|-3i + j|
when i = 1, j = 1; b₁₁ = \(\frac{1}{2}|-3+1|=1\)
when i = 1, j = 2; b₁₂ = \(\frac{1}{2}|-3+2|=\frac{1}{2}\)
when i = 2, j = 1; b₂₁ = \(\frac{1}{2}|-6+1|=\frac{5}{2}\)
when i = 2, j = 2; b₂₂ = \(\frac{1}{2}|-6+2|=\frac{4}{2}\) = 2
Thus, B = \(\left[\begin{array}{ll}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{array}\right]=\left[\begin{array}{cc}
1 & 1 / 2 \\
5 / 2 & 2
\end{array}\right]\)

Question 5.
Construct a 3 x 4 matrix whose elements are:
(a) a_ij = i – j
(ii) a_ij = ij
(iii) a_ij = \(\frac { i }{ j }\)
Solution:
(i) Let A = [a_ij]_3×4;
1 ≤ i ≤ 3; 1 ≤ j ≤ 4;
where a_ij = i – j
when i = 1, j = 1; b₁₁ = 1 – 1 = 0
when i = 1, j = 2; b₁₂ = 1 – 2 = – 1
when i = 1, j = 3; b₁₃ = 1 – 3 = – 2
when i = 1, j = 4; b₁₄ = 1 – 4 = – 3
Similarly
a₂₁ = 2 – 1 = 1; a₂₂ = 2 – 2 = 0;
a₂₃ = 2 – 3 = – 1; a₂₄ = 2 – 4 = – 2;
a₃₁ = 3 – 1 = 2; a₃₂ = 3 – 2 = 1;
a₃₃ = 3 – 3 = 0; a₃₄ = 3 – 4 = – 1
∴ A = \(\left[\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right]\)
= \(\left[\begin{array}{cccc}
0 & -1 & -2 & -3 \\
1 & 0 & -1 & -2 \\
2 & 1 & 0 & -1
\end{array}\right]\)

(ii) Let A = [a_ij]_3×4;
1 ≤ i ≤ 3; 1 ≤ i ≤ 4;
where a_ij = ij
a₁₁ = 1 x 1 = 1, a₁₁ = 1 x 2 = 2;
a₁₃ = 1 x 3 = 1, a₁₄ = 1 x 4 = 4;
a₂₁ = 2 x 1 = 2, a ₂₂ = 2 x 2 = 4;
a₂₃ = 2 x 3 = 6 a₂₄ = 2 x 4 = 8;
a₃₁ = 3 x 1 = 2; a₃₂ = 3 x 2 = 6;
a₃₃ = 3 x 3 = 9; a₃₄ = 3 x 4 = 12
∴ A = \(\left[\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right]\)
= \(\left[\begin{array}{cccc}
1 & 2 & 3 & 4 \\
2 & 4 & 6 & 8 \\
3 & 6 & 9 & 12
\end{array}\right]\)

(iii) det A = \(\left[a_{i j}\right]_{3 \times 4} ; 1 \leq i \leq 3 ; 1 \leq j \leq 4\)

Question 6.
(a) Construct a 2 x 2 matrix whose elements are given by
(i) a_ij = \(\frac{3 i-j}{2}\)
(ii) a_ij = \(\frac{i+3 j}{2}\)
(b) Construct a 3 x 2 matrix whose elements in the ith row and jth column are given by
(i) a_ij = \(\frac{i+3 j}{2}\)
(ii) a_ij = \(\frac{(i+2 j)^2}{2}\)
Solution:
(a) (i) Let A = [a_ij]_2×3;
1 ≤ i ≤ 2; 1 ≤ j ≤ 3;

(b) (i) Let A = [a_ij]_3×2;
1 ≤ i ≤ 3; 1 ≤ j ≤ 2;

Question 7.
If A = \(\left[\begin{array}{rrrrr}
5 & -2 & 1 & 0 & 3 \\
7 & 6 & 4 & 2 & -1 \\
0 & 8 & 3 & 5 & 6
\end{array}\right]\), then State (i) the order of A ; (ii) The number of elements (iii) Write down the entries of the second row of A; (iv) Write down the entries of the third column of A ; (v) State the entries a₁₂, a₂₂, a₃₄, a₁₄ of the above matrix ; (vi) If a_ij = 4, find i, j.
Solution:
Given A = \(\left[\begin{array}{ccccc}
5 & -2 & 1 & 0 & 3 \\
7 & 6 & 4 & 2 & -1 \\
0 & 8 & 3 & 5 & 6
\end{array}\right]\)
(i) Since given matrix contains 3 rows and 5 columns order of A be 3 x 5.
(ii) We know that a matrix of order m x n contains mn elements. Thus given matrix A contains 3×5 i.e. 15 elements.
(iii) The entries of second row of A case 7, 6, 4, 2 & -1.
(iv) The entries of third column of A case 1, 4 & 3.
(v) a₁₂ = – 2; a₂₃ = 4; a₃₄ = 5; a₁₅ = 3;
(vi) Given a_ij = 4 = a₂₃
∴ i = 2 & j = 3

Question 8.
Find x and y such that
(i) \(\left[\begin{array}{rr}
x & y \\
-1 & 5
\end{array}\right]=\left[\begin{array}{ll}
-2 & 0 \\
-1 & 5
\end{array}\right]\)
(ii) \(\left[\begin{array}{ll}
x & 3
\end{array}\right]=\left[\begin{array}{ll}
-1 & y
\end{array}\right]\)
(iii) \(\left[\begin{array}{r}
x+1 \\
-3+y
\end{array}\right]=\left[\begin{array}{r}
-2 \\
0
\end{array}\right]\)
Solution:
(i) \(\left[\begin{array}{rr}
x & y \\
-1 & 5
\end{array}\right]=\left[\begin{array}{ll}
-2 & 0 \\
-1 & 5
\end{array}\right]\)
so their cortesfounding t ments are equal
∴ x = – 2 & y = 0

(ii) \(\left[\begin{array}{ll}
x & 3
\end{array}\right]=\left[\begin{array}{ll}
-1 & y
\end{array}\right]\)
∴ x = – 1 & y = 3
[since their correspounding entries are equal]

(iii) \(\left[\begin{array}{r}
x+1 \\
-3+y
\end{array}\right]=\left[\begin{array}{r}
-2 \\
0
\end{array}\right]\)
∴ there correspounding entries are equal
∴ x + 1 = – 2 ⇒ x = – 3
& – 3+ y = 0 ⇒ y = 3

Question 9.
If \(\left[\begin{array}{rr}
x+y & y-z \\
z-2 x & y-x
\end{array}\right]=\left[\begin{array}{rr}
3 & -1 \\
1 & 1
\end{array}\right]\), find x, y, z.
Solution:
Given \(\left[\begin{array}{rr}
x+y & y-z \\
z-2 x & y-x
\end{array}\right]=\left[\begin{array}{rr}
3 & -1 \\
1 & 1
\end{array}\right]\)
Thus, their correspounding entries are equal.
∴ x + y = 3 … (1)
y – z = – 1 … (2)
z – 2x = 1 … (3)
y – x = 1 … (4)
on adding (1) & (4); we have
2y = 4 ⇒ y = 2
from (1); x = 3 – 2 = 1
∴ from (2); z = y + 1 = 2 + 1 = 3
Thus, x = 1; y = 2 & z = 3

Question 10.
(i) If matrix \(\left[\begin{array}{rr}
x-y & 2 x+z \\
2 x-y & 3 z+w
\end{array}\right]=\left[\begin{array}{rr}
-1 & 5 \\
0 & 13
\end{array}\right]\), find x, y, z, w.
(ii) If matrix \(\left[\begin{array}{cr}
a+b & 2 \\
5 & a b
\end{array}\right]=\left[\begin{array}{ll}
6 & 2 \\
5 & 8
\end{array}\right]\), find the values of a and b.
Solution:
(i) Given \(\left[\begin{array}{rr}
x-y & 2 x+z \\
2 x-y & 3 z+w
\end{array}\right]=\left[\begin{array}{rr}
-1 & 5 \\
0 & 13
\end{array}\right]\)
Thus their correspounding elements are equal.
∴ x – y = – 1 …(1)
2x – y = 0 …(2)
2x + z = 5 …(3)
& 3z + w = 13 …(4)
eqn. (1) -eqn. (2) gives;
– x = – 1 ⇒ x = 1
∴ from (2); 2 – y = 0 ⇒ y = 2
∴ from (3); 2 + z = 5 ⇒ z = 3
from (4); 9 + w = 13 ⇒ w = 4
Hence, x = 1; y = 2; z = 3, w = 4

(ii) Given \(\left[\begin{array}{cr}
a+b & 2 \\
5 & a b
\end{array}\right]=\left[\begin{array}{ll}
6 & 2 \\
5 & 8
\end{array}\right]\)
Then their correspounding elements are equal.
a + b = 6 …(1)
ab = 8 …(2)
Now, (a – b)² = (a + b)² – 4ab = 6² – 4 x 8 = 4
⇒ a – b = ± 2
Case-I
when a – b = 2 …(3)
on adding eqn. (1) & (3); we have
2a = 8 ⇒ a = 4
∴ from (3); b = 4 – 2 = 2
Case-II
when a – b = – 2 …(4)
on solving eqn. (1) & (4); we have a = 2 and b = 4

Question 11.
Find the values of x, and z from the following equations:
\(\left[\begin{array}{r}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{l}
9 \\
5 \\
7
\end{array}\right]\)
Solution:
Given
\(\left[\begin{array}{r}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{l}
9 \\
5 \\
7
\end{array}\right]\)
This their correspounding entries are equal.
x + y + z = 9 …(1)
x + z = 5 …(2)
y + z = 1 …(3)
∴ from (1) & (2); 5 + y = 9 ⇒ y = 4
∴ from (1) & (3); x + 7 = 9 ⇒ x = 2
∴ from (1) z = 9 – 6 = 3

Question 12.
Find the values of a, b, c and d, if
\(\left[\begin{array}{rr}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{rr}
-1 & 5 \\
0 & 13
\end{array}\right]\)
Solution:
Given \(\left[\begin{array}{rr}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{rr}
-1 & 5 \\
0 & 13
\end{array}\right]\)
Thus their correspounding elements are equal.
a – b = – 1 …(1)
2a – b = 0 …(2)
2a + c = 5 …(3)
3o + d = 13 …(4)
on solving eqn. (1) & eqn. (2); we have
a = 1; b = 2
∴ from (3); 2 + c = 5 ⇒ c = 3
∴ from (4); 9 + d = 13 ⇒ d = 4
Hence, a = 1; b = 2; c = 3 & d = A.