AP Inter 2nd Year Maths Exercise 3a Solutions

Referring to the AP Inter 2nd Year Maths Study Material Chapter 3 Matrices Exercise 3a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Matrices Solutions Exercise 3a

Question 1.
In the matrix A = \(\left[\begin{array}{cccc}
2 & 5 & 19 & -7 \\
35 & -2 & \frac{5}{2} & 12 \\
\sqrt{3} & 1 & -5 & 17
\end{array}\right]\),write the order of the matrix
Solution:
There are 3 rows and 4 columns in the given matrix.
∴ Order is 3 × 4.

Question 2.
In the matrix A = \(\left[\begin{array}{cccc}
2 & 5 & 19 & -7 \\
35 & -2 & \frac{5}{2} & 12 \\
\sqrt{3} & 1 & -5 & 17
\end{array}\right]\), write the number of elements.
Solution:
Order of the matrix is 3 × 4
∴ Number of elements is 3 × 4 = 12 elements.

Question 3.
In the matrix A \(\left[\begin{array}{cccc}
2 & 5 & 19 & -7 \\
35 & -2 & \frac{5}{2} & 12 \\
\sqrt{3} & 1 & -5 & 17
\end{array}\right]\), write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃.
Solution:
a₁₃ = 19, a₂₁ = 35, a₃₃ = -5, a₂₄ = 12, a₂₃ = 5/2

Question 4.
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Solution:
We know that if a matrix A is of the order m × n, then A has mn elements.
Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the
ordered pairs of natural numbers whose product is 24.
The ordered pairs are: (1, 24),(24, 1),(2, 12),( 12, 2),(3, 8),(8, 3),(4, 6), (6, 4)
Hence, the possible orders of a matrix having 24 elements are: :
(1×24),(24×1),(2×12),(12×2),(3 ×8),(8×3),(4×6) , (6×4).
13 is a prime, so we get only 2 ordered pairs with product 13. They are (1×13) and(13×1)

Question 5.
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Solution:
We know that if a matrix A is of the order m×n , then A has mn elements.
Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.
The ordered pairs are: (1,18),(18,1),(2,9),(9,2),(3,6), (6,3)
Hence, the possible orders of a matrix having 18 elements are:
(1×18),(18×1),(2×9),(9×2),(3×6),(6×3)
5 is a prime, so we get only two ordered pairs with product 5. They are (1 × 5) and (5×1)

Question 6.
Construct a 2 × 2 matrix, A = [a_ij], whose elements are given by a_ij = \(\frac{(i+j)^2}{2}\)
Solution:
A 2×2 matrix is given by A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]\)
Given that a_ij = \(\frac{(i+j)^2}{2}\); i, j = 1,2
∴ a₁₁ = \(\frac{(1+1)^2}{2}=\frac{4}{2}\) = 2;
a₁₂ = \(\frac{(1+2)^2}{2}=\frac{9}{2}\)
a₂₁ = \(\frac{(2+1)^2}{2}=\frac{9}{2}\)
a₂₂ = \(\frac{(2+2)^2}{2}=\frac{16}{2}\) = 8
Thus, the required matrix is A = \(\left[\begin{array}{cc}
2 & \frac{9}{2} \\
\frac{9}{2} & 8
\end{array}\right]\)

Question 7.
Construct a 2 × 2 matrix, A = [a_ij], whose elements are given by a_ij = \(\frac{i}{j}\)
Solution:
A 2×2 matrix is given by A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]\)
Given that a_ij = \(\frac{i}{j}\); i, j = 1, 2
∴ a₁₁ = \(\frac{1}{1}\) = 1;
a₁₂ = \(\frac{1}{2}\)
a₂₁ = \(\frac{2}{1}\) = 2
a₂₂ = \(\frac{2}{2}\) = 1
Thus, the required matrix is A = \(\left[\begin{array}{ll}
1 & \frac{1}{2} \\
2 & 1
\end{array}\right]\)

Question 8.
Construct a 2 × 2 matrix, A = [a_ij], whose elements are given by a_ij = \(\frac{(i+2j)^2}{2}\)
Solution:
A 2×2 matrix is given by A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]\)
Given that a_ij = \(\frac{(i+2j)^2}{2}\); i, j = 1,2
∴ a₁₁ = \(\frac{(1+2)^2}{2}=\frac{9}{2}\)
a₁₂ = \(\frac{(1+4)^2}{2}=\frac{25}{2}\)
a₂₁ = \(\frac{(2+2)^2}{2}\) = 8
a₂₂ = \(\frac{(2+4)^2}{2}\) = 18
Thus, the required matrix is A = \(\left[\begin{array}{cc}
\frac{9}{2} & \frac{25}{2} \\
8 & 18
\end{array}\right]\)

Question 9.
Construct a 3 × 4 matrix, whose elements are given by a_ij = \(\frac{1}{2}\)|-3i + j|
Solution:
A 3 × 4 matrix is given by A = \(\left[\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right]\)
Given a_ij = \(\frac{1}{2}\)|-3i + j|; i = 1,2,3; j = 1,2,3,4
a₁₁ = \(\frac{1}{2}\)|-3(1) + 1| = \(\frac{1}{2}\)|-3 + 1| = \(\frac{1}{2}\)|-2| = \(\frac{2}{2}\) = 1;
a₂₁ = \(\frac{1}{2}\)|-3(2) + 1| = \(\frac{1}{2}\)|-6 + 1| = \(\frac{1}{2}\)|-5| = \(\frac{5}{2}\)
a₃₁ = \(\frac{1}{2}\)|-3(3) + 1| = \(\frac{1}{2}\)|-9 + 1| = \(\frac{1}{2}\)|-8| = \(\frac{8}{2}\) = 4;
a₁₂ = \(\frac{1}{2}\)|-3(1) + 2| = \(\frac{1}{2}\)|-3 + 2| = \(\frac{1}{2}\)|-1| = \(\frac{1}{2}\) ;
a₂₂ = \(\frac{1}{2}\)|-3(2) + 2| = \(\frac{1}{2}\)|-6 + 2| = \(\frac{1}{2}\)|-4| = \(\frac{4}{2}\) = 2;
a₃₂ = \(\frac{1}{2}\)|-3(3) + 2| = \(\frac{1}{2}\)|-9 + 2| = \(\frac{1}{2}\)|-7| = \(\frac{7}{2}\) ;
a₁₃ = \(\frac{1}{2}\)|-3(1) + 3| = \(\frac{1}{2}\)|-3 + 3| = 0 ;
a₂₃ = \(\frac{1}{2}\)|-3(2) + 3| = \(\frac{1}{2}\)|-6 + 3| = \(\frac{1}{2}\)|-3| = \(\frac{3}{2}\) ;
a₃₃ = \(\frac{1}{2}\)|-3(3) + 3| = \(\frac{1}{2}\)|-9 + 3| = \(\frac{1}{2}\)|-6| = \(\frac{6}{2}\) = 3;
a₁₄ = \(\frac{1}{2}\)|-3(1) + 4| = \(\frac{1}{2}\)|-3 + 4| = \(\frac{1}{2}\)|1| = \(\frac{1}{2}\) ;
a₂₄ = \(\frac{1}{2}\)|-3(2) + 4| = \(\frac{1}{2}\)|-6 + 4| = \(\frac{1}{2}\)|-2| = \(\frac{2}{2}\) = 1;
a₃₄ = \(\frac{1}{2}\)|-3(3) + 4| = \(\frac{1}{2}\)|-9 + 4| = \(\frac{1}{2}\)|-5| = \(\frac{5}{2}\) ;
Thu, the required matrix is A = \(\left[\begin{array}{cccc}
1 & \frac{1}{2} & 0 & \frac{1}{2} \\
\frac{5}{2} & 2 & \frac{3}{2} & 1 \\
4 & \frac{7}{2} & 3 & \frac{5}{2}
\end{array}\right]\)

Question 10.
Construct a 3 × 4 matrix, whose elements are given by a_ij = 2i – j
Solution:
A 3 × 4 matrix is given by A = \(\left[\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right]\)
Given a_ij = 2i – j; i = 1,2,3; j = 1,2,3,4
a₁₁ = 2(1) – 1 = 2 – 1 = 1;
a₂₁ = 2(2) – 1 = 4 – 1 = 3
a₃₁ = 2(3) – 1 = 6 – 1 = 5;
a₁₂= 2(1) – 2 = 2 – 2 = 0
a₂₂ = 2(2) – 2= 4 – 2 = 2;
a₃₂ = 2(3) – 2 = 6 – 2 = 4
a₁₃ = 2(1) – 3 = 2 – 3 = -1;
a₂₃ = 2(2) – 3 = 4 – 3 = 1
a₃₃ = 2(3) – 3 = 6 – 3 = 3;
a₁₄ = 2(1) – 4 = 2 – 4 = -2;
a₂₄ = 2(2) – 4 = 4 – 4 = 0;
a₃₄ = 2(3) – 4 = 6 – 4 = 2
Thus, the required matrix is A = \(\left[\begin{array}{cccc}
1 & 0 & -1 & -2 \\
3 & 2 & 1 & 0 \\
5 & 4 & 3 & 2
\end{array}\right]\)

Question 11.
Find the values of x,y and z from \(\left[\begin{array}{ll}
4 & 3 \\
x & 5
\end{array}\right]=\left[\begin{array}{ll}
y & z \\
1 & 5
\end{array}\right]\)
Solution:
Given \(\left[\begin{array}{ll}
4 & 3 \\
x & 5
\end{array}\right]=\left[\begin{array}{ll}
y & z \\
1 & 5
\end{array}\right]\)
As the two matrices are equal, equating the corresponding elements, we get x = 1, y = 4 and z = 3

Question 12.
Find the values of x,y and z from \(\left[\begin{array}{cc}
x+y & 2 \\
5+z & x y
\end{array}\right]=\left[\begin{array}{cc}
6 & 2 \\
5 & 8
\end{array}\right]\)
Solution:
Given \(\left[\begin{array}{cc}
x+y & 2 \\
5+z & x y
\end{array}\right]=\left[\begin{array}{cc}
6 & 2 \\
5 & 8
\end{array}\right]\)
As the two matrices are equal, equating the corresponding elements,
we get x + y = 6; xy = 8;5 + z = 5 ⇒ z = 0
Now (x – y)² = (x + y)² – 4xy ⇒ (x – y )² = 6² – 4(8) = 36 – 32 = 4
⇒ (x – y)² = 4 ⇒ (x – y) = ±2 ⇒ x – y = 2 or x – y = -2
When x – y = 2 and x + y = 6 we get x = 4, y = 2
When x – y = – 2 and x + y = 6 we get x = 2, y = 4
Thus, x = 4, y = 2, z = 0 or x = 2, y = 4, z = 0

Question 13.
Find the values of x,y and z from \(\left[\begin{array}{c}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{c}
9 \\
5 \\
7
\end{array}\right]\)
Solution:
Given \(\left[\begin{array}{c}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{c}
9 \\
5 \\
7
\end{array}\right]\)
As the two matrices are equal, equating the corresponding elements, we get
x + y + z = 9 ………. (1); x + z = 5 …………..(2); y + z = 7 …………… (3)
From (1) and (2), we have y + 5 = 9 ⇒ y = 4
From (3), we have 4 + z = 7 ⇒ z = 3
Now x + z = 5 ⇒ x + 3 = 5 ⇒ x = 2.
Thus, x = 2, y = 4, z = 3

Question 14.
Find the values of a,b, c and d from the equation \(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\)
Solution:
Given \(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\)
As the two matrices are equal, equating the corresponding elements, we get
a – b = – 1 …………… (1)
2a – b = 0 …………. (2)
2a + c = 5 ……….. (3)
3c + d = 13 …………… (4)
From (2), b = 2a
Putting this value in (1), ⇒ a – 2a = – 1 ⇒ a = 1 Hence, b = 2
Putting a = 1 in (3) ⇒ 2(1) + c = 5 ⇒ c = 3
Putting c = 3 in(4) ⇒ 3(3) + d = 13 ⇒ d = 4
Thus, a = 1, b = 2, c = 3 and d = 4