NCERT Solutions for Class 12 Maths Chapter 3 Matrices

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Exercise 3.1

Question 1:
In the matrix
\[ A = \begin{bmatrix}
2 & 5 & 19 & -7 \\
35 & -2 & \frac{5}{2} & 12\\
\sqrt{3} & 1 & -5 & 17
\end{bmatrix},
\]write:
(i) The order of the matrix, (ii) The number of elements, (iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃.
Answer
(i) Order of the matrix is 3 x 4.
(ii) The number of elements in the matrix A is 3 x 4 = 12.
(iii) a ₁₃= 19; a ₂₁= 35; a ₃₃= -5; a ₂₄= 12; a ₂₃= \(\frac{5}{2}\)

Question 2:
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Answer
We know that if a matrix is of the order m × n, it has mn elements.
The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and (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, and 6 × 4
(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.
Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.

Question 3:
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Answer
We know that if a matrix is of the order m × n, it has mn elements.
The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6,), and (6, 3)
Hence, the possible orders of a matrix having 18 elements are: 1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3
(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.
Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

Question 4:
Construct a 2 × 2 matrix, A = [a_ij], whose elements are given by:
\((i) \; a_{ij} = \frac{(i + j)^2}{2} \quad (ii) \; a_{ij} = \frac{i}{j} \quad (iii) \; a_{ij} = \frac{(i + 2j)^2}{2}\)
Answer
\((i) \; a_{ij} = \frac{(i + j)^2}{2} \)
\(\quad\) a₁₁ = \(\frac{(1 + 1)^2}{2} \) = 2; \(\quad\) a₁₂ = \(\frac{(1 + 2)^2}{2} \) = \(\frac{9}{2}\)
\(\quad\) a₂₁ = \(\frac{(2 + 1)^2}{2} \) = \(\frac{9}{2}\); \(\quad\) a₂₂ = \(\frac{(2 + 2)^2}{2} \) = 8
\(\quad\) \( A = \begin{bmatrix}
2 & \frac{9}{2} \\
\frac{9}{2} & 8\\
\end{bmatrix}
\)

\((ii) \; a_{ij} = \frac{i}{j} \)
\(\quad\) a₁₁ = \(\frac{1}{1} \) = 1; \(\quad\) a₁₂ = \(\frac{1}{2} \);
\(\quad\) a₂₁ = \(\frac{2}{1} \) = 2; \(\quad\) a₂₂ = \(\frac{2}{2} \) = 1
\(\quad\) \( A = \begin{bmatrix}
1 & \frac{1}{2} \\
2 & 1\\
\end{bmatrix}
\)

\((iii) \; a_{ij} = \frac{(i + 2j)^2}{2}\)
\(\quad\) a₁₁ = \(\frac{(1 + 2(1))^2}{2} \) = \(\frac{9}{2}\); \(\quad\) a₁₂ = \(\frac{(1 + 2(2))^2}{2} \) = \(\frac{25}{2}\);
\(\quad\) a₂₁ = \(\frac{(2 + 2(1))^2}{2} \) = 8; \(\quad\) a₂₂ = \(\frac{(2 + 2(2))^2}{2} \) = 18
\(\quad\) \( A = \begin{bmatrix}
\frac{9}{2} & \frac{25}{2} \\
8 & 18\\
\end{bmatrix}
\)

Question 5:
Construct a 3 × 4 matrix, whose elements are given by:
\((i)\; a_{ij} = \frac{1}{2} |-3i + j| \quad (ii) \; a_{ij} = 2i – j\)
Answer
\((i)\; a_{ij} = \frac{1}{2} |-3i + j|\)
\(\quad\) a₁₁ = \(\frac{1}{2} |-3(1) + 1|\) = 1; \(\quad\) a₁₂ = \(\frac{1}{2} |-3(1) + 2)|\) = \(\frac{1}{2}\); \(\quad\) a₁₃ = \(\frac{1}{2} |-3(1) + 3)|\) = 0; \(\quad\) a₁₄ = \(\frac{1}{2} |-3(1) + 4|\) = \(\frac{1}{2}\);
\(\quad\) a₂₁ = \(\frac{1}{2} |-3(2) + 1)|\) = \(\frac{5}{2}\); \(\quad\) a₂₂ = \(\frac{1}{2} |-3(2) + 2)|\) = 2; \(\quad\) a₂₃ = \(\frac{1}{2} |-3(2) + 3)|\) = \(\frac{3}{2}\); \(\quad\) a₂₄ = \(\frac{1}{2} |-3(2) + 4|\) = 1;
\(\quad\) a₃₁ = \(\frac{1}{2} |-3(3) + (1)|\) = 4; \(\quad\) a₃₂ = \(\frac{1}{2} |-3(3) + 2)|\) = \(\frac{7}{2}\); \(\quad\) a₃₃ = \(\frac{1}{2} |-3(3) + 3)|\) = 3; \(\quad\) a₃₄ = \(\frac{1}{2} |-3(3) + 4|\) = \(\frac{5}{2}\);
\(\quad\) \( A = \begin{bmatrix}
1 & \frac{1}{2} & 0 & \frac{1}{2} \\
\frac{5}{2} & 2 & \frac{3}{2} & 1\\
4 & \frac{7}{2} & 3 & \frac{5}{2}
\end{bmatrix}
\)

\((ii) \; a_{ij} = 2i – j\)
\(\quad\) a₁₁ = 2(1) – 1 = 2 – 1 = 1; \(\quad\) a₁₂ = 2(1) – 2 = 2 – 2 = 0; \(\quad\) a₁₃ = 2(1) – 3 = 2 – 3 = -1 \(\quad\) a₁₄ = 2(1) – 4 = 2 – 4 = -2;
\(\quad\) a₂₁ = 2(2) – 1 = 4 – 1 = 3, \(\quad\) a₂₂ = 2(2) – 2 = 4 – 2 = 2; \(\quad\) a₂₃ = 2(2) – 3 = 4 – 3 = 1; \(\quad\) a₂₄ = 2(2) – 4 = 4 – 4 = 0;
\(\quad\) a₃₁ = 2(3) – 1 = 6 – 1 = 5; \(\quad\) a₃₂ = 2(3) -2 = 6 – 2 = 4; \(\quad\) a₃₃ = 2(3) – 3 = 6 – 3 = 3; \(\quad\) a₃₄ = 2(3) – 4 = 6 – 4 = 2;
\(\quad\) \( A = \begin{bmatrix}
1 & 0& -1 & -2 \\ 3& 2 & 1 & 0\\
5& 4 & 3 & 2
\end{bmatrix}
\)

Question 6:
Find the values of x, y and z from the following equations:
\((i)\; \begin {bmatrix}
4 & 3 \\ x & 5
\end {bmatrix} = \begin {bmatrix}
y & z \\ 1 & 5
\end {bmatrix} \quad
(ii)\; \begin {bmatrix}
x + y & 2 \\ 5 + z & xy
\end {bmatrix} = \begin {bmatrix}
6 & 2 \\ 5 & 5
\end {bmatrix} \quad
(iii)\; \begin {bmatrix}
x + y + z \\ x + z \\y + z \\
\end {bmatrix}= \begin {bmatrix}
9 \\5 \\7
\end {bmatrix}
\)
Answer
\((i)\; \begin {bmatrix}
4 & 3 \\ x & 5
\end {bmatrix} = \begin {bmatrix}
y & z \\ 1 & 5
\end {bmatrix}\)
So, x = 1; y = 4; z = 3

\((ii)\; \begin {bmatrix}
x + y & 2 \\ 5 + z & xy
\end {bmatrix} = \begin {bmatrix}
6 & 2 \\ 5 & 5
\end {bmatrix}\)
So, x + y = 6 .. (1) and 5 + z = 5 \(\Rightarrow\) z = 0
and, xy = 5 …(2)
solving (1) and (2), y= 2 or 4 and x = 4 or 2
Hence, x = 2, y = 4, z =0 or x = 4, y =2, z = 0

\((iii)\; \begin {bmatrix}
x + y + z \\ x + z \\y + z \\
\end {bmatrix}= \begin {bmatrix}
9 \\5 \\7
\end {bmatrix}\)
So, x + y + z = 9 …(1)
x + z = 5 \(\Rightarrow\) x = 5 – z …(2)
and y + z = 7 \(\Rightarrow\) y = 7 – z …(3)
Putting the values of x and y in (1), 5 – z + 7 – z + z = 9 \(\Rightarrow\) 12 – z = 9 \(\Rightarrow\) z = 3
Putting the value of z in (2) & (3), x = 2 and y = 4
Hence, x = 2, y = 4 and z = 3

Question 7:
Find the value of a, b, c and d from the equation:
\[\begin {bmatrix}
a-b & 2a + c \\ 2a – b & 3c + d
\end {bmatrix}
=
\begin {bmatrix}
-1 & 5 \\ 0 & 13
\end {bmatrix}
\]
Solution
\(\begin {bmatrix} a-b & 2a + c \\ 2a – b & 3c + d \end {bmatrix}
=
\begin {bmatrix} -1 & 5 \\ 0 & 13 \end {bmatrix} \)
a – b = -1 …(1) and 2a – b = 0 …(2) & 2a + c = 5 …(3) and 3c + d = 13 …(4)
Solving (1) & (2), a = 1 and b = 2
Putting the value of a in (3), c =3
Putting the value of c in (4), d = 3

Question 8:
A = [a_ij]_{m × n} is a square matrix, if
(A) m < n (B) m > n (C) m = n (D) None of these
Solution
In a square matrix the number of column is same as the number of rows, so the options (c) is correct.

Question 9:
Which of the given values of x and y make the following pair of matrices equal
\[\begin {bmatrix} 3x + 7 & 5 \\ y + 1 & 2 – 3x \end {bmatrix},
\begin {bmatrix} 0 & y – 2 \\ 8 & 4 \end {bmatrix} \]
\((A)\; x =\frac{-1}{3} \quad (B) \;\) Not possible to find \(\quad (C)\; y = 7, x = \frac{-2}{3} \quad (D) \;x = \frac{-1}{3}, y = \frac{-2}{3} \)
Solution
\(\begin {bmatrix} 3x + 7 & 5 \\ y + 1 & 2 – 3x \end {bmatrix},
\begin {bmatrix} 0 & y – 2 \\ 8 & 4 \end {bmatrix} \)
3x + 7 = 0 \(\Rightarrow\) x = \(\frac{-7}{3}\) & y – 2 = 5 \(\Rightarrow\) y = 7
y + 1 = 8 \(\Rightarrow\) y = 7 & 2 – 3x = 4 \(\Rightarrow\) x = \(\frac {-2}{3}\)
So, the option (B) is correct.

Question 10:
The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
(A) 27 (B) 18 (C) 81 (D) 512
Solution
Total number of elements in a matrix of order 3 × 3 is 9
If each entry is 0, or 1, then total number of each element = 2
So, total permutation for 9 elements = 2⁹ = 512 (D)

NCERT Solutions for Class 12 Maths Chapter 3 Matrices

Exercise 3.2

Question 1:
Let \(A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix},
B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix},
C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} \)
Find each of the following:
(i) A + B (ii) A – B (iii) 3A – C (iv) AB (v) BA
Solution
(i) A + B = \(\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}\)
\(\quad\quad\quad\;\) = \(\begin{bmatrix} 2 + 1 & 4 + 3 \\ 3 – 2 & 2 + 5 \end{bmatrix}\) = \(\begin{bmatrix} 3 & 7\\ 1 & 7 \end{bmatrix}\)

(ii) A – B = \(\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} – \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}\)
\(\quad\quad\quad\;\) = \(\begin{bmatrix} 2 – 1 & 4 – 3 \\ 3 + 2 & 2 – 5 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 1\\ 5 & -3 \end{bmatrix}\)

(iii) 3A – C = 3\(\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} – \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}\)
\(\quad\quad\quad\;\) = \(\begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} – \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}\)
\(\quad\quad\quad\;\) = \(\begin{bmatrix} 6 + 2 & 12 – 5 \\ 9 – 3 & 6 – 4 \end{bmatrix}\) = \(\begin{bmatrix} 8 & 7\\ 6 & 2 \end{bmatrix}\)

(iv) AB = \(\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}\)
\(\quad\quad\quad\) = \(\begin{bmatrix} 2 – 8 & 6 + 20 \\ 3 – 4 & 9 + 10 \end{bmatrix}\) = \(\begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}\)

(v) BA = \(\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}\)
\(\quad\quad\quad\) = \(\begin{bmatrix} 2 + 9 & 4 + 6 \\ -4 + 15 & -8 + 10 \end{bmatrix}\) = \(\begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix}\)

Question 2:
Compute the following:
(i) \(\begin{bmatrix} a & b \\ -b & a \end{bmatrix}
+
\begin{bmatrix} a & b \\ b & a \end{bmatrix} \\\)
(ii) \(\begin{bmatrix} a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2 \end{bmatrix}
+
\begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix} \\ \)
(iii) \(\begin{bmatrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{bmatrix}
+
\begin{bmatrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{bmatrix} \\\)
(iv) \(\begin{bmatrix} \cos^{2} x & \sin^{2} x \\ \sin^{2} x & \cos^{2} x \end{bmatrix}
+
\begin{bmatrix} \sin^{2} x & \cos^{2} x \\ \cos^{2} x & \sin^{2} x \end{bmatrix} \)
Solution
(i) \(\begin{bmatrix} a & b \\ -b & a \end{bmatrix}
+
\begin{bmatrix} a & b \\ b & a \end{bmatrix} \) = \(\begin{bmatrix} a+ a & b + b \\ -b + b & a + a \end{bmatrix}\) = \(\begin{bmatrix} 2a & 2b \\ 0 & 2a \end{bmatrix}\)

(ii) \(\begin{bmatrix} a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2 \end{bmatrix}
+
\begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix} \) = \(\begin{bmatrix} a^2 + b^2 + 2ab& b^2 + c^2 +2bc\\ a^2 + c^2 – 2ac & a^2 + b^2 – 2ab \end{bmatrix}\) = \(\begin{bmatrix} (a + b)^2 & (b + c)^2 \\ (a – c)^2 & (a – b)^2 \end{bmatrix}\)

(iii) \(\begin{bmatrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{bmatrix}
+
\begin{bmatrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{bmatrix} \)
= \(\begin{bmatrix} -1 + 12 & 4 + 7& -6 + 6\\ 8 + 8& 5 + 0& 16 + 5\\ 2 + 3& 8 + 2 & 5 + 4 \end{bmatrix}\)
= \(\begin{bmatrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix}\)

(iv) \(\begin{bmatrix} \cos^{2} x & \sin^{2} x \\ \sin^{2} x & \cos^{2} x \end{bmatrix}
+
\begin{bmatrix} \sin^{2} x & \cos^{2} x \\ \cos^{2} x & \sin^{2} x \end{bmatrix} \)
= \(\begin{bmatrix} \cos^{2} x + \sin^{2} x & \sin^{2} x + \cos^{2} x\\ \sin^{2} x + \cos^{2}x & \cos^{2} x + \sin^{2} x\end{bmatrix}\)
= \(\begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix}\)

Question 3:
Compute the indicated products.
(i) \(\begin{bmatrix} a & b \\ -b & a \end{bmatrix}
\begin{bmatrix} a & -b \\ b & a \end{bmatrix} \quad\quad\) (ii) \(\begin{bmatrix} 1 \\ 2\\ 3\\ \end{bmatrix}
\begin{bmatrix} 2 & 3 & 4 \\ \end{bmatrix} \\\)
(iii) \(\begin{bmatrix} 1 &-2 \\ 2 & 3 \end{bmatrix}
\begin{bmatrix} 1 & 2&3 \\ 2 & 3&1 \end{bmatrix} \quad\quad\) (iv) \(\begin{bmatrix} 2 & 3 &4 \\ 3 &4&5\\ 4&5&6 \end{bmatrix}
\begin{bmatrix} 1 & -3&5 \\ 0 &2&4\\ 3&0&5 \end{bmatrix} \\\)
(v) \(\begin{bmatrix} 2 &1 \\ 3 &2\\ -1&1 \end{bmatrix}
\begin{bmatrix} 1 & 0&1 \\ -1 &2&1 \end{bmatrix} \quad \quad\) (vi) \(\begin{bmatrix} 3 & -1 &3\\ -1 &0&2 \end{bmatrix}
\begin{bmatrix} 2 &-3 \\ 1 &0\\ 3&1 \end{bmatrix} \)
Solution
(i) \(\begin{bmatrix} a & b \\ -b & a \end{bmatrix}
\begin{bmatrix} a & -b \\ b & a \end{bmatrix} \)
= \(\begin{bmatrix} a×a+b×b &a×(- b)+b×a \\ -b×a+a×b & -b×(-b)+ a×a \end{bmatrix}\)
= \(\begin{bmatrix} a^2+b^2 & 0 \\ 0 & b^2 + a^2 \end{bmatrix}\)

(ii) \(\begin{bmatrix} 1 \\ 2\\ 3\\ \end{bmatrix}
\begin{bmatrix} 2 & 3 & 4 \\ \end{bmatrix} \)
= \(\begin{bmatrix} 1×2&1×3&1×4\\ 2×2& 2×3&2×4\\3×2 &3×3&3×4 \end{bmatrix}\)
= \(\begin{bmatrix} 2&3&4\\ 4& 6&8\\6 &9&12 \end{bmatrix}\)

(iii) \(\begin{bmatrix} 1 &-2 \\ 2 & 3 \end{bmatrix}
\begin{bmatrix} 1 & 2&3 \\ 2 & 3&1 \end{bmatrix} \)
= \(\begin{bmatrix} 1×1+(-2)×2 &1×2+ (-2)×3&1×3+(-2)×1\\2×1+3×2 &2×2+3×3&2×3+3×1 \end{bmatrix}\)
= \(\begin{bmatrix} -3&-4&1\\ 8& 13&9 \end{bmatrix}\)

(iv) \(\begin{bmatrix} 2 & 3 &4 \\ 3 &4&5\\ 4&5&6 \end{bmatrix}
\begin{bmatrix} 1 & -3&5 \\ 0 &2&4\\ 3&0&5 \end{bmatrix} \)
= \(\begin{bmatrix} 2×1 + 3 × 0 + 4 × 3&2×(-3) + 3 × 2 + 4 × 0&2×5 + 3 × 4 + 4 × 5\\ 3×1 + 4 × 0 + 5 × 3 & 3×(-3) + 4 × 2 + 5 × 0 & 3×5 + 4 × 4 + 5 × 5\\4×1 + 5 × 0 + 6 × 3 &4×(-3)+5×2+6×0&4×5+5×4+6×5 \end{bmatrix}\)
= \(\begin{bmatrix} 14&0&42\\ 18&-1&56\\22 &-2&70 \end{bmatrix}\)

(v) \(\begin{bmatrix} 2 &1 \\ 3 &2\\ -1&1 \end{bmatrix}
\begin{bmatrix} 1 & 0&1 \\ -1 &2&1 \end{bmatrix} \)
= \(\begin{bmatrix} 2×1+1×(-1)&2×0+1×2&2×1+1×1\\3×1+2×(-1)&3×0+2×2&3×1+2×1\\-1×1 +1×(-1)&-1×0+1×2&-1×1+1×1 \end{bmatrix}\)
= \(\begin{bmatrix} 1&2&3\\1&4&5\\-2&2&0 \end{bmatrix}\)

(vi) \(\begin{bmatrix} 3 & -1 &3\\ -1 &0&2 \end{bmatrix}
\begin{bmatrix} 2 &-3 \\ 1 &0\\ 3&1 \end{bmatrix} \)
= \(\begin{bmatrix} 3×2+(-1)+3×3&3×(-3)+(-1)×0+3×1\\ -1×2+0×1+2×3&-1×(-3)+0×0+2×1\end{bmatrix}\)
= \(\begin{bmatrix} 14&-6\\4&5 \end{bmatrix}\)

Question 4:
If A = \(\begin{bmatrix} 1 & 2 &-3 \\ 5 &0&2\\ 1&-1&1 \end{bmatrix}\), B = \(\begin{bmatrix}
3&-1&2 \\ 4&2&5\\ 2&0&3 \end{bmatrix}\) and C= \(\begin{bmatrix} 4&1&2 \\ 0&3&2\\ 1&-2&3 \end{bmatrix}\),
then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.
Solution
A + B =\(\begin{bmatrix} 1 & 2 &-3 \\ 5 &0&2\\ 1&-1&1 \end{bmatrix}\) + \(\begin{bmatrix} 3&-1&2 \\ 4&2&5\\ 2&0&3 \end{bmatrix}\)
= \(\begin{bmatrix} 4 & 1 &-1 \\9 &2&7\\ 3&-1&4 \end{bmatrix}\)

B – C = B = \(\begin{bmatrix} 3&-1&2 \\ 4&2&5\\ 2&0&3 \end{bmatrix}\) – \(\begin{bmatrix} 4&1&2 \\ 0&3&2\\ 1&-2&3 \end{bmatrix}\)
= \(\begin{bmatrix} -1&-2&0 \\ 4&-1&3\\ 1&2&0 \end{bmatrix}\)

LHS
= A + (B – C) = \(\begin{bmatrix} 1 & 2 &-3 \\ 5 &0&2\\ 1&-1&1 \end{bmatrix}\) + \(\begin{bmatrix} -1&-2&0 \\ 4&-1&3\\ 1&2&0 \end{bmatrix}\)
= \(\begin{bmatrix} 0&0&-3 \\9&-1&5\\ 2&1&1 \end{bmatrix}\)

RHS
= (A + B) – C = \(\begin{bmatrix} 4 & 1 &-1 \\9 &2&7\\ 3&-1&4 \end{bmatrix}\) – \(\begin{bmatrix} 4&1&2 \\ 0&3&2\\ 1&-2&3 \end{bmatrix}\)
= \(\begin{bmatrix} 0&0&-3 \\9&-1&5\\ 2&1&1 \end{bmatrix}\)
Hence, LHS = RHS

Question 5:
If A = \(\begin{bmatrix}
\frac{2}{3}&1 &\frac{5}{3} \\\frac{1}{3}&\frac{2}{3}&\frac{4}{3}\\\frac{7}{3}&2&\frac{2}{3}\end{bmatrix}\), B = \(\begin{bmatrix}\frac{2}{5}&\frac{5}{3}&1 \\\frac{1}{5}&\frac{2}{5}&\frac{4}{5}\\\frac{7}{5}&\frac{6}{5}&\frac{2}{5}
\end{bmatrix}
\), then compute 3A – 5B.
Solution
3A – 5B = 3\(\begin{bmatrix}
\frac{2}{3}&1 &\frac{5}{3} \\\frac{1}{3}&\frac{2}{3}&\frac{4}{3}\\\frac{7}{3}&2&\frac{2}{3}
\end{bmatrix}\) – 5\(\begin{bmatrix}\frac{2}{5}&\frac{5}{3}&1 \\\frac{1}{5}&\frac{2}{5}&\frac{4}{5}\\\frac{7}{5}&\frac{6}{5}&\frac{2}{5}
\end{bmatrix}\) = \(\begin{bmatrix}2&3&5\\1&2&4\\7&6&2 \end{bmatrix}\) – \(\begin{bmatrix}2&3&5 \\1&2&4\\7&6&2 \end{bmatrix}\) = \(\begin{bmatrix}0&0&0 \\0&0&0\\0&0&0 \end{bmatrix} \) = 0

Question 6:
Simplify: \(\cos \theta\) \(\begin{bmatrix}
\cos \theta &\sin \theta \\
-\sin \theta &\cos \theta
\end{bmatrix}\) + \(\sin \theta \begin{bmatrix}
\sin \theta &-\cos \theta \\
\cos \theta &\sin \theta
\end{bmatrix}
\)
Solution
\(\cos \theta\) \(\begin{bmatrix}
\cos \theta &\sin \theta \\
-\sin \theta &\cos \theta
\end{bmatrix}\) + \(\sin \theta \begin{bmatrix}
\sin \theta &-\cos \theta \\
\cos \theta &\sin \theta
\end{bmatrix}
\) = \(\begin{bmatrix}
\cos^{2} \theta &\sin \theta \cos \theta \\ -\sin \theta \cos \theta &\cos^{2} \theta
\end{bmatrix}\) + \( \begin{bmatrix}
\sin^{2} \theta &-\sin \theta \cos \theta \\ \sin \theta \cos \theta &\sin^{2} \theta
\end{bmatrix}\)
= \(\begin{bmatrix}
\cos^{2} \theta+sin^{2} \theta &\sin \theta \cos \theta – \sin \theta \cos \theta\\ -\sin \theta \cos \theta+ \sin \theta \cos \theta&\cos^{2} \theta + \sin^{2} \theta
\end{bmatrix}\) = \(\begin{bmatrix}1&0\\0&1\end{bmatrix}\) = I

Question 7:
Find X and Y, if
(i) X + Y = \(\begin{bmatrix}
7 & 0 \\2 &5
\end{bmatrix}\), and X – Y = \(\begin{bmatrix}
3&0 \\0&3\\
\end{bmatrix}\\\)
(ii) 2X + 3Y = \(\begin{bmatrix}
2 &3 \\4 &0
\end{bmatrix}\), and X – Y = \(\begin{bmatrix}
2&-2 \\-1&5\\
\end{bmatrix}\)
Solution
(i) X + Y = \(\begin{bmatrix}
7 & 0 \\2 &5
\end{bmatrix}\) …(1)
and X – Y = \(\begin{bmatrix}
3&0 \\0&3\\
\end{bmatrix}\) …(2)
Adding (1) & (2),
2X = \(\begin{bmatrix}
10&0 \\2&8\\
\end{bmatrix}\) \(\Rightarrow\) X = \(\begin{bmatrix}
5&0 \\1&4\\
\end{bmatrix}\)
Putting the value of X in (1), \(\begin{bmatrix}
5&0 \\1&4\\
\end{bmatrix}\) + Y = \(\begin{bmatrix}
7&0 \\2&5\\
\end{bmatrix}\) \(\Rightarrow\) Y = \(\begin{bmatrix}
2&0 \\1&1\\
\end{bmatrix}\)

(ii) 2X + 3Y = \(\begin{bmatrix}
2 &3 \\4 &0
\end{bmatrix}\) … (1)
and 3X – 2Y = \(\begin{bmatrix}
2&-2 \\-1&5\\
\end{bmatrix}\) …(2)
Mutltiply (1) by 3 and (2) by 2, on subtracting,
3(2X + 3Y) – 2(3X – 2Y) =3 \(\begin{bmatrix}
2&3 \\4&0\\
\end{bmatrix}\) – 2 \(\begin{bmatrix}
2&-2 \\-1&5\\
\end{bmatrix}\)
\(\Rightarrow\) 6X + 9Y – 6X – 4Y = \(\begin{bmatrix}
6&9 \\10&0\\
\end{bmatrix}\) – \(\begin{bmatrix}
4&-4 \\-2&10\\
\end{bmatrix}\)
\(\Rightarrow\) 5Y = \(\begin{bmatrix}
2&13 \\14&-10\\
\end{bmatrix}\) \(\Rightarrow\) Y = \(\begin{bmatrix}
\frac{2}{5}&\frac{13}{5} \\\frac{14}{5}&-2\\
\end{bmatrix}\)
Putting the value of X in (1), 2X + \(\begin{bmatrix}
\frac{2}{5}&\frac{13}{5} \\\frac{14}{5}&-2\\
\end{bmatrix}\) = \(\begin{bmatrix}
2&3 \\4&0\\
\end{bmatrix}\) \(\Rightarrow\) 2X= \(\begin{bmatrix}
2&0 \\1&1\\
\end{bmatrix}\)

Question 8:
Find X, if Y = \(\begin{bmatrix}
3 &2 \\
1 &4
\end{bmatrix}\) and 2X + Y = \(\begin{bmatrix}
1 &0 \\
-3 &2
\end{bmatrix}\)

Question 9:
Find x and y, if 2 \(\begin{bmatrix}
1 &3 \\0&x
\end{bmatrix}\) + \(\begin{bmatrix}
y &0 \\1 &2
\end{bmatrix}\) = \(\begin{bmatrix}
5 &6 \\1 &8
\end{bmatrix}\)
Answer
2 \(\begin{bmatrix}
1 &3 \\0&x
\end{bmatrix}\) + \(\begin{bmatrix}
y &0 \\1 &2
\end{bmatrix}\) = \(\begin{bmatrix}
5 &6 \\1 &8
\end{bmatrix}\)
\(\Rightarrow\) \(\begin{bmatrix}
2 &6 \\0&2x
\end{bmatrix}\) + \(\begin{bmatrix}
y &0 \\1 &2
\end{bmatrix}\) = \(\begin{bmatrix}
5 &6 \\1 &8
\end{bmatrix}\)
\(\Rightarrow\) \(\begin{bmatrix}
2+y &6 \\1&2x+2
\end{bmatrix}\) = \(\begin{bmatrix}
5 &6 \\1 &8
\end{bmatrix}\)
\(\Rightarrow\) 2 + y = 1 \(\Rightarrow\) y = 1 – 2 = -1
and 2x + 2 = 8 \(\Rightarrow\) 2x = 6 \(\Rightarrow\) x = 3
Hence, x = 3 and y = -1

Question 10:
Solve the equation for x, y, z and t, if 2 \(\begin{bmatrix}
x &z \\
y &t
\end{bmatrix}\) + 3 \(\begin{bmatrix}
1 &-1 \\
0 &2
\end{bmatrix}\) = 3 \(\begin{bmatrix}
3 &5 \\
4 &6
\end{bmatrix}\)

Question 11:
If x \(\begin{bmatrix}
2 \\
3
\end{bmatrix}\) + y \(\begin{bmatrix}
-1 \\
1
\end{bmatrix}\) = \(\begin{bmatrix}
10 \\
5
\end{bmatrix}\), find the values of x and y.

Question 12:
Given 3 \(\begin{bmatrix}
x &y \\
z &w
\end{bmatrix}\) = \(\begin{bmatrix}
x &6 \\
-1 &2w
\end{bmatrix}\) + \(\begin{bmatrix}
4 &x + y \\
z + w &3
\end{bmatrix}\), find the values of x, y, z and w.

Question 13:
If F(x) = \(\begin{bmatrix}
\cos x &-\sin x & 0\\
\sin x &\cos x & 0\\
0 & 0&1
\end{bmatrix}\), show that F(x) F(y) = F(x + y)

Question 14:
Show that
(i) \(\begin{bmatrix}
5 &-1 \\
6 &7
\end{bmatrix}\) \(\begin{bmatrix}
2 &1 \\
3 &4
\end{bmatrix}\) \(\neq\) \(\begin{bmatrix}
2 &1 \\
3 &4
\end{bmatrix}\) \(\begin{bmatrix}
5 &-1 \\
6 &7
\end{bmatrix}\)

(ii) \(\begin{bmatrix}
1 &2&3 \\
0 &1&0\\
1&1&0
\end{bmatrix}\) \(\begin{bmatrix}
-1 &1&0 \\
0 &-1&1 \\
2&3&4
\end{bmatrix}\) \(\neq\) \(\begin{bmatrix}
-1 &1&0 \\
0 &-1&1\\
2&3&4
\end{bmatrix}\) \(\begin{bmatrix}
1 &2&3\\
0 &1&0\\
1&1&0
\end{bmatrix}\)

Question 15:
Find A² – 5A + 6I, if A = \(\begin{bmatrix}
2 &0&1 \\
2 &1&3\\
1&-1&0
\end{bmatrix}\)

Question 16:
If A = \(\begin{bmatrix}
1&0&2 \\
0 &2&1\\
2&0&3
\end{bmatrix}\), prove that A³ – 6A² +7A + 2I = 0.

Question 17:
If A = \(\begin{bmatrix}
3 &-2 \\
4 &-2
\end{bmatrix}\) and I = \(\begin{bmatrix}
1 &0 \\
0 &1
\end{bmatrix}\), find k so that A² = kA – 2I.

Question 18:
If A = \(\begin{bmatrix}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2}&0
\end{bmatrix}\) and I is the identity matrix of order 2, show that I + A = (I – A) \(\begin{bmatrix}
\cos \alpha & -\sin \alpha \\
\sin \alpha &\cos \alpha
\end{bmatrix}\)

Question 19:
A trust fund has ₹ 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:
(a) ₹ 1800 (b) ₹ 2000

Question 20:
The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹80, ₹60 and ₹ 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct answer in Exercises 21 and 22.

Question 21:
The restriction on n, k and p so that PY + WY will be defined are:
(A) k = 3 , p = n \(\quad\) (B) k is arbitrary, p = 2 \(\quad\) (C) p is arbitrary, k = 3 \(\quad\) (D) k = 2, p = 3

Question 22:
If n = p, then the order of the matrix 7X – 5Z is:
(A) p × 2 \(\quad\) (B) 2 × n \(\quad\) (C) n × 3 \(\quad\) (D) p × n

Exercise 3.3

Question 1:
Find the transpose of each of the following matrices:
(i) \(\begin{bmatrix}
5\\
\frac{1}{2}\\
-1
\end{bmatrix}\) , \(\quad\) \(\quad\) (ii) \(\begin{bmatrix}
1&-1\\
2&3\\
\end{bmatrix}\), \(\quad\) \(\quad\) (iii) \(\begin{bmatrix}
-1&5&6\\
\sqrt{3}&5&6\\
2&3&-1
\end{bmatrix}\)

Question 2:
If A = \(\begin{bmatrix}
-1&2&3\\
5&7&9\\
-2&1&1
\end{bmatrix}\) and B = \(\begin{bmatrix}
-4&1&-5\\
1&2&0\\
1&3&1
\end{bmatrix}\), then verify that (i) (A + B)’ = A’ + B’, \(\quad\) (ii) (A – B)’ = A’ – B’

Question 3:
If A’ = \(\begin{bmatrix}
3&3\\
-1&2\\
0&1
\end{bmatrix}\) and B = \(\begin{bmatrix}
-1&2&1\\
1&2&3
\end{bmatrix}\), then verify that (i) (A + B)’ = A’ + B’, \(\quad\) (ii) (A – B)’ = A’ – B’

Question 4:
If A’ = \(\begin{bmatrix}
-2&3\\
1&2
\end{bmatrix}\) and B = \(\begin{bmatrix}
-1&0\\
1&2
\end{bmatrix}\), then find (A + 2B)’.

Question 5:
For the matrices A and B, verify that (AB)’ = B’A’, where
(i) A = \(\begin{bmatrix}
1\\ -4\\ 3
\end{bmatrix}\), B= \(\begin{bmatrix}
-1&2&1
\end{bmatrix}\)
(ii) A = \(\begin{bmatrix}
0\\ 1\\2
\end{bmatrix}\), B= \(\begin{bmatrix}
5&5&7
\end{bmatrix}\)

Question 6:
If (i) A = \(\begin{bmatrix}
\cos \alpha & \sin \alpha\\-\sin \alpha & \cos \alpha
\end{bmatrix}\), then verify that A’A = I.
\(\quad\)(ii) A = \(\begin{bmatrix}
\sin \alpha & \cos \alpha\\-\cos \alpha & \sin \alpha
\end{bmatrix}\), then verify that A’A = I.

Question 7:
(i) Show that the matrix A = \(\begin{bmatrix}
1& -1&5\\-1&2&1& \\5&1&3
\end{bmatrix}\) is a symmetric matrix.
(ii)Show that the matrix A = \(\begin{bmatrix}
0&1&-1\\-1&0&1 \\1&-1&0
\end{bmatrix}\) is a skew symmetric matrix.

Question 8:
For the matrix a = \(\begin{bmatrix}
1&5\\6&7
\end{bmatrix}\) , verify that
(i) (A + A′) is a symmetric matrix
(ii) (A – A′) is a skew symmetric matrix

Question 9:
Find \(\frac{1}{2}\) (A + A’) and \(\frac{1}{2}\) (A – A’), when A = \(\begin{bmatrix}
0&a&b\\-a&0&c \\-b&-c&0 \end{bmatrix}\)

Question 10:
Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
(i) \(\begin{bmatrix}
3&5\\1&-1 \end{bmatrix}\) \(\quad\)(ii) \(\begin{bmatrix}
6&-2&2\\-2&3&-1 \\2&-1&3 \end{bmatrix}\)\(\quad\) (iii) \(\begin{bmatrix}
3&3&-1\\-2&-2&1 \\-4&-5&2 \end{bmatrix}\) \(\quad\)(iv) \(\begin{bmatrix}
1&5\\-1& 2 \end{bmatrix}\)

Choose the correct answer in the Exercises 11 and 12.

Question 11:
If A, B are symmetric matrices of same order, then AB – BA is a
(A) Skew symmetric matrix \(\quad\)(B) Symmetric matrix \(\quad\)(C) Zero matrix \(\quad\)(D) Identity matrix

Question 12:
If A = \(\begin{bmatrix}
\cos \alpha&-\sin \alpha\\\sin \alpha&\cos \alpha \end{bmatrix}\) , and a + A’ = I, then the value of \(\alpha\) is
(A) \(\frac{\pi}{2}\) \(\quad\) (B) \(\frac{\pi}{3}\) \(\quad\) (C) \(\pi\) \(\quad\) (D) \(\frac{3\pi}{2}\)

Exercise 3.4

Question 1:
Matrices A and B will be inverse of each other only if
(A) AB = BA \(\quad\)(B) AB = BA = 0 \(\quad\)(C) AB = 0, BA = I \(\quad\)(D) AB = BA = I

Miscellaneous Exercise on Chapter 3

Question 1:
If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.

Question 2:
Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Question 3:
Find the values of x, y, z if the matrix A = \(\begin{bmatrix}
0&2y&z\\x&y&-z \\x&-y&z \end{bmatrix}\) satisfy the equation A’A = I.

Question 4:
For what value of x: \(\begin{bmatrix}
1&2&1 \end{bmatrix}\)\(\begin{bmatrix}
1&2&0\\2&0&1 \\1&0&2 \end{bmatrix}\)\(\begin{bmatrix}
0\\2\\x\end{bmatrix}\) = 0?

Question 5:
If A = \(\begin{bmatrix}
3&1\\-1&2 \\-1&2 \end{bmatrix}\), show that A² – 5A + 7I = 0.

Question 6:
Find x, if \(\begin{bmatrix}
x&-5&-1 \end{bmatrix}\)\(\begin{bmatrix}
1&0&2\\0&2&1 \\2&0&3 \end{bmatrix}\) \(\begin{bmatrix}
x\\4 \\1 \end{bmatrix}\) = 0.

Question 7:
A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:
\(\quad\) \(\quad\) \(\quad\)Market \(\quad\) \(\quad\) \(\quad\) \(\quad\)\(\quad\) Product
\(\quad\)\(\quad\)\(\quad\) \(\quad\)I \(\quad\)\(\quad\)\(\quad\)\(\quad\)\(\quad\)10, 000 \(\quad\) \(\quad\)2,000 \(\quad\) 18,000
\(\quad\)\(\quad\)\(\quad\) \(\quad\)II \(\quad\)\(\quad\)\(\quad\)\(\quad\)\(\quad\)6,000\(\quad\)\(\quad\)\(\;\)20,000 \(\quad\)8,000
(a) If unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00, respectively, find the total revenue in each market with the help of matrix algebra.
(b) If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.

Question 8:
Find the matrix X so that X = \(\begin{bmatrix}
1&2&3\\4&5&6 \end{bmatrix}\) = \(\begin{bmatrix}
-7&-8&-9\\2&4&6 \end{bmatrix}\)

Choose the correct answer in the following questions:

Question 9:
If A = \(\begin{bmatrix}
\alpha &\beta\\\gamma&-\alpha \end{bmatrix}\) is such that A² = I, then
(A) 1 +\(\alpha^{2}\) +\(\beta\gamma\) = 0 (B) 1 -\(\alpha^{2}\) +\(\beta\gamma\) = 0
(C) 1 -\(\alpha^{2}\) -\(\beta\gamma\) = 0 (D) 1 +\(\alpha^{2}\) -\(\beta\gamma\) = 0

Question 10:
If the matrix A is both symmetric and skew symmetric, then
(A) A is a diagonal matrix (B) A is a zero matrix (C) A is a square matrix (D) None of these

Question 11:
If A is square matrix such that A² = A, then (I + A)³ – 7 A is equal to
(A) A (B) I – A (C) I (D) 3A

NCERT Solutions for Class 12 Maths Chapter 3