Matrices — Questions and Answers
Welcome to HSLC Guru. This lesson gives complete, step-by-step answers to every question in ASSEB Class 12 Mathematics Chapter 3, Matrices (মৌলকক্ষ). It covers Exercise 3.1, Exercise 3.2, Exercise 3.3, Exercise 3.4 and the Miscellaneous Exercise on Chapter 3, with each answer fully worked out so you can follow the reasoning and prepare confidently for your examination.
Summary
A matrix is an ordered rectangular array of numbers or functions. A matrix with $m$ rows and $n$ columns is said to be of order $m \times n$, and its element in the $i$-th row and $j$-th column is written $a_{ij}$. Special matrices include the row matrix ($1 \times n$), the column matrix ($m \times 1$), the square matrix ($m = n$), the diagonal matrix ($a_{ij} = 0$ when $i \neq j$), the scalar matrix (diagonal with equal diagonal entries), the identity matrix $I$ (diagonal entries $1$, others $0$) and the zero matrix $O$. Two matrices are equal when they have the same order and equal corresponding elements.
Matrices of the same order can be added and subtracted element by element, and any matrix can be multiplied by a scalar $k$, giving $kA = [k a_{ij}]$. The product $AB$ is defined only when the number of columns of $A$ equals the number of rows of $B$; if $A$ is $m \times n$ and $B$ is $n \times p$, then $AB$ is $m \times p$ with entry $c_{ik} = \sum_{j=1}^{n} a_{ij} b_{jk}$. Matrix multiplication is associative and distributive but not commutative in general, and $AB = O$ does not force $A = O$ or $B = O$.
The transpose $A^{\prime}$ (also written $A^{T}$) interchanges rows and columns, and satisfies $(A^{\prime})^{\prime} = A$, $(kA)^{\prime} = kA^{\prime}$, $(A + B)^{\prime} = A^{\prime} + B^{\prime}$ and $(AB)^{\prime} = B^{\prime} A^{\prime}$. A square matrix is symmetric if $A^{\prime} = A$ and skew symmetric if $A^{\prime} = -A$; every square matrix can be written as the sum of a symmetric part $\tfrac{1}{2}(A + A^{\prime})$ and a skew symmetric part $\tfrac{1}{2}(A – A^{\prime})$. A square matrix $A$ is invertible if there is a matrix $B$ of the same order with $AB = BA = I$; the inverse, when it exists, is unique and $(AB)^{-1} = B^{-1} A^{-1}$.
Summary: ASSEB Class 12 Mathematics Chapter 3 Matrices explains the order and types of matrices, equality, addition, scalar multiplication, matrix multiplication, transpose, symmetric and skew symmetric matrices and invertible matrices, with complete worked answers to Exercise 3.1, Exercise 3.2, Exercise 3.3, Exercise 3.4 and the Miscellaneous Exercise for Assam Board (ASSEB) Class 12 students.
Textbook Questions and Answers
Exercise 3.1
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 $a_{13}, a_{21}, a_{33}, a_{24}, a_{23}$.
Answer: (i) The matrix has $3$ rows and $4$ columns, so its order is $3 \times 4$.
(ii) The number of elements is $3 \times 4 = 12$.
(iii) Reading $a_{ij}$ as the entry in the $i$-th row and $j$-th column: $a_{13} = 19$, $\;a_{21} = 35$, $\;a_{33} = -5$, $\;a_{24} = 12$, $\;a_{23} = \dfrac{5}{2}$.
2. If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Answer: A matrix of order $m \times n$ has $mn$ elements, so we list all pairs of natural numbers whose product is the given number.
For $24 = mn$, the ordered pairs $(m, n)$ are $(1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), (6, 4)$. Hence the possible orders are $1 \times 24,\; 24 \times 1,\; 2 \times 12,\; 12 \times 2,\; 3 \times 8,\; 8 \times 3,\; 4 \times 6,\; 6 \times 4$ — that is $8$ orders.
Since $13$ is prime, $13 = mn$ only for $(1, 13)$ and $(13, 1)$. Hence the possible orders are $1 \times 13$ and $13 \times 1$ — that is $2$ orders.
3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Answer: For $18 = mn$, the ordered pairs $(m, n)$ are $(1, 18), (18, 1), (2, 9), (9, 2), (3, 6), (6, 3)$. Hence the possible orders are $1 \times 18,\; 18 \times 1,\; 2 \times 9,\; 9 \times 2,\; 3 \times 6,\; 6 \times 3$ — that is $6$ orders.
Since $5$ is prime, the only ordered pairs are $(1, 5)$ and $(5, 1)$, giving the orders $1 \times 5$ and $5 \times 1$ — that is $2$ orders.
4. Construct a $2 \times 2$ matrix, $A = [a_{ij}]$, whose elements are given by:
(i) $a_{ij} = \dfrac{(i + j)^2}{2}$
Answer: $a_{11} = \dfrac{(1+1)^2}{2} = \dfrac{4}{2} = 2$, $\;a_{12} = \dfrac{(1+2)^2}{2} = \dfrac{9}{2}$, $\;a_{21} = \dfrac{(2+1)^2}{2} = \dfrac{9}{2}$, $\;a_{22} = \dfrac{(2+2)^2}{2} = \dfrac{16}{2} = 8$.
$$A = \begin{bmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix}$$
(ii) $a_{ij} = \dfrac{i}{j}$
Answer: $a_{11} = \dfrac{1}{1} = 1$, $\;a_{12} = \dfrac{1}{2}$, $\;a_{21} = \dfrac{2}{1} = 2$, $\;a_{22} = \dfrac{2}{2} = 1$.
$$A = \begin{bmatrix} 1 & \frac{1}{2} \\ 2 & 1 \end{bmatrix}$$
(iii) $a_{ij} = \dfrac{(i + 2j)^2}{2}$
Answer: $a_{11} = \dfrac{(1+2)^2}{2} = \dfrac{9}{2}$, $\;a_{12} = \dfrac{(1+4)^2}{2} = \dfrac{25}{2}$, $\;a_{21} = \dfrac{(2+2)^2}{2} = \dfrac{16}{2} = 8$, $\;a_{22} = \dfrac{(2+4)^2}{2} = \dfrac{36}{2} = 18$.
$$A = \begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix}$$
5. Construct a $3 \times 4$ matrix, whose elements are given by:
(i) $a_{ij} = \dfrac{1}{2}\,|-3i + j|$
Answer: Computing $\dfrac{1}{2}|-3i + j|$ for $i = 1, 2, 3$ and $j = 1, 2, 3, 4$:
Row $1$: $\tfrac{1}{2}|-3+1| = 1$, $\;\tfrac{1}{2}|-3+2| = \tfrac{1}{2}$, $\;\tfrac{1}{2}|-3+3| = 0$, $\;\tfrac{1}{2}|-3+4| = \tfrac{1}{2}$. Row $2$: $\tfrac{1}{2}|-6+1| = \tfrac{5}{2}$, $\;\tfrac{1}{2}|-6+2| = 2$, $\;\tfrac{1}{2}|-6+3| = \tfrac{3}{2}$, $\;\tfrac{1}{2}|-6+4| = 1$. Row $3$: $\tfrac{1}{2}|-9+1| = 4$, $\;\tfrac{1}{2}|-9+2| = \tfrac{7}{2}$, $\;\tfrac{1}{2}|-9+3| = 3$, $\;\tfrac{1}{2}|-9+4| = \tfrac{5}{2}$.
$$A = \begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix}$$
(ii) $a_{ij} = 2i – j$
Answer: Computing $2i – j$: Row $1$: $1, 0, -1, -2$; Row $2$: $3, 2, 1, 0$; Row $3$: $5, 4, 3, 2$.
$$A = \begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}$$
6. Find the values of $x, y$ and $z$ from the following equations:
(i) $\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}$
Answer: Equal matrices have equal corresponding elements. Comparing entries: $y = 4$, $\;z = 3$, $\;x = 1$. Hence $x = 1,\; y = 4,\; z = 3$.
(ii) $\begin{bmatrix} x + y & 2 \\ 5 + z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$
Answer: Comparing corresponding elements: $x + y = 6$, $\;5 + z = 5 \Rightarrow z = 0$, $\;xy = 8$. From $x + y = 6$ and $xy = 8$, $x$ and $y$ are the roots of $t^2 – 6t + 8 = 0$, i.e. $(t – 2)(t – 4) = 0$, so $t = 2$ or $t = 4$. Hence $x = 2, y = 4$ (or $x = 4, y = 2$) and $z = 0$.
(iii) $\begin{bmatrix} x + y + z \\ x + z \\ y + z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}$
Answer: Comparing elements: $x + y + z = 9$, $\;x + z = 5$, $\;y + z = 7$. Subtracting the second from the first, $y = 9 – 5 = 4$. Then from $y + z = 7$, $z = 7 – 4 = 3$, and from $x + z = 5$, $x = 5 – 3 = 2$. Hence $x = 2,\; y = 4,\; z = 3$.
7. Find the value of $a, b, c$ and $d$ from the equation:
$$\begin{bmatrix} a – b & 2a + c \\ 2a – b & 3c + d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$$
Answer: Equating corresponding elements gives $a – b = -1$, $\;2a + c = 5$, $\;2a – b = 0$, $\;3c + d = 13$. From $2a – b = 0$, $b = 2a$; substituting into $a – b = -1$ gives $a – 2a = -1$, so $-a = -1$, i.e. $a = 1$ and $b = 2$. Then $2a + c = 5 \Rightarrow 2 + c = 5 \Rightarrow c = 3$, and $3c + d = 13 \Rightarrow 9 + d = 13 \Rightarrow d = 4$. Hence $a = 1,\; b = 2,\; c = 3,\; d = 4$.
8. $A = [a_{ij}]_{m \times n}$ is a square matrix, if
(A) $m < n$ (B) $m > n$ (C) $m = n$ (D) None of these
Answer: (C) $m = n$. A matrix is square exactly when the number of rows equals the number of columns.
9. Which of the given values of $x$ and $y$ make the following pair of matrices equal
$$\begin{bmatrix} 3x + 7 & 5 \\ y + 1 & 2 – 3x \end{bmatrix}, \quad \begin{bmatrix} 0 & y – 2 \\ 8 & 4 \end{bmatrix}$$
(A) $x = -\dfrac{1}{3},\; y = 7$ (B) Not possible to find (C) $y = 7,\; x = -\dfrac{2}{3}$ (D) $x = -\dfrac{1}{3},\; y = -\dfrac{2}{3}$
Answer: (B) Not possible to find. Equating corresponding elements: $3x + 7 = 0$ gives $x = -\dfrac{7}{3}$, while $2 – 3x = 4$ gives $x = -\dfrac{2}{3}$. These two values of $x$ are different, so no single $x$ can make the matrices equal. Hence it is not possible to find such values.
10. The number of all possible matrices of order $3 \times 3$ with each entry $0$ or $1$ is:
(A) 27 (B) 18 (C) 81 (D) 512
Answer: (D) 512. A $3 \times 3$ matrix has $9$ entries, and each entry can be chosen in $2$ ways ($0$ or $1$). By the multiplication principle the number of such matrices is $2^9 = 512$.
Exercise 3.2
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) $A + B$
Answer: Adding corresponding elements, $$A + B = \begin{bmatrix} 2+1 & 4+3 \\ 3-2 & 2+5 \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix}$$
(ii) $A – B$
Answer: $$A – B = \begin{bmatrix} 2-1 & 4-3 \\ 3-(-2) & 2-5 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix}$$
(iii) $3A – C$
Answer: $3A = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}$, so $$3A – C = \begin{bmatrix} 6-(-2) & 12-5 \\ 9-3 & 6-4 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}$$
(iv) $AB$
Answer: Multiplying rows of $A$ by columns of $B$, $$AB = \begin{bmatrix} 2(1)+4(-2) & 2(3)+4(5) \\ 3(1)+2(-2) & 3(3)+2(5) \end{bmatrix} = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}$$
(v) $BA$
Answer: $$BA = \begin{bmatrix} 1(2)+3(3) & 1(4)+3(2) \\ -2(2)+5(3) & -2(4)+5(2) \end{bmatrix} = \begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix}$$ Note that $AB \neq BA$.
2. Compute the following:
(i) $\begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix}$
Answer: $$\begin{bmatrix} a+a & b+b \\ -b+b & a+a \end{bmatrix} = \begin{bmatrix} 2a & 2b \\ 0 & 2a \end{bmatrix}$$
(ii) $\begin{bmatrix} a^2 + b^2 & b^2 + c^2 \\ a^2 + c^2 & a^2 + b^2 \end{bmatrix} + \begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix}$
Answer: Adding corresponding elements and using the identities $a^2 + b^2 + 2ab = (a+b)^2$, etc., $$\begin{bmatrix} a^2+b^2+2ab & b^2+c^2+2bc \\ a^2+c^2-2ac & a^2+b^2-2ab \end{bmatrix} = \begin{bmatrix} (a+b)^2 & (b+c)^2 \\ (a-c)^2 & (a-b)^2 \end{bmatrix}$$
(iii) $\begin{bmatrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{bmatrix}$
Answer: $$\begin{bmatrix} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{bmatrix} = \begin{bmatrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix}$$
(iv) $\begin{bmatrix} \cos^2 x & \sin^2 x \\ \sin^2 x & \cos^2 x \end{bmatrix} + \begin{bmatrix} \sin^2 x & \cos^2 x \\ \cos^2 x & \sin^2 x \end{bmatrix}$
Answer: Each entry becomes $\sin^2 x + \cos^2 x = 1$, so $$\begin{bmatrix} \cos^2 x + \sin^2 x & \sin^2 x + \cos^2 x \\ \sin^2 x + \cos^2 x & \cos^2 x + \sin^2 x \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$
3. Compute the indicated products.
(i) $\begin{bmatrix} a & b \\ -b & a \end{bmatrix}\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$
Answer: $$\begin{bmatrix} a(a)+b(b) & a(-b)+b(a) \\ -b(a)+a(b) & -b(-b)+a(a) \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & 0 \\ 0 & a^2 + b^2 \end{bmatrix}$$
(ii) $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \end{bmatrix}$
Answer: A $3 \times 1$ matrix times a $1 \times 3$ matrix gives a $3 \times 3$ matrix: $$\begin{bmatrix} 1(2) & 1(3) & 1(4) \\ 2(2) & 2(3) & 2(4) \\ 3(2) & 3(3) & 3(4) \end{bmatrix} = \begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}$$
(iii) $\begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}$
Answer: A $2 \times 2$ times a $2 \times 3$ gives a $2 \times 3$ matrix: $$\begin{bmatrix} 1(1)-2(2) & 1(2)-2(3) & 1(3)-2(1) \\ 2(1)+3(2) & 2(2)+3(3) & 2(3)+3(1) \end{bmatrix} = \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix}$$
(iv) $\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}\begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix}$
Answer: Taking each row of the first matrix against each column of the second, $$\begin{bmatrix} 2+0+12 & -6+6+0 & 10+12+20 \\ 3+0+15 & -9+8+0 & 15+16+25 \\ 4+0+18 & -12+10+0 & 20+20+30 \end{bmatrix} = \begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix}$$
(v) $\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix}$
Answer: A $3 \times 2$ times a $2 \times 3$ gives a $3 \times 3$ matrix: $$\begin{bmatrix} 2-1 & 0+2 & 2+1 \\ 3-2 & 0+4 & 3+2 \\ -1-1 & 0+2 & -1+1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{bmatrix}$$
(vi) $\begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix}$
Answer: A $2 \times 3$ times a $3 \times 2$ gives a $2 \times 2$ matrix: $$\begin{bmatrix} 6-1+9 & -9+0+3 \\ -2+0+6 & 3+0+2 \end{bmatrix} = \begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}$$
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 $A + (B – C) = (A + B) – C$.
Answer: $$A + B = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix}, \quad B – C = \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix}$$
Now $$A + (B – C) = \begin{bmatrix} 1-1 & 2-2 & -3+0 \\ 5+4 & 0-1 & 2+3 \\ 1+1 & -1+2 & 1+0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}$$
and $$(A + B) – C = \begin{bmatrix} 4-4 & 1-1 & -1-2 \\ 9-0 & 2-3 & 7-2 \\ 3-1 & -1+2 & 4-3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}$$ Since the two results are identical, $A + (B – C) = (A + B) – C$ is verified.
5. If $A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ and $B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$, then compute $3A – 5B$.
Answer: Multiplying clears the denominators: $$3A = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}, \quad 5B = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}$$
Since $3A = 5B$, $$3A – 5B = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$$
6. Simplify $\cos\theta\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix}$.
Answer: Multiplying each matrix by its scalar and adding, $$\begin{bmatrix} \cos^2\theta & \sin\theta\cos\theta \\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix} + \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix}$$
The off-diagonal terms cancel and the diagonal terms give $\cos^2\theta + \sin^2\theta = 1$: $$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$
7. Find $X$ and $Y$, if
(i) $X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}$ and $X – Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$
Answer: Adding the two equations, $2X = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix}$, so $$X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$$ Subtracting, $2Y = \begin{bmatrix} 4 & 0 \\ 2 & 2 \end{bmatrix}$, so $$Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$$
(ii) $2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}$ and $3X + 2Y = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}$
Answer: Call the two right-hand matrices $P$ and $Q$. Then $3Q – 2P = 5X$: $$5X = 3\begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix} – 2\begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} = \begin{bmatrix} 2 & -12 \\ -11 & 15 \end{bmatrix}$$ so $X = \begin{bmatrix} \frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3 \end{bmatrix}$. Also $3P – 2Q = 5Y$: $$5Y = 3\begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} – 2\begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix} = \begin{bmatrix} 2 & 13 \\ 14 & -10 \end{bmatrix}$$ so $Y = \begin{bmatrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix}$.
8. Find $X$, if $Y = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$ and $2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$.
Answer: From $2X = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} – \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix}$, so $$X = \frac{1}{2}\begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix}$$
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: The left side is $\begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix}$. Equating with the right side: $2 + y = 5 \Rightarrow y = 3$, and $2x + 2 = 8 \Rightarrow 2x = 6 \Rightarrow x = 3$. Hence $x = 3,\; y = 3$.
10. Solve the equation for $x, y, z$ and $t$, if $2\begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$.
Answer: The equation becomes $\begin{bmatrix} 2x + 3 & 2z – 3 \\ 2y & 2t + 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$. Comparing entries: $2x + 3 = 9 \Rightarrow x = 3$; $\;2z – 3 = 15 \Rightarrow z = 9$; $\;2y = 12 \Rightarrow y = 6$; $\;2t + 6 = 18 \Rightarrow t = 6$. Hence $x = 3,\; y = 6,\; z = 9,\; t = 6$.
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 $y$.
Answer: The left side is $\begin{bmatrix} 2x – y \\ 3x + y \end{bmatrix}$, giving the equations $2x – y = 10$ and $3x + y = 5$. Adding them, $5x = 15$, so $x = 3$. Then $y = 2x – 10 = 6 – 10 = -4$. Hence $x = 3,\; y = -4$.
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 $w$.
Answer: The equation is $\begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix} = \begin{bmatrix} x + 4 & 6 + x + y \\ -1 + z + w & 2w + 3 \end{bmatrix}$. Comparing entries:
$3x = x + 4 \Rightarrow 2x = 4 \Rightarrow x = 2$. $\;3w = 2w + 3 \Rightarrow w = 3$. $\;3y = 6 + x + y \Rightarrow 2y = 6 + 2 = 8 \Rightarrow y = 4$. $\;3z = -1 + z + w \Rightarrow 2z = -1 + 3 = 2 \Rightarrow z = 1$. Hence $x = 2,\; y = 4,\; z = 1,\; w = 3$.
13. If $F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$, show that $F(x)\,F(y) = F(x + y)$.
Answer: Multiplying the two matrices, $$F(x)F(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
The $(1,1)$ entry is $\cos x \cos y – \sin x \sin y = \cos(x + y)$; the $(1,2)$ entry is $-\cos x \sin y – \sin x \cos y = -\sin(x + y)$; the $(2,1)$ entry is $\sin x \cos y + \cos x \sin y = \sin(x + y)$; the $(2,2)$ entry is $\cos x \cos y – \sin x \sin y = \cos(x + y)$. Hence $$F(x)F(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} = F(x + y)$$
14. Show that
(i) $\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \neq \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}$
Answer: The left product is $$\begin{bmatrix} 10-3 & 5-4 \\ 12+21 & 6+28 \end{bmatrix} = \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$$ while the right product is $$\begin{bmatrix} 10+6 & -2+7 \\ 15+24 & -3+28 \end{bmatrix} = \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$$ The two products are different, so the matrices do not commute.
(ii) $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$
Answer: The left product is $$\begin{bmatrix} -1+0+6 & 1-2+9 & 0+2+12 \\ 0+0+0 & 0-1+0 & 0+1+0 \\ -1+0+0 & 1-1+0 & 0+1+0 \end{bmatrix} = \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}$$ while the right product is $$\begin{bmatrix} -1+0+0 & -2+1+0 & -3+0+0 \\ 0+0+1 & 0-1+1 & 0+0+0 \\ 2+0+4 & 4+3+4 & 6+0+0 \end{bmatrix} = \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix}$$ The two products are clearly different, so the matrices do not commute.
15. Find $A^2 – 5A + 6I$, if $A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$.
Answer: First $$A^2 = A \cdot A = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}$$ (for example, the $(2,1)$ entry is $2(2)+1(2)+3(1) = 9$). Then $5A = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix}$ and $6I = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$. Therefore $$A^2 – 5A + 6I = \begin{bmatrix} 5-10+6 & -1 & 2-5 \\ 9-10 & -2-5+6 & 5-15 \\ -5 & -1+5 & -2+6 \end{bmatrix} = \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$$
16. If $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$, prove that $A^3 – 6A^2 + 7A + 2I = 0$.
Answer: Computing step by step, $$A^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}, \quad A^3 = A^2 \cdot A = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}$$
Then $6A^2 = \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix}$, $\;7A = \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{bmatrix}$ and $2I = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$. Adding $A^3 – 6A^2 + 7A + 2I$ entry by entry (for instance the $(1,1)$ entry is $21 – 30 + 7 + 2 = 0$ and the $(3,3)$ entry is $55 – 78 + 21 + 2 = 0$), every entry is zero: $$A^3 – 6A^2 + 7A + 2I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$$
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 $A^2 = kA – 2I$.
Answer: $$A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$$ Also $kA – 2I = \begin{bmatrix} 3k – 2 & -2k \\ 4k & -2k – 2 \end{bmatrix}$. Comparing the $(1,1)$ entries, $3k – 2 = 1 \Rightarrow k = 1$; every other entry then agrees (e.g. $-2k = -2$, $4k = 4$, $-2k – 2 = -4$). Hence $k = 1$.
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 $I + A = (I – A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$.
Answer: Let $t = \tan\dfrac{\alpha}{2}$. Then $I + A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$ and $I – A = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$. Using the half-angle identities $\cos\alpha = \dfrac{1 – t^2}{1 + t^2}$ and $\sin\alpha = \dfrac{2t}{1 + t^2}$, compute the right-hand side:
$$(I – A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} \cos\alpha + t\sin\alpha & -\sin\alpha + t\cos\alpha \\ -t\cos\alpha + \sin\alpha & t\sin\alpha + \cos\alpha \end{bmatrix}$$
Now $\cos\alpha + t\sin\alpha = \dfrac{1 – t^2 + 2t^2}{1 + t^2} = \dfrac{1 + t^2}{1 + t^2} = 1$, and $-\sin\alpha + t\cos\alpha = \dfrac{-2t + t(1 – t^2)}{1 + t^2} = \dfrac{-t – t^3}{1 + t^2} = -t$. Similarly $-t\cos\alpha + \sin\alpha = t$ and $t\sin\alpha + \cos\alpha = 1$. Hence the right-hand side equals $\begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = I + A$, which proves the result.
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.
Answer: Let ₹$x$ be invested in the first bond and ₹$(30000 – x)$ in the second. The annual interest is obtained by the matrix product $$\begin{bmatrix} x & 30000 – x \end{bmatrix}\begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix} = \frac{5x}{100} + \frac{7(30000 – x)}{100}$$
(a) Setting the interest equal to $1800$: $\dfrac{5x + 7(30000 – x)}{100} = 1800$, i.e. $5x + 210000 – 7x = 180000$, so $-2x = -30000$ and $x = 15000$. Thus ₹$15000$ should be invested in each of the two bonds.
(b) Setting the interest equal to $2000$: $5x + 210000 – 7x = 200000$, so $-2x = -10000$ and $x = 5000$. Thus ₹$5000$ should be invested in the first bond and ₹$25000$ in the second.
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.
Answer: The numbers of books are $10 \times 12 = 120$ chemistry, $8 \times 12 = 96$ physics and $10 \times 12 = 120$ economics. Writing the quantities as a row matrix and the prices as a column matrix, $$\begin{bmatrix} 120 & 96 & 120 \end{bmatrix}\begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix} = 120(80) + 96(60) + 120(40)$$ $$= 9600 + 5760 + 4800 = 20160$$ So the total amount received is ₹$20160$.
Assume $X, Y, Z, W$ and $P$ are matrices of order $2 \times n$, $3 \times k$, $2 \times p$, $n \times 3$ and $p \times k$, respectively. Choose the correct answer in Exercises 21 and 22.
21. The restriction on $n, k$ and $p$ so that $PY + WY$ will be defined are:
(A) $k = 3, p = n$ (B) $k$ is arbitrary, $p = 2$ (C) $p$ is arbitrary, $k = 3$ (D) $k = 2, p = 3$
Answer: (A) $k = 3, p = n$. Here $P$ is $p \times k$ and $Y$ is $3 \times k$; for $PY$ to be defined the number of columns of $P$ must equal the number of rows of $Y$, so $k = 3$, and then $PY$ is $p \times k$. Also $W$ is $n \times 3$ and $Y$ is $3 \times k$, so $WY$ is $n \times k$. For the sum $PY + WY$ the two products must have the same order, which needs $p = n$. Hence $k = 3$ and $p = n$.
22. If $n = p$, then the order of the matrix $7X – 5Z$ is:
(A) $p \times 2$ (B) $2 \times n$ (C) $n \times 3$ (D) $p \times n$
Answer: (B) $2 \times n$. $X$ is $2 \times n$ and $Z$ is $2 \times p$. When $n = p$, both $7X$ and $5Z$ are of order $2 \times n$, so their difference $7X – 5Z$ is also of order $2 \times n$.
Exercise 3.3
1. Find the transpose of each of the following matrices:
(i) $\begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix}$
Answer: The transpose of a $3 \times 1$ column matrix is a $1 \times 3$ row matrix: $$\begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix}^{\prime} = \begin{bmatrix} 5 & \frac{1}{2} & -1 \end{bmatrix}$$
(ii) $\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$
Answer: Interchanging rows and columns, $$\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}^{\prime} = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$$
(iii) $\begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix}$
Answer: Each row becomes a column: $$\begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix}^{\prime} = \begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix}$$
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) $(A + B)^{\prime} = A^{\prime} + B^{\prime}$, (ii) $(A – B)^{\prime} = A^{\prime} – B^{\prime}$.
Answer: First, $A^{\prime} = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}$ and $B^{\prime} = \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix}$.
(i) $A + B = \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix}$, so $(A + B)^{\prime} = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$. Also $A^{\prime} + B^{\prime} = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$. The two are equal, so $(A + B)^{\prime} = A^{\prime} + B^{\prime}$.
(ii) $A – B = \begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix}$, so $(A – B)^{\prime} = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$. Also $A^{\prime} – B^{\prime} = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$. The two are equal, so $(A – B)^{\prime} = A^{\prime} – B^{\prime}$.
3. If $A^{\prime} = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}$, then verify that (i) $(A + B)^{\prime} = A^{\prime} + B^{\prime}$, (ii) $(A – B)^{\prime} = A^{\prime} – B^{\prime}$.
Answer: Since $A^{\prime}$ is given, $A = (A^{\prime})^{\prime} = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}$, and $B^{\prime} = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}$.
(i) $A + B = \begin{bmatrix} 2 & 1 & 1 \\ 5 & 4 & 4 \end{bmatrix}$, so $(A + B)^{\prime} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$. Also $A^{\prime} + B^{\prime} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$. Hence $(A + B)^{\prime} = A^{\prime} + B^{\prime}$.
(ii) $A – B = \begin{bmatrix} 4 & -3 & -1 \\ 3 & 0 & -2 \end{bmatrix}$, so $(A – B)^{\prime} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$. Also $A^{\prime} – B^{\prime} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$. Hence $(A – B)^{\prime} = A^{\prime} – B^{\prime}$.
4. If $A^{\prime} = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$, then find $(A + 2B)^{\prime}$.
Answer: Since $A^{\prime} = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$, we have $A = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}$. Then $2B = \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$, so $A + 2B = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}$. Therefore $$(A + 2B)^{\prime} = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$$
5. For the matrices $A$ and $B$, verify that $(AB)^{\prime} = B^{\prime} A^{\prime}$, where
(i) $A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$, $B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$
Answer: $AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}\begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}$, so $(AB)^{\prime} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$.
Also $B^{\prime} A^{\prime} = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}\begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$. The two are equal, so $(AB)^{\prime} = B^{\prime} A^{\prime}$.
(ii) $A = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}$
Answer: $AB = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\begin{bmatrix} 1 & 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}$, so $(AB)^{\prime} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$.
Also $B^{\prime} A^{\prime} = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}\begin{bmatrix} 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$. The two are equal, so $(AB)^{\prime} = B^{\prime} A^{\prime}$.
6. (i) If $A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$, then verify that $A^{\prime} A = I$.
Answer: $A^{\prime} = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$, so $$A^{\prime} A = \begin{bmatrix} \cos^2\alpha + \sin^2\alpha & \cos\alpha\sin\alpha – \sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha – \cos\alpha\sin\alpha & \sin^2\alpha + \cos^2\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$
(ii) If $A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$, then verify that $A^{\prime} A = I$.
Answer: $A^{\prime} = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}$, so $$A^{\prime} A = \begin{bmatrix} \sin^2\alpha + \cos^2\alpha & \sin\alpha\cos\alpha – \cos\alpha\sin\alpha \\ \cos\alpha\sin\alpha – \sin\alpha\cos\alpha & \cos^2\alpha + \sin^2\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$
7. (i) Show that the matrix $A = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$ is a symmetric matrix.
Answer: Taking the transpose (rows become columns), $$A^{\prime} = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix} = A$$ Since $A^{\prime} = A$, the matrix $A$ is symmetric.
(ii) Show that the matrix $A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}$ is a skew symmetric matrix.
Answer: Taking the transpose, $$A^{\prime} = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix} = -A$$ Since $A^{\prime} = -A$ (and all diagonal entries are $0$), the matrix $A$ is skew symmetric.
8. For the matrix $A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$, verify that (i) $(A + A^{\prime})$ is a symmetric matrix, (ii) $(A – A^{\prime})$ is a skew symmetric matrix.
Answer: Here $A^{\prime} = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$.
(i) $A + A^{\prime} = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$. Its transpose is $\begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$, which equals $A + A^{\prime}$, so $(A + A^{\prime})$ is symmetric.
(ii) $A – A^{\prime} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. Its transpose is $\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = -(A – A^{\prime})$, so $(A – A^{\prime})$ is skew symmetric.
9. Find $\frac{1}{2}(A + A^{\prime})$ and $\frac{1}{2}(A – A^{\prime})$, when $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$.
Answer: Here $A^{\prime} = \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}$. Then $A + A^{\prime} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$, so $$\frac{1}{2}(A + A^{\prime}) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$$ Also $A – A^{\prime} = \begin{bmatrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{bmatrix}$, so $$\frac{1}{2}(A – A^{\prime}) = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} = A$$ (This is expected, since $A$ is already skew symmetric.)
10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
In each case, if $A$ is a square matrix, then $A = P + Q$ where $P = \frac{1}{2}(A + A^{\prime})$ is symmetric and $Q = \frac{1}{2}(A – A^{\prime})$ is skew symmetric.
(i) $\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$
Answer: $A^{\prime} = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$. So $P = \frac{1}{2}\begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$ and $Q = \frac{1}{2}\begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$. Hence $$\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$$
(ii) $\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$
Answer: This matrix is already symmetric ($A^{\prime} = A$), so $P = A$ and $Q = O$: $$\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$
(iii) $\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}$
Answer: $A^{\prime} = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}$. Then $P = \frac{1}{2}(A + A^{\prime}) = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix}$ and $Q = \frac{1}{2}(A – A^{\prime}) = \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{bmatrix}$.
Thus $$\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{bmatrix}$$
(iv) $\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}$
Answer: $A^{\prime} = \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}$. So $P = \frac{1}{2}\begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix}$ and $Q = \frac{1}{2}\begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$. Hence $$\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$$
11. If $A, B$ are symmetric matrices of same order, then $AB – BA$ is a
(A) Skew symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrix
Answer: (A) Skew symmetric matrix. Since $A^{\prime} = A$ and $B^{\prime} = B$, $$(AB – BA)^{\prime} = (AB)^{\prime} – (BA)^{\prime} = B^{\prime}A^{\prime} – A^{\prime}B^{\prime} = BA – AB = -(AB – BA)$$ so $AB – BA$ is skew symmetric.
12. If $A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$, and $A + A^{\prime} = I$, then the value of $\alpha$ is
(A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{3}$ (C) $\pi$ (D) $\dfrac{3\pi}{2}$
Answer: (B) $\dfrac{\pi}{3}$. Here $A^{\prime} = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$, so $A + A^{\prime} = \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix}$. Setting this equal to $I$ gives $2\cos\alpha = 1$, i.e. $\cos\alpha = \dfrac{1}{2}$, so $\alpha = \dfrac{\pi}{3}$.
Exercise 3.4
1. Matrices $A$ and $B$ will be inverse of each other only if
(A) $AB = BA$ (B) $AB = BA = 0$ (C) $AB = 0, BA = I$ (D) $AB = BA = I$
Answer: (D) $AB = BA = I$. By definition, a square matrix $B$ is the inverse of a square matrix $A$ of the same order precisely when both products $AB$ and $BA$ equal the identity matrix $I$. Options (A), (B) and (C) do not guarantee this, so they are not sufficient.
Miscellaneous Exercise on Chapter 3
1. If $A$ and $B$ are symmetric matrices, prove that $AB – BA$ is a skew symmetric matrix.
Answer: Since $A$ and $B$ are symmetric, $A^{\prime} = A$ and $B^{\prime} = B$. Using $(XY)^{\prime} = Y^{\prime}X^{\prime}$, $$(AB – BA)^{\prime} = (AB)^{\prime} – (BA)^{\prime} = B^{\prime}A^{\prime} – A^{\prime}B^{\prime} = BA – AB = -(AB – BA)$$ Since $(AB – BA)^{\prime} = -(AB – BA)$, the matrix $AB – BA$ is skew symmetric.
2. Show that the matrix $B^{\prime} A B$ is symmetric or skew symmetric according as $A$ is symmetric or skew symmetric.
Answer: Using $(XYZ)^{\prime} = Z^{\prime}Y^{\prime}X^{\prime}$ and $(B^{\prime})^{\prime} = B$, $$(B^{\prime}AB)^{\prime} = B^{\prime}A^{\prime}(B^{\prime})^{\prime} = B^{\prime}A^{\prime}B$$ If $A$ is symmetric ($A^{\prime} = A$), then $(B^{\prime}AB)^{\prime} = B^{\prime}AB$, so $B^{\prime}AB$ is symmetric. If $A$ is skew symmetric ($A^{\prime} = -A$), then $(B^{\prime}AB)^{\prime} = B^{\prime}(-A)B = -B^{\prime}AB$, so $B^{\prime}AB$ is skew symmetric.
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 $A^{\prime} A = I$.
Answer: The condition $A^{\prime}A = I$ means that the columns of $A$ are mutually orthogonal and each has unit length. Taking the three columns:
Column 1 is $(0, x, x)$, with $0 + x^2 + x^2 = 2x^2 = 1$, so $x = \pm\dfrac{1}{\sqrt{2}}$. Column 2 is $(2y, y, -y)$, with $4y^2 + y^2 + y^2 = 6y^2 = 1$, so $y = \pm\dfrac{1}{\sqrt{6}}$. Column 3 is $(z, -z, z)$, with $z^2 + z^2 + z^2 = 3z^2 = 1$, so $z = \pm\dfrac{1}{\sqrt{3}}$.
The three columns are automatically orthogonal (for example column 1 $\cdot$ column 2 $= 0 + xy – xy = 0$). Hence $$x = \pm\frac{1}{\sqrt{2}}, \quad y = \pm\frac{1}{\sqrt{6}}, \quad z = \pm\frac{1}{\sqrt{3}}$$
4. For what values of $x$ : $\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = O$?
Answer: Multiplying the first two factors, $$\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 1+4+1 & 2+0+0 & 0+2+2 \end{bmatrix} = \begin{bmatrix} 6 & 2 & 4 \end{bmatrix}$$ Then $$\begin{bmatrix} 6 & 2 & 4 \end{bmatrix}\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = 6(0) + 2(2) + 4(x) = 4 + 4x$$ Setting $4 + 4x = 0$ gives $x = -1$.
5. If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 – 5A + 7I = 0$.
Answer: $$A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$$ Then $5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$ and $7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$, so $$A^2 – 5A + 7I = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0$$
6. Find $x$, if $\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = O$.
Answer: Multiplying the first two factors, $$\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} x-2 & -10 & 2x-8 \end{bmatrix}$$ Then $$\begin{bmatrix} x-2 & -10 & 2x-8 \end{bmatrix}\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = (x-2)x + (-10)(4) + (2x-8)(1) = x^2 – 48$$ Setting $x^2 – 48 = 0$ gives $x^2 = 48$, so $x = \pm 4\sqrt{3}$.
7. A manufacturer produces three products $x, y, z$ which he sells in two markets. Annual sales are indicated below:
| Market | Products ($x$) | Products ($y$) | Products ($z$) |
|---|---|---|---|
| I | 10,000 | 2,000 | 18,000 |
| II | 6,000 | 20,000 | 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.
Answer: Write the sales as a $2 \times 3$ matrix $S = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}$.
(a) Multiplying by the price column, $$S\begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix} = \begin{bmatrix} 25000 + 3000 + 18000 \\ 15000 + 30000 + 8000 \end{bmatrix} = \begin{bmatrix} 46000 \\ 53000 \end{bmatrix}$$ So the total revenue is ₹$46000$ in Market I and ₹$53000$ in Market II.
(b) The total cost in each market is $$S\begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix} = \begin{bmatrix} 20000 + 2000 + 9000 \\ 12000 + 20000 + 4000 \end{bmatrix} = \begin{bmatrix} 31000 \\ 36000 \end{bmatrix}$$ Gross profit = revenue − cost = $\begin{bmatrix} 46000 – 31000 \\ 53000 – 36000 \end{bmatrix} = \begin{bmatrix} 15000 \\ 17000 \end{bmatrix}$. So the gross profit is ₹$15000$ in Market I and ₹$17000$ in Market II, giving a total gross profit of ₹$32000$.
8. Find the matrix $X$ so that $X\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$.
Answer: The matrix on the right is $2 \times 3$ and the known factor is $2 \times 3$, so $X$ must be $2 \times 2$. Let $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Then $$X\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d \end{bmatrix}$$
Comparing the first row: $a + 4b = -7$ and $2a + 5b = -8$. Solving, $3b = -6$ so $b = -2$ and $a = -7 – 4(-2) = 1$. Comparing the second row: $c + 4d = 2$ and $2c + 5d = 4$. Solving, $3d = 0$ so $d = 0$ and $c = 2$. Hence $$X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$$
9. If $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ is such that $A^2 = I$, then
(A) $1 + \alpha^2 + \beta\gamma = 0$ (B) $1 – \alpha^2 + \beta\gamma = 0$ (C) $1 – \alpha^2 – \beta\gamma = 0$ (D) $1 + \alpha^2 – \beta\gamma = 0$
Answer: (C) $1 – \alpha^2 – \beta\gamma = 0$. Computing $$A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} = \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \alpha^2 + \beta\gamma \end{bmatrix}$$ Setting $A^2 = I$ gives $\alpha^2 + \beta\gamma = 1$, i.e. $1 – \alpha^2 – \beta\gamma = 0$.
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
Answer: (B) $A$ is a zero matrix. Being symmetric means $A^{\prime} = A$ and being skew symmetric means $A^{\prime} = -A$. Therefore $A = -A$, so $2A = 0$, which gives $A = O$.
11. If $A$ is square matrix such that $A^2 = A$, then $(I + A)^3 – 7A$ is equal to
(A) $A$ (B) $I – A$ (C) $I$ (D) $3A$
Answer: (C) $I$. Expanding, $(I + A)^3 = I + 3A + 3A^2 + A^3$. Since $A^2 = A$, we get $A^3 = A^2 \cdot A = A \cdot A = A^2 = A$. Hence $$(I + A)^3 = I + 3A + 3A + A = I + 7A$$ Therefore $(I + A)^3 – 7A = I + 7A – 7A = I$.
Additional Questions and Answers
Multiple Choice Questions
1. If a matrix has $12$ elements, the number of possible orders it can have is
(A) 4 (B) 6 (C) 8 (D) 12
Answer: (B) 6. The ordered pairs $(m, n)$ with $mn = 12$ are $(1,12),(12,1),(2,6),(6,2),(3,4),(4,3)$, giving $6$ possible orders.
2. If $A$ is of order $3 \times 4$ and $B$ is of order $4 \times 2$, then the order of $AB$ is
(A) $3 \times 2$ (B) $4 \times 4$ (C) $2 \times 3$ (D) $3 \times 4$
Answer: (A) $3 \times 2$. When $A$ is $m \times n$ and $B$ is $n \times p$, the product $AB$ is $m \times p$; here $m = 3$ and $p = 2$.
3. The number of all possible matrices of order $2 \times 2$ with each entry chosen from $\{0, 1, 2\}$ is
(A) 16 (B) 64 (C) 81 (D) 12
Answer: (C) 81. A $2 \times 2$ matrix has $4$ entries and each entry can be chosen in $3$ ways, so the number of matrices is $3^4 = 81$.
4. If $A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$, then $A A^{\prime}$ equals
(A) $O$ (B) $I$ (C) $2A$ (D) $A$
Answer: (B) $I$. Since $A^{\prime} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$, the product $A A^{\prime}$ has diagonal entries $\cos^2\theta + \sin^2\theta = 1$ and off-diagonal entries $0$, so $A A^{\prime} = I$.
5. A diagonal matrix in which all the diagonal elements are equal is called a
(A) row matrix (B) scalar matrix (C) zero matrix (D) column matrix
Answer: (B) scalar matrix. A diagonal matrix with all diagonal entries equal to a constant $k$ is a scalar matrix.
6. The diagonal elements of a skew symmetric matrix are all
(A) $1$ (B) equal (C) $0$ (D) positive
Answer: (C) $0$. For a skew symmetric matrix $a_{ji} = -a_{ij}$; putting $i = j$ gives $a_{ii} = -a_{ii}$, so $2a_{ii} = 0$ and every diagonal entry is $0$.
7. If $A$ and $B$ are symmetric matrices of the same order, then $A + B$ is
(A) skew symmetric (B) symmetric (C) a zero matrix (D) not defined
Answer: (B) symmetric. $(A + B)^{\prime} = A^{\prime} + B^{\prime} = A + B$, so $A + B$ is symmetric.
8. If $A$ is a square matrix such that $A^2 = A$, then $A$ is called
(A) a nilpotent matrix (B) an idempotent matrix (C) an orthogonal matrix (D) a scalar matrix
Answer: (B) an idempotent matrix. A matrix satisfying $A^2 = A$ is called idempotent.
Fill in the Blanks
1. A matrix having only one column is called a ________ matrix.
Answer: column
2. The transpose of the transpose of a matrix $A$ is ________.
Answer: $A$ (that is, $(A^{\prime})^{\prime} = A$)
3. A square matrix $A$ for which $A^{\prime} = A$ is called a ________ matrix.
Answer: symmetric
4. Matrix multiplication is not ________ in general, that is $AB \neq BA$.
Answer: commutative
5. The inverse of a square matrix, if it exists, is ________.
Answer: unique
True or False
1. Every square matrix is a diagonal matrix.
Answer: False. A diagonal matrix must have all off-diagonal entries zero; a general square matrix need not.
2. If $AB = O$, then either $A = O$ or $B = O$.
Answer: False. For example $\begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$, yet neither factor is a zero matrix.
3. For matrices of the same order, $(A + B)^{\prime} = A^{\prime} + B^{\prime}$.
Answer: True. The transpose of a sum equals the sum of the transposes.
4. Every scalar matrix is a diagonal matrix.
Answer: True. A scalar matrix has equal diagonal entries and all off-diagonal entries zero, so it is a special diagonal matrix.
5. Matrix addition is commutative.
Answer: True. For matrices of the same order, $A + B = B + A$.
Short Answer Questions
1. Construct the $2 \times 2$ matrix $A = [a_{ij}]$ whose elements are given by $a_{ij} = i + j$.
Answer: $a_{11} = 2, a_{12} = 3, a_{21} = 3, a_{22} = 4$, so $$A = \begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix}$$
2. If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, find $A + A^{\prime}$ and state whether it is symmetric.
Answer: $A^{\prime} = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$, so $A + A^{\prime} = \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix}$. Since $(A + A^{\prime})^{\prime} = A + A^{\prime}$, it is symmetric.
3. State whether the matrix $\begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}$ is symmetric or skew symmetric, with reason.
Answer: Its transpose is $\begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$, which is the negative of the matrix. Since $A^{\prime} = -A$ and the diagonal entries are zero, the matrix is skew symmetric.
4. Find the values of $a$ and $b$ if $\begin{bmatrix} a + b & 2 \\ 5 & ab \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$.
Answer: Equating corresponding elements, $a + b = 6$ and $ab = 8$. Then $a$ and $b$ are the roots of $t^2 – 6t + 8 = 0$, i.e. $(t-2)(t-4)=0$, so $\{a, b\} = \{2, 4\}$.
Key Terms
| Term | Meaning |
|---|---|
| Matrix (মৌলকক্ষ) | An ordered rectangular array of numbers or functions, enclosed in brackets. |
| Order of a matrix | The description $m \times n$, where $m$ is the number of rows and $n$ the number of columns. |
| Row / Column matrix | A matrix with a single row ($1 \times n$) or a single column ($m \times 1$). |
| Square matrix | A matrix with equal numbers of rows and columns ($m = n$). |
| Diagonal matrix | A square matrix in which every non-diagonal element is zero ($a_{ij} = 0$ for $i \neq j$). |
| Scalar matrix | A diagonal matrix in which all diagonal elements are equal. |
| Identity matrix ($I$) | A diagonal matrix whose diagonal elements are all $1$; it satisfies $AI = IA = A$. |
| Zero matrix ($O$) | A matrix all of whose elements are zero; the additive identity for matrices. |
| Transpose ($A^{\prime}$) | The matrix obtained by interchanging the rows and columns of $A$. |
| Symmetric matrix | A square matrix with $A^{\prime} = A$. |
| Skew symmetric matrix | A square matrix with $A^{\prime} = -A$; its diagonal elements are all zero. |
| Invertible matrix | A square matrix $A$ for which there is a matrix $B$ with $AB = BA = I$; then $B = A^{-1}$. |