This section covers proofs of matrix properties, solving matrix equations, and applying matrix algebra to real-world problems.
Table of Contents
Question 1.
If $A$ and $B$ are symmetric matrices, prove that $AB – BA$ is a skew symmetric matrix.
Solution:
Given that $A$ and $B$ are symmetric matrices, by definition:
$$A’ = A \quad \text{and} \quad B’ = B$$
Let $P = AB – BA$. We need to show that $P’ = -P$.
We take the transpose of $P$:
$$P’ = (AB – BA)’$$
Using the property $(M – N)’ = M’ – N’$:
$$P’ = (AB)’ – (BA)’$$
Using the Reversal Law for transpose, $(MN)’ = N’M’$:
$$P’ = B’A’ – A’B’$$
Substituting the given condition $A’=A$ and $B’=B$:
$$P’ = BA – AB$$
We can factor out $-1$:
$$P’ = -(AB – BA)$$
Since $P = AB – BA$, we have:
$$P’ = -P$$
Therefore, $AB – BA$ is a skew symmetric matrix. $\checkmark$
Question 2.
Show that the matrix $B’AB$ is symmetric or skew symmetric according as $A$ is symmetric or skew symmetric.
Solution:
Let $P = B’AB$. We examine $P’$ in both cases.
Case 1: $A$ is Symmetric ($A’ = A$)
We take the transpose of $P$:
$$P’ = (B’AB)’$$
Using the Reversal Law $(MNP)’ = P’N’M’$:
$$P’ = (AB)'(B’)’$$
Using $(AB)’ = B’A’$ and $(B’)’ = B$:
$$P’ = (B’A’)(B)$$
$$P’ = B’A’B$$
Substituting the condition $A’ = A$:
$$P’ = B’AB$$
Since $P’ = P$, $B’AB$ is symmetric when $A$ is symmetric. $\checkmark$
Case 2: $A$ is Skew Symmetric ($A’ = -A$)
We take the transpose of $P$:
$$P’ = (B’AB)’$$
Following the same steps as above:
$$P’ = B’A’B$$
Substituting the condition $A’ = -A$:
$$P’ = B'(-A)B$$
$$P’ = -(B’AB)$$
Since $P’ = -P$, $B’AB$ is skew symmetric when $A$ is skew symmetric. $\checkmark$
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 $A’A = I$.
Solution:
Given $A’A = I$, where $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
First, find $A’$:
$$A’ = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix}$$
Now compute $A’A$:
$$A’A = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix} \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} = \begin{bmatrix} 0+x^2+x^2 & 0+xy-xy & 0-xz+xz \\ 0+xy-xy & 4y^2+y^2+y^2 & 2yz-yz-yz \\ 0-xz+xz & 2yz-yz-yz & z^2+z^2+z^2 \end{bmatrix}$$
$$A’A = \begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix}$$
Equating $A’A = I$:
$$\begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
Equating the diagonal elements:
- $2x^2 = 1 \implies x^2 = 1/2 \implies x = \pm 1/\sqrt{2}$
- $6y^2 = 1 \implies y^2 = 1/6 \implies y = \pm 1/\sqrt{6}$
- $3z^2 = 1 \implies z^2 = 1/3 \implies z = \pm 1/\sqrt{3}$
$$\mathbf{x = \pm \frac{1}{\sqrt{2}}, \quad y = \pm \frac{1}{\sqrt{6}}, \quad z = \pm \frac{1}{\sqrt{3}}}$$
Question 4.
For what values of $x$: $\begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = O$?
Solution:
We perform matrix multiplication sequentially, from left to right.
Step 1: Multiply the first two matrices ($1 \times 3$ by $3 \times 3 \to 1 \times 3$)
$$P = \begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix}$$
$$P = \begin{bmatrix} 1(1)+2(2)+1(1) & 1(2)+2(0)+1(0) & 1(0)+2(1)+1(2) \end{bmatrix}$$
$$P = \begin{bmatrix} 1+4+1 & 2+0+0 & 0+2+2 \end{bmatrix} = \begin{bmatrix} 6 & 2 & 4 \end{bmatrix}$$
Step 2: Multiply the result $P$ by the column matrix ($1 \times 3$ by $3 \times 1 \to 1 \times 1$)
$$P \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = \begin{bmatrix} 6 & 2 & 4 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = O$$
$$[ 6(0) + 2(2) + 4(x) ] = [ 0 ]$$
$$[ 0 + 4 + 4x ] = [ 0 ]$$
Step 3: Solve for $x$
$$4 + 4x = 0$$
$$4x = -4 \implies \mathbf{x = -1}$$
Question 5.
If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 – 5A + 7I = 0$.
Solution:
We calculate $A^2$, $5A$, and $7I$ separately.
- Calculate $A^2 = A \cdot A$$$A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3(3)+1(-1) & 3(1)+1(2) \\ -1(3)+2(-1) & -1(1)+2(2) \end{bmatrix}$$$$A^2 = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$$
- Calculate $5A$ and $7I$$$5A = 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$$$$7I = 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$$
- Compute $A^2 – 5A + 7I$$$A^2 – 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} – \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$$$$= \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5-(-5)+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0$$The equation is shown. $\checkmark$
Question 6.
Find $x$, if $\begin{bmatrix} x & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 1 \\ -1 \end{bmatrix} = O$.
Solution:
We perform matrix multiplication sequentially, from left to right.
Step 1: Multiply the first two matrices ($1 \times 3$ by $3 \times 3 \to 1 \times 3$)
$$P = \begin{bmatrix} x & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$$
$$P = \begin{bmatrix} x(1)-1(0)+1(2) & x(0)-1(2)+1(0) & x(2)-1(1)+1(3) \end{bmatrix}$$
$$P = \begin{bmatrix} x+2 & -2 & 2x-1+3 \end{bmatrix} = \begin{bmatrix} x+2 & -2 & 2x+2 \end{bmatrix}$$
Step 2: Multiply the result $P$ by the column matrix ($1 \times 3$ by $3 \times 1 \to 1 \times 1$)
$$P \begin{bmatrix} x \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} x+2 & -2 & 2x+2 \end{bmatrix} \begin{bmatrix} x \\ 1 \\ -1 \end{bmatrix} = O$$
$$[ (x+2)x + (-2)(1) + (2x+2)(-1) ] = [ 0 ]$$
$$[ x^2 + 2x – 2 – 2x – 2 ] = [ 0 ]$$
$$[ x^2 – 4 ] = [ 0 ]$$
Step 3: Solve for $x$
$$x^2 – 4 = 0$$
$$x^2 = 4 \implies \mathbf{x = \pm 2}$$
Question 7.
A manufacturer produces three products $x, y, z$ in two markets.
Sales Matrix ($S$): $S = \begin{matrix} \text{I} \\ \text{II} \end{matrix} \begin{bmatrix} 10,000 & 2,000 & 18,000 \\ 6,000 & 20,000 & 8,000 \end{bmatrix}$ (Order $2 \times 3$)
(a) Total Revenue
Unit sale prices ($P_S$) are ₹ $2.50$ (x), ₹ $1.50$ (y), and ₹ $1.00$ (z).
The Price Matrix ($P_S$) should be $3 \times 1$ for $S P_S$ multiplication:
$$P_S = \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix}$$
Revenue Matrix ($R$) = $S P_S$:
$$R = \begin{bmatrix} 10,000 & 2,000 & 18,000 \\ 6,000 & 20,000 & 8,000 \end{bmatrix} \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix}$$
$$R = \begin{bmatrix} 10000(2.5) + 2000(1.5) + 18000(1.0) \\ 6000(2.5) + 20000(1.5) + 8000(1.0) \end{bmatrix}$$
$$R = \begin{bmatrix} 25,000 + 3,000 + 18,000 \\ 15,000 + 30,000 + 8,000 \end{bmatrix} = \begin{bmatrix} 46,000 \\ 53,000 \end{bmatrix}$$
$$\mathbf{\text{Total Revenue: Market I = ₹ 46,000, Market II = ₹ 53,000}}$$
(b) Gross Profit
Unit costs ($P_C$) are ₹ $2.00$ (x), ₹ $1.00$ (y), and ₹ $0.50$ (z).
The Cost Matrix ($P_C$) is:
$$P_C = \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix}$$
Total Cost Matrix ($C$) = $S P_C$:
$$C = \begin{bmatrix} 10,000 & 2,000 & 18,000 \\ 6,000 & 20,000 & 8,000 \end{bmatrix} \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix}$$
$$C = \begin{bmatrix} 10000(2.0) + 2000(1.0) + 18000(0.5) \\ 6000(2.0) + 20000(1.0) + 8000(0.5) \end{bmatrix}$$
$$C = \begin{bmatrix} 20,000 + 2,000 + 9,000 \\ 12,000 + 20,000 + 4,000 \end{bmatrix} = \begin{bmatrix} 31,000 \\ 36,000 \end{bmatrix}$$
Gross Profit Matrix ($P$) = $R – C$:
$$P = \begin{bmatrix} 46,000 \\ 53,000 \end{bmatrix} – \begin{bmatrix} 31,000 \\ 36,000 \end{bmatrix} = \begin{bmatrix} 46000 – 31000 \\ 53000 – 36000 \end{bmatrix} = \begin{bmatrix} 15,000 \\ 17,000 \end{bmatrix}$$
$$\mathbf{\text{Gross Profit: Market I = ₹ 15,000, Market II = ₹ 17,000}}$$
Question 8.
Find the matrix $X$ so that $X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$.
Solution:
Let the orders of the matrices be:
- $X$: $m \times n$
- $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$: $2 \times 3$
- $B = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$: $2 \times 3$
For $X A = B$ to be possible:
- The number of columns in $X$ must equal the number of rows in $A$: $n = 2$.
- The number of rows in $X$ must equal the number of rows in $B$: $m = 2$.So, $X$ is a $2 \times 2$ matrix. Let $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$$
Perform the multiplication:
$$\begin{bmatrix} a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$$
Equating the first row elements:
- $a + 4b = -7$
- $2a + 5b = -8$Multiply (1) by 2: $2a + 8b = -14$. Subtract this from (2):$$(2a + 5b) – (2a + 8b) = -8 – (-14)$$$$-3b = 6 \implies \mathbf{b = -2}$$Substitute $b=-2$ into (1): $a + 4(-2) = -7 \implies a – 8 = -7 \implies \mathbf{a = 1}$
Equating the second row elements:
3. $c + 4d = 2$
4. $2c + 5d = 4$
Multiply (3) by 2: $2c + 8d = 4$. Subtract (4) from this:
$$(2c + 8d) – (2c + 5d) = 4 – 4$$
$$3d = 0 \implies \mathbf{d = 0}$$
Substitute $d=0$ into (3): $c + 4(0) = 2 \implies \mathbf{c = 2}$
$$\mathbf{X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}}$$
Question 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) $1 + \alpha^2 – \beta\gamma = 0$
Solution:
Calculate $A^2$:
$$A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} = \begin{bmatrix} \alpha^2 + \beta\gamma & \alpha\beta – \beta\alpha \\ \gamma\alpha – \alpha\gamma & \beta\gamma + (-\alpha)^2 \end{bmatrix}$$
$$A^2 = \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \alpha^2 + \beta\gamma \end{bmatrix}$$
Given $A^2 = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Equating the elements:
$$\alpha^2 + \beta\gamma = 1$$
Rearranging the terms to match the options:
$$1 – \alpha^2 – \beta\gamma = 0$$
$$\text{The correct option is } \mathbf{(C)}$$
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
Solution:
- Symmetric: $A’ = A \implies a_{ij} = a_{ji}$
- Skew Symmetric: $A’ = -A \implies a_{ij} = -a_{ji}$
Combining the two conditions: $a_{ij} = -a_{ij}$
This implies $2a_{ij} = 0$, so $a_{ij} = 0$ for all elements $i$ and $j$.
Therefore, $A$ must be a matrix where every element is zero, which is the zero matrix (also called the null matrix).
$$\text{The correct option is } \mathbf{(B)}$$
Question 11.
If $A$ is a square matrix such that $A^2 = A$, then $(I + A)^3 – 7A$ is equal to:
(A) $A$ (B) $I – A$ (C) $I$ (D) $3A$
Solution:
We use the binomial expansion for $(I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$.
Since $I$ is the identity matrix, $I^n = I$.
We are given the condition $\mathbf{A^2 = A}$. This implies $A^3 = A^2 \cdot A = A \cdot A = A^2 = A$.
In general, $A^n = A$ for $n \ge 1$.
Expand $(I + A)^3$:
$$(I + A)^3 = I + 3A + 3A^2 + A^3$$
Substitute $A^2 = A$ and $A^3 = A$:
$$(I + A)^3 = I + 3A + 3A + A$$
$$(I + A)^3 = I + 7A$$
Now substitute this back into the required expression:
$$(I + A)^3 – 7A = (I + 7A) – 7A$$
$$(I + A)^3 – 7A = I$$
$$\text{The correct option is } \mathbf{(C)}$$