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Math Notes For Class 11 Chapter 2 Ex 2.1 Solved & Quiz
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Math Notes For Class 11 Chapter 2 Ex 2.1 Solved & Quiz
A matrix is a rectangular array of numbers, functions, or complex expressions arranged in horizontal rows and vertical columns, enclosed within square brackets. In FBISE and Punjab Boards Class 11 Mathematics (New 2025–2026 Syllabus), matrices are studied extensively over real and complex number fields.
Key Matrix Classifications & Definitions:
| Matrix Property / Concept | Mathematical Condition / Formula | Key Domain Rules |
|---|---|---|
| Transpose Rule for Sum | $(A + B)^T = A^T + B^T$ | $A$ and $B$ must have the same order. |
| Reversal Rule for Transpose | $(AB)^T = B^T A^T$ | Number of columns of $A$ equals rows of $B$. |
| Symmetric Decomposition | $A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$ | $A+A^T$ is symmetric; $A-A^T$ is skew-symmetric. |
| Conjugate Transpose (Hermitian) | $A^\theta = (\bar{A})^T = A$ | Complex matrices; diagonal entries are real. |
| Conjugate Transpose (Skew-Hermitian) | $A^\theta = (\bar{A})^T = -A$ | Complex matrices; diagonal entries are zero or imaginary. |
Question 1 (Part i): Find the values of $x$ and $y$ if $\begin{bmatrix} x + 3 & 1 \\ 2y & -4 \end{bmatrix} = \begin{bmatrix} 5 & 1 \\ 6 & -4 \end{bmatrix}$.
Solution:
Since the two matrices are equal, their corresponding elements must be equal:
1) Equating entry $(1,1)$:
$$x + 3 = 5 \implies x = 5 - 3 \implies x = 2$$
2) Equating entry $(2,1)$:
$$2y = 6 \implies y = \frac{6}{2} \implies y = 3$$
Answer: $$x = 2, \quad y = 3$$
Question 1 (Part ii): If $A = \begin{bmatrix} 2 & 1 \\ -1 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 4 \\ 2 & -1 \end{bmatrix}$, compute $3A - 2B$.
Solution:
First, calculate scalar multiples $3A$ and $2B$:
$$3A = 3 \begin{bmatrix} 2 & 1 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 6 & 3 \\ -3 & 9 \end{bmatrix}$$
$$2B = 2 \begin{bmatrix} 0 & 4 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 8 \\ 4 & -2 \end{bmatrix}$$
Now, subtract $2B$ from $3A$:
$$3A - 2B = \begin{bmatrix} 6 & 3 \\ -3 & 9 \end{bmatrix} - \begin{bmatrix} 0 & 8 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 6-0 & 3-8 \\ -3-4 & 9-(-2) \end{bmatrix}$$
$$3A - 2B = \begin{bmatrix} 6 & -5 \\ -7 & 11 \end{bmatrix}$$
Answer: $$\begin{bmatrix} 6 & -5 \\ -7 & 11 \end{bmatrix}$$
Question 2 (Part i): Compute the matrix product $AB$ where $A = \begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 0 \\ -1 & 5 \end{bmatrix}$.
Solution:
Using matrix row-by-column multiplication:
$$AB = \begin{bmatrix} (1)(2) + (-2)(-1) & (1)(0) + (-2)(5) \\ (3)(2) + (4)(-1) & (3)(0) + (4)(5) \end{bmatrix}$$
$$AB = \begin{bmatrix} 2 + 2 & 0 - 10 \\ 6 - 4 & 0 + 20 \end{bmatrix} = \begin{bmatrix} 4 & -10 \\ 2 & 20 \end{bmatrix}$$
Answer: $$\begin{bmatrix} 4 & -10 \\ 2 & 20 \end{bmatrix}$$
Question 2 (Part ii): Verify whether $AB = BA$ for $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}$.
Solution:
First, calculate $AB$:
$$AB = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} (1)(2)+(2)(1) & (1)(1)+(2)(3) \\ (0)(2)+(1)(1) & (0)(1)+(1)(3) \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ 1 & 3 \end{bmatrix}$$
Next, calculate $BA$:
$$BA = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} (2)(1)+(1)(0) & (2)(2)+(1)(1) \\ (1)(1)+(3)(0) & (1)(2)+(3)(1) \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 1 & 5 \end{bmatrix}$$
Comparing $AB$ and $BA$:
$$\begin{bmatrix} 4 & 7 \\ 1 & 3 \end{bmatrix} \neq \begin{bmatrix} 2 & 5 \\ 1 & 5 \end{bmatrix}$$
Answer: $$AB \neq BA \quad \text{(Matrix multiplication is non-commutative)}$$
Question 3 (Part i): For $A = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}$, prove that $A + A^T$ is a symmetric matrix.
Solution:
Find $A^T$ by interchanging rows and columns:
$$A^T = \begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}$$
Compute $S = A + A^T$:
$$S = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 3+3 & 4+1 \\ 1+4 & 2+2 \end{bmatrix} = \begin{bmatrix} 6 & 5 \\ 5 & 4 \end{bmatrix}$$
Now take the transpose of $S$:
$$S^T = \begin{bmatrix} 6 & 5 \\ 5 & 4 \end{bmatrix}^T = \begin{bmatrix} 6 & 5 \\ 5 & 4 \end{bmatrix} = S$$
Since $S^T = S$, $A + A^T$ is symmetric.
Answer: $$A + A^T = \begin{bmatrix} 6 & 5 \\ 5 & 4 \end{bmatrix} \quad \text{is symmetric because } (A+A^T)^T = A+A^T$$
Question 3 (Part ii): For $A = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}$, prove that $A - A^T$ is a skew-symmetric matrix.
Solution:
We have $A = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}$ and $A^T = \begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}$.
Compute $K = A - A^T$:
$$K = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 3-3 & 4-1 \\ 1-4 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$$
Take the transpose of $K$:
$$K^T = \begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix} = - \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix} = -K$$
Since $K^T = -K$, $A - A^T$ is skew-symmetric.
Answer: $$A - A^T = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix} \quad \text{is skew-symmetric because } (A-A^T)^T = -(A-A^T)$$
Question 4 (Part i): Show that the matrix $A = \begin{bmatrix} 2 & 1+i \\ 1-i & 3 \end{bmatrix}$ is Hermitian.
Solution:
Step 1: Take the complex conjugate $\bar{A}$ of matrix $A$ by replacing $i$ with $-i$:
$$\bar{A} = \begin{bmatrix} 2 & 1-i \\ 1+i & 3 \end{bmatrix}$$
Step 2: Take the transpose of $\bar{A}$:
$$(\bar{A})^T = \begin{bmatrix} 2 & 1+i \\ 1-i & 3 \end{bmatrix}$$
Comparing $(\bar{A})^T$ with $A$:
$$(\bar{A})^T = A$$
Since $(\bar{A})^T = A$, matrix $A$ is Hermitian.
Answer: $$(\bar{A})^T = A \implies A \text{ is Hermitian}$$
Question 4 (Part ii): Show that the matrix $A = \begin{bmatrix} 2i & 3+i \\ -3+i & 0 \end{bmatrix}$ is Skew-Hermitian.
Solution:
Step 1: Compute the complex conjugate $\bar{A}$:
$$\bar{A} = \begin{bmatrix} -2i & 3-i \\ -3-i & 0 \end{bmatrix}$$
Step 2: Take the transpose of $\bar{A}$:
$$(\bar{A})^T = \begin{bmatrix} -2i & -3-i \\ 3-i & 0 \end{bmatrix}$$
Step 3: Factor out $-1$ from $(\bar{A})^T$:
$$(\bar{A})^T = - \begin{bmatrix} 2i & 3+i \\ -3+i & 0 \end{bmatrix} = -A$$
Since $(\bar{A})^T = -A$, matrix $A$ is Skew-Hermitian.
Answer: $$(\bar{A})^T = -A \implies A \text{ is Skew-Hermitian}$$
Question 5 (Part i): Find the $2 \times 2$ matrix $X$ such that $X + \begin{bmatrix} -1 & 2 \\ 3 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 1 & 2 \end{bmatrix}$.
Solution:
Let $A = \begin{bmatrix} -1 & 2 \\ 3 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 0 \\ 1 & 2 \end{bmatrix}$.
The equation is $X + A = B \implies X = B - A$.
$$X = \begin{bmatrix} 4 & 0 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} -1 & 2 \\ 3 & 5 \end{bmatrix}$$
$$X = \begin{bmatrix} 4 - (-1) & 0 - 2 \\ 1 - 3 & 2 - 5 \end{bmatrix} = \begin{bmatrix} 5 & -2 \\ -2 & -3 \end{bmatrix}$$
Answer: $$X = \begin{bmatrix} 5 & -2 \\ -2 & -3 \end{bmatrix}$$
Question 5 (Part ii): Find matrix $X$ if $\begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} X = \begin{bmatrix} 1 & 3 \\ 0 & 2 \end{bmatrix}$.
Solution:
Let $M = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}$ and $C = \begin{bmatrix} 1 & 3 \\ 0 & 2 \end{bmatrix}$. Let $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
Then $M X = C$:
$$\begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 2a - c & 2b - d \\ a & b \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 2 \end{bmatrix}$$
Equating corresponding elements:
1) From row 2: $a = 0$ and $b = 2$.
2) From row 1, column 1: $2a - c = 1 \implies 2(0) - c = 1 \implies c = -1$.
3) From row 1, column 2: $2b - d = 3 \implies 2(2) - d = 3 \implies 4 - d = 3 \implies d = 1$.
Therefore, $X = \begin{bmatrix} 0 & 2 \\ -1 & 1 \end{bmatrix}$.
Answer: $$X = \begin{bmatrix} 0 & 2 \\ -1 & 1 \end{bmatrix}$$
Question 6 (Part i): Verify that $(AB)^T = B^T A^T$ for $A = \begin{bmatrix} 1 & 3 \\ -2 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 5 \\ 0 & 1 \end{bmatrix}$.
Solution:
Left Hand Side (LHS):
First compute $AB$:
$$AB = \begin{bmatrix} 1 & 3 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} 2 & 5 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} (1)(2)+(3)(0) & (1)(5)+(3)(1) \\ (-2)(2)+(4)(0) & (-2)(5)+(4)(1) \end{bmatrix} = \begin{bmatrix} 2 & 8 \\ -4 & -6 \end{bmatrix}$$
Taking the transpose $(AB)^T$:
$$(AB)^T = \begin{bmatrix} 2 & -4 \\ 8 & -6 \end{bmatrix}$$
Right Hand Side (RHS):
Compute $A^T$ and $B^T$:
$$A^T = \begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}, \quad B^T = \begin{bmatrix} 2 & 0 \\ 5 & 1 \end{bmatrix}$$
Compute $B^T A^T$:
$$B^T A^T = \begin{bmatrix} 2 & 0 \\ 5 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} (2)(1)+(0)(3) & (2)(-2)+(0)(4) \\ (5)(1)+(1)(3) & (5)(-2)+(1)(4) \end{bmatrix} = \begin{bmatrix} 2 & -4 \\ 8 & -6 \end{bmatrix}$$
Since $\text{LHS} = \text{RHS} = \begin{bmatrix} 2 & -4 \\ 8 & -6 \end{bmatrix}$, the property is verified.
Answer: $$(AB)^T = B^T A^T = \begin{bmatrix} 2 & -4 \\ 8 & -6 \end{bmatrix}$$
Question 6 (Part ii): Verify that $(A + B)^T = A^T + B^T$ for $A = \begin{bmatrix} 1 & 3 \\ -2 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 5 \\ 0 & 1 \end{bmatrix}$.
Solution:
LHS:
$$A + B = \begin{bmatrix} 1+2 & 3+5 \\ -2+0 & 4+1 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ -2 & 5 \end{bmatrix}$$
$$(A + B)^T = \begin{bmatrix} 3 & -2 \\ 8 & 5 \end{bmatrix}$$
RHS:
$$A^T = \begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}, \quad B^T = \begin{bmatrix} 2 & 0 \\ 5 & 1 \end{bmatrix}$$
$$A^T + B^T = \begin{bmatrix} 1+2 & -2+0 \\ 3+5 & 4+1 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ 8 & 5 \end{bmatrix}$$
Since $\text{LHS} = \text{RHS}$, the property is verified.
Answer: $$(A + B)^T = A^T + B^T = \begin{bmatrix} 3 & -2 \\ 8 & 5 \end{bmatrix}$$
Q1: If $A$ is a square matrix of order $n \times n$, then $A - A^T$ is always:
Q2: For any real skew-symmetric matrix, what must be true about its main diagonal elements?