Class 10 Mathematics - Ch 3: Matrices and Determinants
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📘 Comprehensive Syllabus & Examination Guide
Class 10 Mathematics - Ch 3: Matrices and Determinants
Official curriculum roadmap, subject/topic distribution, negative marking rules, pacing guidelines, and solved sample questions.
🎯 Question Types & Curriculum Breakdown
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68 Questions
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Multiple Choice (MCQs)
38%
26 MCQs
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32%
22 Questions
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8 Questions
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True / False
9%
6 Questions
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4 Questions
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2 Questions
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📊 Question Pool Structure
68 Solved Questions (MCQs, Short & Long Questions, Blanks, True/False).
⚡ Recommended Pacing
1 to 3 minutes per question depending on question type (MCQ, Short, Long).
⚖️ Scoring & Negative Marking
1 to 5 marks per question aligned with official board examination rubrics.
💡 Strategic Preparation & Exam Hall Guidelines
To maximize your score on Class 10 Mathematics - Ch 3: Matrices and Determinants, candidates are advised to follow a structured three-pass approach. In the First Pass, solve all direct recall and formula-based questions within 30 seconds each to secure foundational marks. In the Second Pass, tackle multi-step analytical and quantitative reasoning problems. In the Third Pass, review marked questions and verify calculations.
Practice with the interactive player below to evaluate your speed and accuracy under real exam pressure. Every question features full mathematical formulas, step-by-step worked solutions, and conceptual explanations vetted by Apex Rankers Academy subject matter specialists.
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📝 Pre-Rendered Solved Sample Questions & Detailed Solutions
Review the solved problems below to understand question phrasing, answer choices, and step-by-step solution logic prior to starting the full interactive practice drill:
Sample Question 1
Exercise 3.1 • Introduction & Order of Matrices
MEDIUM • Short Question
Find the number of rows, number of columns, and the order of the following matrices:
(a) $A = \begin{bmatrix} 2 & 3 \\ 0 & 5 \end{bmatrix}$
(b) $B = \begin{bmatrix} 1 & -1 & 2 \end{bmatrix}$
(c) $C = \begin{bmatrix} 4 \\ 2 \\ -1 \end{bmatrix}$
(d) $D = \begin{bmatrix} 2 & 1 & 0 \\ 3 & -1 & 4 \end{bmatrix}$
(a) $A = \begin{bmatrix} 2 & 3 \\ 0 & 5 \end{bmatrix}$
(b) $B = \begin{bmatrix} 1 & -1 & 2 \end{bmatrix}$
(c) $C = \begin{bmatrix} 4 \\ 2 \\ -1 \end{bmatrix}$
(d) $D = \begin{bmatrix} 2 & 1 & 0 \\ 3 & -1 & 4 \end{bmatrix}$
✓ Correct Answer: (a) Rows = 2, Cols = 2, Order = 2x2; (b) Rows = 1, Cols = 3, Order = 1x3; (c) Rows = 3, Cols = 1, Order = 3x1; (d) Rows = 2, Cols = 3, Order = 2x3
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Resolution:**
- **(a)** Matrix $A = \begin{bmatrix} 2 & 3 \\ 0 & 5 \end{bmatrix}$:
- Number of horizontal rows ($m$) = $2$
- Number of vertical columns ($n$) = $2$
- **Order of Matrix $A$:** $m \times n = 2 \times 2$.
- **(b)** Matrix $B = \begin{bmatrix} 1 & -1 & 2 \end{bmatrix}$:
- Number of horizontal rows ($m$) = $1$
- Number of vertical columns ($n$) = $3$
- **Order of Matrix $B$:** $m \times n = 1 \times 3$.
- **(c)** Matrix $C = \begin{bmatrix} 4 \\ 2 \\ -1 \end{bmatrix}$:
- Number of horizontal rows ($m$) = $3$
- Number of vertical columns ($n$) = $1$
- **Order of Matrix $C$:** $m \times n = 3 \times 1$.
- **(d)** Matrix $D = \begin{bmatrix} 2 & 1 & 0 \\ 3 & -1 & 4 \end{bmatrix}$:
- Number of horizontal rows ($m$) = $2$
- Number of vertical columns ($n$) = $3$
- **Order of Matrix $D$:** $m \times n = 2 \times 3$.
- **(a)** Matrix $A = \begin{bmatrix} 2 & 3 \\ 0 & 5 \end{bmatrix}$:
- Number of horizontal rows ($m$) = $2$
- Number of vertical columns ($n$) = $2$
- **Order of Matrix $A$:** $m \times n = 2 \times 2$.
- **(b)** Matrix $B = \begin{bmatrix} 1 & -1 & 2 \end{bmatrix}$:
- Number of horizontal rows ($m$) = $1$
- Number of vertical columns ($n$) = $3$
- **Order of Matrix $B$:** $m \times n = 1 \times 3$.
- **(c)** Matrix $C = \begin{bmatrix} 4 \\ 2 \\ -1 \end{bmatrix}$:
- Number of horizontal rows ($m$) = $3$
- Number of vertical columns ($n$) = $1$
- **Order of Matrix $C$:** $m \times n = 3 \times 1$.
- **(d)** Matrix $D = \begin{bmatrix} 2 & 1 & 0 \\ 3 & -1 & 4 \end{bmatrix}$:
- Number of horizontal rows ($m$) = $2$
- Number of vertical columns ($n$) = $3$
- **Order of Matrix $D$:** $m \times n = 2 \times 3$.
Sample Question 2
Exercise 3.1 • Equal Matrices
MEDIUM • Short Question
Which of the following pairs of matrices are equal?
(a) $A = \begin{bmatrix} 2 & 3 \\ 1 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 2+1 \\ 1 & 4+1 \end{bmatrix}$
(b) $C = \begin{bmatrix} 3 & 4 \end{bmatrix}$ and $D = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$
(c) $E = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $F = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$
(d) $G = \begin{bmatrix} 5-2 & 3 \\ 2+1 & 4+2 \end{bmatrix}$ and $H = \begin{bmatrix} 3 & 3 \\ 3 & 6 \end{bmatrix}$
(a) $A = \begin{bmatrix} 2 & 3 \\ 1 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 2+1 \\ 1 & 4+1 \end{bmatrix}$
(b) $C = \begin{bmatrix} 3 & 4 \end{bmatrix}$ and $D = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$
(c) $E = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $F = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$
(d) $G = \begin{bmatrix} 5-2 & 3 \\ 2+1 & 4+2 \end{bmatrix}$ and $H = \begin{bmatrix} 3 & 3 \\ 3 & 6 \end{bmatrix}$
✓ Correct Answer: (a) Equal ($A=B$); (b) Not Equal; (c) Not Equal; (d) Equal ($G=H$)
📖 Step-by-Step Solution & Conceptual Rationale:
**Two matrices are equal if and only if:**
1. They have the same order ($m \times n$).
2. Their corresponding entries are identical.
- **(a)** $A = \begin{bmatrix} 2 & 3 \\ 1 & 5 \end{bmatrix}$, $B = \begin{bmatrix} 2 & 2+1 \\ 1 & 4+1 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 1 & 5 \end{bmatrix}$.
- Both have order $2 \times 2$ and all corresponding entries match ($2=2, 3=3, 1=1, 5=5$).
- **Result:** **$A = B$ (Equal)**.
- **(b)** $C = \begin{bmatrix} 3 & 4 \end{bmatrix}$ (order $1 \times 2$), $D = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$ (order $2 \times 1$).
- Since orders are different ($1 \times 2 \neq 2 \times 1$), they cannot be equal.
- **Result:** **$C \neq D$ (Not Equal)**.
- **(c)** $E = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, $F = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$.
- Both have order $2 \times 2$, but entry $e_{22} = 1 \neq f_{22} = 2$.
- **Result:** **$E \neq F$ (Not Equal)**.
- **(d)** $G = \begin{bmatrix} 5-2 & 3 \\ 2+1 & 4+2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & 6 \end{bmatrix}$, $H = \begin{bmatrix} 3 & 3 \\ 3 & 6 \end{bmatrix}$.
- Both have order $2 \times 2$ and all corresponding entries are equal.
- **Result:** **$G = H$ (Equal)**.
1. They have the same order ($m \times n$).
2. Their corresponding entries are identical.
- **(a)** $A = \begin{bmatrix} 2 & 3 \\ 1 & 5 \end{bmatrix}$, $B = \begin{bmatrix} 2 & 2+1 \\ 1 & 4+1 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 1 & 5 \end{bmatrix}$.
- Both have order $2 \times 2$ and all corresponding entries match ($2=2, 3=3, 1=1, 5=5$).
- **Result:** **$A = B$ (Equal)**.
- **(b)** $C = \begin{bmatrix} 3 & 4 \end{bmatrix}$ (order $1 \times 2$), $D = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$ (order $2 \times 1$).
- Since orders are different ($1 \times 2 \neq 2 \times 1$), they cannot be equal.
- **Result:** **$C \neq D$ (Not Equal)**.
- **(c)** $E = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, $F = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$.
- Both have order $2 \times 2$, but entry $e_{22} = 1 \neq f_{22} = 2$.
- **Result:** **$E \neq F$ (Not Equal)**.
- **(d)** $G = \begin{bmatrix} 5-2 & 3 \\ 2+1 & 4+2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & 6 \end{bmatrix}$, $H = \begin{bmatrix} 3 & 3 \\ 3 & 6 \end{bmatrix}$.
- Both have order $2 \times 2$ and all corresponding entries are equal.
- **Result:** **$G = H$ (Equal)**.
Sample Question 3
Exercise 3.1 • Row & Column Matrices
MEDIUM • Short Question
A student scores $8, 9, 10$ in three mathematics quizzes. Represent these scores as:
(a) A row matrix
(b) A column matrix
(a) A row matrix
(b) A column matrix
✓ Correct Answer: (a) $R = \begin{bmatrix} 8 & 9 & 10 \end{bmatrix}$; (b) $C = \begin{bmatrix} 8 \\ 9 \\ 10 \end{bmatrix}$
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Resolution:**
- **(a) Row Matrix Representation:**
- A row matrix has exactly one horizontal row ($1 \times 3$).
- **Row Matrix:** $R = \begin{bmatrix} 8 & 9 & 10 \end{bmatrix}$.
- **(b) Column Matrix Representation:**
- A column matrix has exactly one vertical column ($3 \times 1$).
- **Column Matrix:** $C = \begin{bmatrix} 8 \\ 9 \\ 10 \end{bmatrix}$.
- **(a) Row Matrix Representation:**
- A row matrix has exactly one horizontal row ($1 \times 3$).
- **Row Matrix:** $R = \begin{bmatrix} 8 & 9 & 10 \end{bmatrix}$.
- **(b) Column Matrix Representation:**
- A column matrix has exactly one vertical column ($3 \times 1$).
- **Column Matrix:** $C = \begin{bmatrix} 8 \\ 9 \\ 10 \end{bmatrix}$.
Sample Question 4
Exercise 3.1 • Classification of Matrix Types
MEDIUM • Short Question
Classify each of the following matrices with the most specific appropriate name:
(a) $A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
(b) $B = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$
(c) $C = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
(d) $D = \begin{bmatrix} 3 & 0 \\ 0 & 7 \end{bmatrix}$
(e) $E = \begin{bmatrix} 4 & 5 & 6 \end{bmatrix}$
(a) $A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
(b) $B = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$
(c) $C = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
(d) $D = \begin{bmatrix} 3 & 0 \\ 0 & 7 \end{bmatrix}$
(e) $E = \begin{bmatrix} 4 & 5 & 6 \end{bmatrix}$
✓ Correct Answer: (a) Null/Zero Matrix; (b) Scalar Matrix; (c) Identity/Unit Matrix; (d) Diagonal Matrix; (e) Row Matrix
📖 Step-by-Step Solution & Conceptual Rationale:
**Classification Guide:**
- **(a)** $A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$: All entries are zero $\implies$ **Null Matrix (or Zero Matrix)** of order $2 \times 2$.
- **(b)** $B = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$: A diagonal matrix whose non-zero diagonal entries are all identical constants ($k=5 \neq 0, 1$) $\implies$ **Scalar Matrix**.
- **(c)** $C = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$: A scalar matrix whose diagonal entries are all $1$ $\implies$ **Identity Matrix (or Unit Matrix)** denoted by $I_2$.
- **(d)** $D = \begin{bmatrix} 3 & 0 \\ 0 & 7 \end{bmatrix}$: Non-diagonal elements are 0, and diagonal elements are non-zero but distinct $\implies$ **Diagonal Matrix**.
- **(e)** $E = \begin{bmatrix} 4 & 5 & 6 \end{bmatrix}$: Contains only 1 horizontal row $\implies$ **Row Matrix** (order $1 \times 3$).
- **(a)** $A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$: All entries are zero $\implies$ **Null Matrix (or Zero Matrix)** of order $2 \times 2$.
- **(b)** $B = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$: A diagonal matrix whose non-zero diagonal entries are all identical constants ($k=5 \neq 0, 1$) $\implies$ **Scalar Matrix**.
- **(c)** $C = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$: A scalar matrix whose diagonal entries are all $1$ $\implies$ **Identity Matrix (or Unit Matrix)** denoted by $I_2$.
- **(d)** $D = \begin{bmatrix} 3 & 0 \\ 0 & 7 \end{bmatrix}$: Non-diagonal elements are 0, and diagonal elements are non-zero but distinct $\implies$ **Diagonal Matrix**.
- **(e)** $E = \begin{bmatrix} 4 & 5 & 6 \end{bmatrix}$: Contains only 1 horizontal row $\implies$ **Row Matrix** (order $1 \times 3$).
Sample Question 5
Exercise 3.1 • Symmetric & Skew-Symmetric Matrices
MEDIUM • Short Question
Determine whether each of the following matrices is Symmetric, Skew-Symmetric, or neither:
(a) $A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$
(b) $B = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$
(c) $C = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix}$
(d) $D = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$
(e) $E = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix}$
(a) $A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$
(b) $B = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$
(c) $C = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix}$
(d) $D = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$
(e) $E = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix}$
✓ Correct Answer: (a) Symmetric; (b) Skew-Symmetric; (c) Skew-Symmetric; (d) Neither; (e) Symmetric
📖 Step-by-Step Solution & Conceptual Rationale:
**Rules:**
- Matrix is **Symmetric** if $M^t = M$.
- Matrix is **Skew-Symmetric** if $M^t = -M$ (and main diagonal entries are all 0).
- **(a)** $A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \implies A^t = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = A$.
- **Conclusion:** **Symmetric Matrix**.
- **(b)** $B = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \implies B^t = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} = -B$.
- **Conclusion:** **Skew-Symmetric Matrix**.
- **(c)** $C = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix} \implies C^t = \begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix} = -C$.
- **Conclusion:** **Skew-Symmetric Matrix**.
- **(d)** $D = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \implies D^t = \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} \neq D \text{ and } \neq -D$.
- **Conclusion:** **Neither Symmetric nor Skew-Symmetric**.
- **(e)** $E = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix} \implies E^t = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix} = E$.
- **Conclusion:** **Symmetric Matrix**.
- Matrix is **Symmetric** if $M^t = M$.
- Matrix is **Skew-Symmetric** if $M^t = -M$ (and main diagonal entries are all 0).
- **(a)** $A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \implies A^t = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = A$.
- **Conclusion:** **Symmetric Matrix**.
- **(b)** $B = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \implies B^t = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} = -B$.
- **Conclusion:** **Skew-Symmetric Matrix**.
- **(c)** $C = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix} \implies C^t = \begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix} = -C$.
- **Conclusion:** **Skew-Symmetric Matrix**.
- **(d)** $D = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \implies D^t = \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} \neq D \text{ and } \neq -D$.
- **Conclusion:** **Neither Symmetric nor Skew-Symmetric**.
- **(e)** $E = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix} \implies E^t = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix} = E$.
- **Conclusion:** **Symmetric Matrix**.
Sample Question 6
Exercise 3.1 • Symmetric & Skew-Symmetric Matrix Construction
MEDIUM • Short Question
Write one example each of a Symmetric matrix and a Skew-Symmetric matrix of order $3 \times 3$.
✓ Correct Answer: Symmetric: e.g. $\begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$; Skew-Symmetric: e.g. $\begin{bmatrix} 0 & e & f \\ -e & 0 & g \\ -f & -g & 0 \end{bmatrix}$
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Construction:**
- **1. Symmetric Matrix of order $3 \times 3$:**
- Must satisfy $A^t = A$ ($a_{ij} = a_{ji}$).
- Let $A = \begin{bmatrix} 1 & 4 & 5 \\ 4 & 2 & 6 \\ 5 & 6 & 3 \end{bmatrix}$.
- Transpose: $A^t = \begin{bmatrix} 1 & 4 & 5 \\ 4 & 2 & 6 \\ 5 & 6 & 3 \end{bmatrix} = A$.
- Hence, $A$ is **Symmetric**.
- **2. Skew-Symmetric Matrix of order $3 \times 3$:**
- Must satisfy $B^t = -B$ ($b_{ij} = -b_{ji}$ and $b_{ii} = 0$).
- Let $B = \begin{bmatrix} 0 & 3 & -4 \\ -3 & 0 & 2 \\ 4 & -2 & 0 \end{bmatrix}$.
- Transpose: $B^t = \begin{bmatrix} 0 & -3 & 4 \\ 3 & 0 & -2 \\ -4 & 2 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & 3 & -4 \\ -3 & 0 & 2 \\ 4 & -2 & 0 \end{bmatrix} = -B$.
- Hence, $B$ is **Skew-Symmetric**.
- **1. Symmetric Matrix of order $3 \times 3$:**
- Must satisfy $A^t = A$ ($a_{ij} = a_{ji}$).
- Let $A = \begin{bmatrix} 1 & 4 & 5 \\ 4 & 2 & 6 \\ 5 & 6 & 3 \end{bmatrix}$.
- Transpose: $A^t = \begin{bmatrix} 1 & 4 & 5 \\ 4 & 2 & 6 \\ 5 & 6 & 3 \end{bmatrix} = A$.
- Hence, $A$ is **Symmetric**.
- **2. Skew-Symmetric Matrix of order $3 \times 3$:**
- Must satisfy $B^t = -B$ ($b_{ij} = -b_{ji}$ and $b_{ii} = 0$).
- Let $B = \begin{bmatrix} 0 & 3 & -4 \\ -3 & 0 & 2 \\ 4 & -2 & 0 \end{bmatrix}$.
- Transpose: $B^t = \begin{bmatrix} 0 & -3 & 4 \\ 3 & 0 & -2 \\ -4 & 2 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & 3 & -4 \\ -3 & 0 & 2 \\ 4 & -2 & 0 \end{bmatrix} = -B$.
- Hence, $B$ is **Skew-Symmetric**.
Sample Question 7
Exercise 3.2 • Finding Unknowns from Matrix Equations
MEDIUM • Long Question
Find the values of the unknown variables in each of the following matrix equality equations:
(a) $\begin{bmatrix} x+3 & 1 \\ 4 & y-2 \end{bmatrix} = \begin{bmatrix} 5 & 1 \\ 4 & 6 \end{bmatrix}$
(b) $\begin{bmatrix} 2x & y+1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ 3 & 4 \end{bmatrix}$
(c) $\begin{bmatrix} x-y & 2 \\ 3 & x+y \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix}$
(d) $\begin{bmatrix} 3x & 2y \\ 4z & 5 \end{bmatrix} = \begin{bmatrix} 9 & -6 \\ 12 & 5 \end{bmatrix}$
(e) $\begin{bmatrix} x+y & 3 \\ 2 & x-y \end{bmatrix} = \begin{bmatrix} 8 & 3 \\ 2 & 2 \end{bmatrix}$
(f) $\begin{bmatrix} 2x+3 & y-4 \\ 7 & 2 \end{bmatrix} = \begin{bmatrix} 11 & 1 \\ 7 & 2 \end{bmatrix}$
(g) $\begin{bmatrix} x^2 & 4 \\ 2 & y+3 \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 2 & 7 \end{bmatrix}$
(h) $\begin{bmatrix} x+2y & 5 \\ 1 & 2x-y \end{bmatrix} = \begin{bmatrix} 7 & 5 \\ 1 & 4 \end{bmatrix}$
(i) $\begin{bmatrix} 3x-y & 4 \\ 0 & x+2y \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 0 & 4 \end{bmatrix}$
(j) $\begin{bmatrix} 2x+y & 3 \\ -1 & x-2y \end{bmatrix} = \begin{bmatrix} 7 & 3 \\ -1 & 1 \end{bmatrix}$
(a) $\begin{bmatrix} x+3 & 1 \\ 4 & y-2 \end{bmatrix} = \begin{bmatrix} 5 & 1 \\ 4 & 6 \end{bmatrix}$
(b) $\begin{bmatrix} 2x & y+1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ 3 & 4 \end{bmatrix}$
(c) $\begin{bmatrix} x-y & 2 \\ 3 & x+y \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix}$
(d) $\begin{bmatrix} 3x & 2y \\ 4z & 5 \end{bmatrix} = \begin{bmatrix} 9 & -6 \\ 12 & 5 \end{bmatrix}$
(e) $\begin{bmatrix} x+y & 3 \\ 2 & x-y \end{bmatrix} = \begin{bmatrix} 8 & 3 \\ 2 & 2 \end{bmatrix}$
(f) $\begin{bmatrix} 2x+3 & y-4 \\ 7 & 2 \end{bmatrix} = \begin{bmatrix} 11 & 1 \\ 7 & 2 \end{bmatrix}$
(g) $\begin{bmatrix} x^2 & 4 \\ 2 & y+3 \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 2 & 7 \end{bmatrix}$
(h) $\begin{bmatrix} x+2y & 5 \\ 1 & 2x-y \end{bmatrix} = \begin{bmatrix} 7 & 5 \\ 1 & 4 \end{bmatrix}$
(i) $\begin{bmatrix} 3x-y & 4 \\ 0 & x+2y \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 0 & 4 \end{bmatrix}$
(j) $\begin{bmatrix} 2x+y & 3 \\ -1 & x-2y \end{bmatrix} = \begin{bmatrix} 7 & 3 \\ -1 & 1 \end{bmatrix}$
✓ Correct Answer: (a) x=2, y=8; (b) x=4, y=4; (c) x=4, y=3; (d) x=3, y=-3, z=3; (e) x=5, y=3; (f) x=4, y=5; (g) x=±3, y=4; (h) x=3, y=2; (i) x=2, y=1; (j) x=3, y=1
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Algebraic Resolutions:**
- **(a)** $\begin{bmatrix} x+3 & 1 \\ 4 & y-2 \end{bmatrix} = \begin{bmatrix} 5 & 1 \\ 4 & 6 \end{bmatrix}$:
- Equating entries: $x + 3 = 5 \implies x = 5 - 3 = 2$.
- $y - 2 = 6 \implies y = 6 + 2 = 8$.
- **Result:** $x = 2, y = 8$.
- **(b)** $\begin{bmatrix} 2x & y+1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ 3 & 4 \end{bmatrix}$:
- $2x = 8 \implies x = 4$.
- $y + 1 = 5 \implies y = 4$.
- **Result:** $x = 4, y = 4$.
- **(c)** $\begin{bmatrix} x-y & 2 \\ 3 & x+y \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix}$:
- $x - y = 1$ ...(1) and $x + y = 7$ ...(2).
- Adding (1) & (2): $2x = 8 \implies x = 4$.
- From (2): $4 + y = 7 \implies y = 3$.
- **Result:** $x = 4, y = 3$.
- **(d)** $\begin{bmatrix} 3x & 2y \\ 4z & 5 \end{bmatrix} = \begin{bmatrix} 9 & -6 \\ 12 & 5 \end{bmatrix}$:
- $3x = 9 \implies x = 3$.
- $2y = -6 \implies y = -3$.
- $4z = 12 \implies z = 3$.
- **Result:** $x = 3, y = -3, z = 3$.
- **(e)** $\begin{bmatrix} x+y & 3 \\ 2 & x-y \end{bmatrix} = \begin{bmatrix} 8 & 3 \\ 2 & 2 \end{bmatrix}$:
- $x + y = 8$ ...(1) and $x - y = 2$ ...(2).
- Adding: $2x = 10 \implies x = 5$.
- Subtraction: $2y = 6 \implies y = 3$.
- **Result:** $x = 5, y = 3$.
- **(f)** $\begin{bmatrix} 2x+3 & y-4 \\ 7 & 2 \end{bmatrix} = \begin{bmatrix} 11 & 1 \\ 7 & 2 \end{bmatrix}$:
- $2x + 3 = 11 \implies 2x = 8 \implies x = 4$.
- $y - 4 = 1 \implies y = 5$.
- **Result:** $x = 4, y = 5$.
- **(g)** $\begin{bmatrix} x^2 & 4 \\ 2 & y+3 \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 2 & 7 \end{bmatrix}$:
- $x^2 = 9 \implies x = \pm \sqrt{9} = \pm 3$.
- $y + 3 = 7 \implies y = 4$.
- **Result:** $x = \pm 3, y = 4$.
- **(h)** $\begin{bmatrix} x+2y & 5 \\ 1 & 2x-y \end{bmatrix} = \begin{bmatrix} 7 & 5 \\ 1 & 4 \end{bmatrix}$:
- $x + 2y = 7$ ...(1) and $2x - y = 4 \implies y = 2x - 4$.
- Substitute into (1): $x + 2(2x - 4) = 7 \implies 5x - 8 = 7 \implies 5x = 15 \implies x = 3$.
- Then $y = 2(3) - 4 = 2$.
- **Result:** $x = 3, y = 2$.
- **(i)** $\begin{bmatrix} 3x-y & 4 \\ 0 & x+2y \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 0 & 4 \end{bmatrix}$:
- $3x - y = 5 \implies y = 3x - 5$.
- $x + 2y = 4 \implies x + 2(3x - 5) = 4 \implies 7x - 10 = 4 \implies 7x = 14 \implies x = 2$.
- $y = 3(2) - 5 = 1$.
- **Result:** $x = 2, y = 1$.
- **(j)** $\begin{bmatrix} 2x+y & 3 \\ -1 & x-2y \end{bmatrix} = \begin{bmatrix} 7 & 3 \\ -1 & 1 \end{bmatrix}$:
- $2x + y = 7$ ...(1) and $x - 2y = 1 \implies x = 2y + 1$.
- Substitute into (1): $2(2y + 1) + y = 7 \implies 5y + 2 = 7 \implies 5y = 5 \implies y = 1$.
- $x = 2(1) + 1 = 3$.
- **Result:** $x = 3, y = 1$.
- **(a)** $\begin{bmatrix} x+3 & 1 \\ 4 & y-2 \end{bmatrix} = \begin{bmatrix} 5 & 1 \\ 4 & 6 \end{bmatrix}$:
- Equating entries: $x + 3 = 5 \implies x = 5 - 3 = 2$.
- $y - 2 = 6 \implies y = 6 + 2 = 8$.
- **Result:** $x = 2, y = 8$.
- **(b)** $\begin{bmatrix} 2x & y+1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ 3 & 4 \end{bmatrix}$:
- $2x = 8 \implies x = 4$.
- $y + 1 = 5 \implies y = 4$.
- **Result:** $x = 4, y = 4$.
- **(c)** $\begin{bmatrix} x-y & 2 \\ 3 & x+y \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix}$:
- $x - y = 1$ ...(1) and $x + y = 7$ ...(2).
- Adding (1) & (2): $2x = 8 \implies x = 4$.
- From (2): $4 + y = 7 \implies y = 3$.
- **Result:** $x = 4, y = 3$.
- **(d)** $\begin{bmatrix} 3x & 2y \\ 4z & 5 \end{bmatrix} = \begin{bmatrix} 9 & -6 \\ 12 & 5 \end{bmatrix}$:
- $3x = 9 \implies x = 3$.
- $2y = -6 \implies y = -3$.
- $4z = 12 \implies z = 3$.
- **Result:** $x = 3, y = -3, z = 3$.
- **(e)** $\begin{bmatrix} x+y & 3 \\ 2 & x-y \end{bmatrix} = \begin{bmatrix} 8 & 3 \\ 2 & 2 \end{bmatrix}$:
- $x + y = 8$ ...(1) and $x - y = 2$ ...(2).
- Adding: $2x = 10 \implies x = 5$.
- Subtraction: $2y = 6 \implies y = 3$.
- **Result:** $x = 5, y = 3$.
- **(f)** $\begin{bmatrix} 2x+3 & y-4 \\ 7 & 2 \end{bmatrix} = \begin{bmatrix} 11 & 1 \\ 7 & 2 \end{bmatrix}$:
- $2x + 3 = 11 \implies 2x = 8 \implies x = 4$.
- $y - 4 = 1 \implies y = 5$.
- **Result:** $x = 4, y = 5$.
- **(g)** $\begin{bmatrix} x^2 & 4 \\ 2 & y+3 \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 2 & 7 \end{bmatrix}$:
- $x^2 = 9 \implies x = \pm \sqrt{9} = \pm 3$.
- $y + 3 = 7 \implies y = 4$.
- **Result:** $x = \pm 3, y = 4$.
- **(h)** $\begin{bmatrix} x+2y & 5 \\ 1 & 2x-y \end{bmatrix} = \begin{bmatrix} 7 & 5 \\ 1 & 4 \end{bmatrix}$:
- $x + 2y = 7$ ...(1) and $2x - y = 4 \implies y = 2x - 4$.
- Substitute into (1): $x + 2(2x - 4) = 7 \implies 5x - 8 = 7 \implies 5x = 15 \implies x = 3$.
- Then $y = 2(3) - 4 = 2$.
- **Result:** $x = 3, y = 2$.
- **(i)** $\begin{bmatrix} 3x-y & 4 \\ 0 & x+2y \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 0 & 4 \end{bmatrix}$:
- $3x - y = 5 \implies y = 3x - 5$.
- $x + 2y = 4 \implies x + 2(3x - 5) = 4 \implies 7x - 10 = 4 \implies 7x = 14 \implies x = 2$.
- $y = 3(2) - 5 = 1$.
- **Result:** $x = 2, y = 1$.
- **(j)** $\begin{bmatrix} 2x+y & 3 \\ -1 & x-2y \end{bmatrix} = \begin{bmatrix} 7 & 3 \\ -1 & 1 \end{bmatrix}$:
- $2x + y = 7$ ...(1) and $x - 2y = 1 \implies x = 2y + 1$.
- Substitute into (1): $2(2y + 1) + y = 7 \implies 5y + 2 = 7 \implies 5y = 5 \implies y = 1$.
- $x = 2(1) + 1 = 3$.
- **Result:** $x = 3, y = 1$.
Sample Question 8
Exercise 3.2 • Additive Inverses of Matrices
MEDIUM • Short Question
Find the additive inverse of each of the following matrices:
$R = \begin{bmatrix} 2 & -3 \\ 4 & 1 \end{bmatrix}, \quad S = \begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 4 \end{bmatrix}, \quad T = \begin{bmatrix} -5 \\ 7 \\ 0 \end{bmatrix}$
$R = \begin{bmatrix} 2 & -3 \\ 4 & 1 \end{bmatrix}, \quad S = \begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 4 \end{bmatrix}, \quad T = \begin{bmatrix} -5 \\ 7 \\ 0 \end{bmatrix}$
✓ Correct Answer: $-R = \begin{bmatrix} -2 & 3 \\ -4 & -1 \end{bmatrix}$, $-S = \begin{bmatrix} -1 & 0 & 2 \\ -3 & 1 & -4 \end{bmatrix}$, $-T = \begin{bmatrix} 5 \\ -7 \\ 0 \end{bmatrix}$
📖 Step-by-Step Solution & Conceptual Rationale:
**Definition:** The additive inverse of matrix $M$ is $-M = (-1) \cdot M$, obtained by reversing the sign of every entry.
- **1. Additive Inverse of $R$:**
$$-R = -\begin{bmatrix} 2 & -3 \\ 4 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 3 \\ -4 & -1 \end{bmatrix}$$
*(Check: $R + (-R) = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$)*
- **2. Additive Inverse of $S$:**
$$-S = -\begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 4 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 2 \\ -3 & 1 & -4 \end{bmatrix}$$
- **3. Additive Inverse of $T$:**
$$-T = -\begin{bmatrix} -5 \\ 7 \\ 0 \end{bmatrix} = \begin{bmatrix} 5 \\ -7 \\ 0 \end{bmatrix}$$
- **1. Additive Inverse of $R$:**
$$-R = -\begin{bmatrix} 2 & -3 \\ 4 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 3 \\ -4 & -1 \end{bmatrix}$$
*(Check: $R + (-R) = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$)*
- **2. Additive Inverse of $S$:**
$$-S = -\begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 4 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 2 \\ -3 & 1 & -4 \end{bmatrix}$$
- **3. Additive Inverse of $T$:**
$$-T = -\begin{bmatrix} -5 \\ 7 \\ 0 \end{bmatrix} = \begin{bmatrix} 5 \\ -7 \\ 0 \end{bmatrix}$$
Sample Question 9
Exercise 3.2 • Linear Combinations of Matrices
MEDIUM • Short Question
If $A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}, B = \begin{bmatrix} 0 & 2 \\ -1 & 4 \end{bmatrix}, C = \begin{bmatrix} 3 & -3 \\ 6 & 0 \end{bmatrix}$, evaluate:
(i) $2A + \frac{1}{2}B - \frac{1}{3}C$
(ii) $A - \frac{1}{2}B$
(i) $2A + \frac{1}{2}B - \frac{1}{3}C$
(ii) $A - \frac{1}{2}B$
✓ Correct Answer: (i) $\begin{bmatrix} 3 & 4 \\ 3.5 & 10 \end{bmatrix}$; (ii) $\begin{bmatrix} 2 & 0 \\ 3.5 & 2 \end{bmatrix}$
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Evaluation:**
- **(i) Evaluate $2A + \frac{1}{2}B - \frac{1}{3}C$:**
- $2A = 2 \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 2 \\ 6 & 8 \end{bmatrix}$
- $\frac{1}{2}B = \frac{1}{2} \begin{bmatrix} 0 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -0.5 & 2 \end{bmatrix}$
- $\frac{1}{3}C = \frac{1}{3} \begin{bmatrix} 3 & -3 \\ 6 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 2 & 0 \end{bmatrix}$
- Combining terms:
$$2A + \frac{1}{2}B - \frac{1}{3}C = \begin{bmatrix} 4+0-1 & 2+1-(-1) \\ 6-0.5-2 & 8+2-0 \end{bmatrix} = \begin{bmatrix} 3 & 4 \\ 3.5 & 10 \end{bmatrix} = \begin{bmatrix} 3 & 4 \\ \frac{7}{2} & 10 \end{bmatrix}$$
- **(ii) Evaluate $A - \frac{1}{2}B$:**
$$A - \frac{1}{2}B = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 0 & 1 \\ -\frac{1}{2} & 2 \end{bmatrix} = \begin{bmatrix} 2-0 & 1-1 \\ 3 - (-\frac{1}{2}) & 4-2 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ \frac{7}{2} & 2 \end{bmatrix}$$
- **(i) Evaluate $2A + \frac{1}{2}B - \frac{1}{3}C$:**
- $2A = 2 \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 2 \\ 6 & 8 \end{bmatrix}$
- $\frac{1}{2}B = \frac{1}{2} \begin{bmatrix} 0 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -0.5 & 2 \end{bmatrix}$
- $\frac{1}{3}C = \frac{1}{3} \begin{bmatrix} 3 & -3 \\ 6 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 2 & 0 \end{bmatrix}$
- Combining terms:
$$2A + \frac{1}{2}B - \frac{1}{3}C = \begin{bmatrix} 4+0-1 & 2+1-(-1) \\ 6-0.5-2 & 8+2-0 \end{bmatrix} = \begin{bmatrix} 3 & 4 \\ 3.5 & 10 \end{bmatrix} = \begin{bmatrix} 3 & 4 \\ \frac{7}{2} & 10 \end{bmatrix}$$
- **(ii) Evaluate $A - \frac{1}{2}B$:**
$$A - \frac{1}{2}B = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 0 & 1 \\ -\frac{1}{2} & 2 \end{bmatrix} = \begin{bmatrix} 2-0 & 1-1 \\ 3 - (-\frac{1}{2}) & 4-2 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ \frac{7}{2} & 2 \end{bmatrix}$$
Sample Question 10
Exercise 3.2 • Matrix Conformability & Arithmetic
MEDIUM • Short Question
Given $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, B = \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix}, C = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}, D = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}$:
(i) Compute $A + B$ and $B + A$
(ii) Explain why $A + C$ is not possible
(iii) Compute $C \cdot D$
(iv) Compute $D \cdot C$
(i) Compute $A + B$ and $B + A$
(ii) Explain why $A + C$ is not possible
(iii) Compute $C \cdot D$
(iv) Compute $D \cdot C$
✓ Correct Answer: (i) $\begin{bmatrix} 3 & 2 \\ 4 & 7 \end{bmatrix}$; (ii) Different orders ($2x2 \neq 1x3$); (iii) $[32]$; (iv) $3x3$ matrix
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Resolution:**
- **(i)** $A + B = \begin{bmatrix} 1+2 & 2+0 \\ 3+1 & 4+3 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 4 & 7 \end{bmatrix}$.
- $B + A = \begin{bmatrix} 2+1 & 0+2 \\ 1+3 & 3+4 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 4 & 7 \end{bmatrix}$. ($A+B = B+A$).
- **(ii) Conformability for Addition:**
- Order of $A = 2 \times 2$. Order of $C = 1 \times 3$.
- Addition is defined only for matrices of the same order. Since $2 \times 2 \neq 1 \times 3$, $A+C$ is **not conformable for addition**.
- **(iii) Product $C \cdot D$:**
- Order: $(1 \times 3) \times (3 \times 1) \implies 1 \times 1$.
$$C \cdot D = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix} = [1(4) + 2(5) + 3(6)] = [4 + 10 + 18] = [32]$$
- **(iv) Product $D \cdot C$:**
- Order: $(3 \times 1) \times (1 \times 3) \implies 3 \times 3$.
$$D \cdot C = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 4(1) & 4(2) & 4(3) \\ 5(1) & 5(2) & 5(3) \\ 6(1) & 6(2) & 6(3) \end{bmatrix} = \begin{bmatrix} 4 & 8 & 12 \\ 5 & 10 & 15 \\ 6 & 12 & 18 \end{bmatrix}$$
- **(i)** $A + B = \begin{bmatrix} 1+2 & 2+0 \\ 3+1 & 4+3 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 4 & 7 \end{bmatrix}$.
- $B + A = \begin{bmatrix} 2+1 & 0+2 \\ 1+3 & 3+4 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 4 & 7 \end{bmatrix}$. ($A+B = B+A$).
- **(ii) Conformability for Addition:**
- Order of $A = 2 \times 2$. Order of $C = 1 \times 3$.
- Addition is defined only for matrices of the same order. Since $2 \times 2 \neq 1 \times 3$, $A+C$ is **not conformable for addition**.
- **(iii) Product $C \cdot D$:**
- Order: $(1 \times 3) \times (3 \times 1) \implies 1 \times 1$.
$$C \cdot D = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix} = [1(4) + 2(5) + 3(6)] = [4 + 10 + 18] = [32]$$
- **(iv) Product $D \cdot C$:**
- Order: $(3 \times 1) \times (1 \times 3) \implies 3 \times 3$.
$$D \cdot C = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 4(1) & 4(2) & 4(3) \\ 5(1) & 5(2) & 5(3) \\ 6(1) & 6(2) & 6(3) \end{bmatrix} = \begin{bmatrix} 4 & 8 & 12 \\ 5 & 10 & 15 \\ 6 & 12 & 18 \end{bmatrix}$$
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