Matrices: Detailed Notes, Solved Examples & Exam Questions | Grade 11 Mathematics Unit 3
Welcome, dear student! In this lesson, we will learn about Matrices — one of the most important topics in Grade 11 Mathematics. Matrices are used everywhere: in computer graphics, solving equations, economics, and even in your phone’s camera filters! So let us understand them step by step. Ready? Let’s begin!
1. What is a Matrix?
A matrix (plural: matrices) is a rectangular arrangement of numbers, symbols, or expressions arranged in rows and columns. We always write matrices inside brackets — either square brackets \([ \ ]\) or parentheses \((\ )\).
Each number inside a matrix is called an element (or entry). We identify each element by its position — which row and which column it sits in.
Example: Consider the matrix:
This matrix has 2 rows and 3 columns. We say it is a “2 by 3” matrix, written as \(2 \times 3\).
- \(a_{11} = 3\) (row 1, column 1)
- \(a_{12} = 5\) (row 1, column 2)
- \(a_{23} = 4\) (row 2, column 3)
Think about it: if a matrix has \(m\) rows and \(n\) columns, its order (or size) is \(m \times n\). Can you tell the order of a matrix with 4 rows and 1 column? Yes — it is \(4 \times 1\). Good!
- A matrix is a rectangular array of numbers in rows and columns.
- The order is written as \(m \times n\) where \(m\) = number of rows, \(n\) = number of columns.
- Element \(a_{ij}\) is in row \(i\), column \(j\).
- A matrix with \(m\) rows and \(n\) columns has \(m \times n\) elements.
Q1. A matrix \(B\) has 5 rows and 2 columns. What is its order? How many elements does it have?
Answer: The order of matrix \(B\) is \(5 \times 2\). The total number of elements is \(5 \times 2 = 10\).
Remember: number of elements = rows × columns. Always!
Q2. In the matrix below, what is the value of \(a_{32}\)?
Answer: \(a_{32}\) means row 3, column 2. Looking at the matrix, the element in the 3rd row and 2nd column is 4.
So \(a_{32} = 4\).
Q3. If a matrix has 24 elements, list all possible orders it can have.
Answer: We need all pairs \((m, n)\) such that \(m \times n = 24\), where \(m\) and \(n\) are positive integers.
The possible orders are: \(1 \times 24\), \(2 \times 12\), \(3 \times 8\), \(4 \times 6\), \(6 \times 4\), \(8 \times 3\), \(12 \times 2\), \(24 \times 1\).
Note: In exams, always list both \(m \times n\) and \(n \times m\) as separate orders (unless stated otherwise)!
2. Types of Matrices
There are several special types of matrices you must know. Let us go through each one carefully with examples.
2.1 Row Matrix
A matrix that has exactly one row is called a row matrix. Its order is \(1 \times n\).
2.2 Column Matrix
A matrix that has exactly one column is called a column matrix. Its order is \(m \times 1\).
2.3 Square Matrix
A matrix in which the number of rows equals the number of columns is called a square matrix. Its order is \(n \times n\).
In a square matrix, the elements \(a_{11}, a_{22}, a_{33}, \ldots\) form the principal diagonal (or main diagonal).
2.4 Diagonal Matrix
A square matrix in which all elements outside the principal diagonal are zero is called a diagonal matrix.
2.5 Scalar Matrix
A diagonal matrix in which all the diagonal elements are equal is called a scalar matrix.
2.6 Identity Matrix (Unit Matrix)
A scalar matrix in which the diagonal elements are all 1 is called an identity matrix, denoted by \(I\).
The identity matrix acts like the number 1 in matrix multiplication — any matrix multiplied by \(I\) stays the same! We will see this later.
2.7 Zero Matrix (Null Matrix)
A matrix in which all elements are zero is called a zero matrix, denoted by \(O\).
2.8 Symmetric Matrix
A square matrix is symmetric if \(A = A^T\), meaning the element in row \(i\), column \(j\) equals the element in row \(j\), column \(i\). In other words, \(a_{ij} = a_{ji}\) for all \(i, j\).
Notice: \(a_{12} = 5 = a_{21}\), \(a_{13} = 2 = a_{31}\), \(a_{23} = 6 = a_{32}\). So this is symmetric!
2.9 Skew-Symmetric Matrix
A square matrix is skew-symmetric if \(A = -A^T\), meaning \(a_{ij} = -a_{ji}\) for all \(i, j\). This also means all diagonal elements must be zero (since \(a_{ii} = -a_{ii} \Rightarrow a_{ii} = 0\)).
- Row matrix: 1 row, \(1 \times n\)
- Column matrix: 1 column, \(m \times 1\)
- Square matrix: rows = columns, \(n \times n\)
- Diagonal matrix: square + off-diagonal elements = 0
- Scalar matrix: diagonal + all diagonal elements equal
- Identity matrix: scalar matrix with all 1s on diagonal
- Zero matrix: all elements are 0
- Symmetric: \(A = A^T\) (i.e., \(a_{ij} = a_{ji}\))
- Skew-symmetric: \(A = -A^T\) (i.e., \(a_{ij} = -a_{ji}\), diagonal = 0)
Q1. Identify the type of each matrix:
(a) \(\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\) (b) \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) (c) \(\begin{pmatrix} 5 \\ 2 \\ 8 \end{pmatrix}\)
Answer:
(a) This is a zero matrix (all elements are 0), and it is also a square matrix and a diagonal matrix.
(b) This is an identity matrix of order \(3 \times 3\), denoted \(I_3\). It is also diagonal, scalar, and symmetric.
(c) This is a column matrix of order \(3 \times 1\).
Q2. Is the matrix \(A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{pmatrix}\) symmetric? Explain.
Answer: Yes, it is symmetric. Let us verify:
\(a_{12} = 2 = a_{21}\) ✓ \(a_{13} = 3 = a_{31}\) ✓ \(a_{23} = 5 = a_{32}\) ✓
Since \(a_{ij} = a_{ji}\) for all \(i, j\), the matrix \(A\) is symmetric. So \(A = A^T\).
Q3. Construct a \(3 \times 3\) skew-symmetric matrix with \(a_{12} = 4\) and \(a_{13} = -1\).
Answer: For a skew-symmetric matrix, \(a_{ij} = -a_{ji}\) and diagonal elements are 0.
Given: \(a_{12} = 4\), so \(a_{21} = -4\).
Given: \(a_{13} = -1\), so \(a_{31} = 1\).
We also need \(a_{23}\) and \(a_{32}\). Since no value is given, let us say \(a_{23} = k\), then \(a_{32} = -k\). For simplicity, let \(k = 0\):
(You could also use any value for \(a_{23}\) — the question only specifies two off-diagonal elements.)
3. Transpose of a Matrix
The transpose of a matrix \(A\), written as \(A^T\) (or \(A’\)), is obtained by interchanging rows and columns. That is, the first row becomes the first column, the second row becomes the second column, and so on.
If \(A\) has order \(m \times n\), then \(A^T\) has order \(n \times m\).
Worked Example:
Find the transpose of:
Solution: Interchange rows and columns:
See? The first row \((2, 5, 1)\) became the first column, and the second row \((3, 7, 4)\) became the second column.
- \((A^T)^T = A\)
- \((A + B)^T = A^T + B^T\)
- \((kA)^T = kA^T\) (where \(k\) is a scalar)
- \((AB)^T = B^T A^T\) — note the reversal of order!
- Transpose = interchange rows and columns.
- Order changes from \(m \times n\) to \(n \times m\).
- \((AB)^T = B^T A^T\) (reverse order — very common exam trap!)
- Symmetric matrix: \(A = A^T\); Skew-symmetric: \(A = -A^T\).
Q1. Find the transpose of \(A = \begin{pmatrix} 1 & -2 & 3 \\ 0 & 5 & 7 \\ -1 & 4 & 6 \end{pmatrix}\) and verify that \(A\) is not symmetric.
Answer:
Compare \(A\) and \(A^T\): \(a_{12} = -2\) but \(a_{21}^T = 0\). Since \(-2 \neq 0\), we have \(A \neq A^T\). Therefore, \(A\) is not symmetric.
Q2. If \(A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\) and \(B = \begin{pmatrix} 0 & 1 \\ 5 & 6 \end{pmatrix}\), find \((A + B)^T\) and verify that it equals \(A^T + B^T\).
Answer:
First, \(A + B = \begin{pmatrix} 1+0 & 2+1 \\ 3+5 & 4+6 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 8 & 10 \end{pmatrix}\)
So \((A + B)^T = \begin{pmatrix} 1 & 8 \\ 3 & 10 \end{pmatrix}\)
Now, \(A^T = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}\) and \(B^T = \begin{pmatrix} 0 & 5 \\ 1 & 6 \end{pmatrix}\)
\(A^T + B^T = \begin{pmatrix} 1 & 8 \\ 3 & 10 \end{pmatrix}\)
Since \((A + B)^T = A^T + B^T\), the property is verified! ✓
Q3. If \(A\) is a \(3 \times 5\) matrix, what is the order of \(A^T\)? Can you multiply \(A\) by \(A^T\)? If yes, what is the order of the product?
Answer:
\(A\) is \(3 \times 5\), so \(A^T\) is \(5 \times 3\).
Can we multiply \(A \times A^T\)? \(A\) is \(3 \times 5\) and \(A^T\) is \(5 \times 3\). The inner dimensions match (5 = 5), so yes!
The product \(AA^T\) will have order \(3 \times 3\). (Outer dimensions: 3 and 3.)
Note: We can also multiply \(A^T \times A\), giving a \(5 \times 5\) matrix.
4. Matrix Addition and Subtraction
Two matrices can be added or subtracted only if they have the same order (same number of rows and same number of columns). We add or subtract corresponding elements.
Worked Example 1:
Find \(A + B\) and \(A – B\).
Solution:
Worked Example 2:
Find \(A + B\).
Solution:
- Commutative: \(A + B = B + A\)
- Associative: \((A + B) + C = A + (B + C)\)
- Existence of zero: \(A + O = A\) (where \(O\) is the zero matrix of same order)
- Existence of additive inverse: \(A + (-A) = O\)
- Matrices must have the same order to be added or subtracted.
- Add/subtract corresponding elements only.
- Addition is commutative and associative.
- \(A – B \neq B – A\) in general (subtraction is NOT commutative).
Q1. If \(A = \begin{pmatrix} 5 & -3 \\ 2 & 7 \end{pmatrix}\) and \(B = \begin{pmatrix} -1 & 4 \\ 3 & -2 \end{pmatrix}\), find \(3A – 2B\).
Answer:
First, find \(3A = \begin{pmatrix} 15 & -9 \\ 6 & 21 \end{pmatrix}\)
Then, find \(2B = \begin{pmatrix} -2 & 8 \\ 6 & -4 \end{pmatrix}\)
Now, \(3A – 2B = \begin{pmatrix} 15-(-2) & -9-8 \\ 6-6 & 21-(-4) \end{pmatrix} = \begin{pmatrix} 17 & -17 \\ 0 & 25 \end{pmatrix}\)
Q2. Can you add a \(2 \times 2\) matrix to a \(2 \times 3\) matrix? Give a reason.
Answer: No. Matrix addition is defined only for matrices of the same order. A \(2 \times 2\) matrix and a \(2 \times 3\) matrix have different numbers of columns (2 vs 3), so they cannot be added.
Q3. If \(A + B = \begin{pmatrix} 6 & 3 \\ 1 & 8 \end{pmatrix}\) and \(A = \begin{pmatrix} 2 & -1 \\ 4 & 5 \end{pmatrix}\), find matrix \(B\).
Answer: \(B = (A + B) – A\)
5. Scalar Multiplication of a Matrix
To multiply a matrix by a scalar (a regular number), we multiply every element of the matrix by that scalar.
Worked Example:
If \(A = \begin{pmatrix} 2 & -1 & 3 \\ 0 & 5 & 4 \end{pmatrix}\), find \(3A\) and \(-2A\).
Solution:
- \(k(A + B) = kA + kB\)
- \((k + l)A = kA + lA\) (where \(k, l\) are scalars)
- \(k(lA) = (kl)A\)
- \(1 \cdot A = A\)
Have you noticed how scalar multiplication is like “distributing” the number to every element? It is straightforward, but be careful with negative signs — they affect every single element!
- Multiply the scalar with every element.
- Watch out for negative scalars — every element’s sign flips.
- Properties follow distributive laws.
Q1. If \(A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\), find \(\frac{1}{2}A\).
Answer:
Each element is simply divided by 2.
Q2. If \(5A = \begin{pmatrix} 10 & -15 \\ 25 & 0 \end{pmatrix}\), find \(A\).
Answer: Divide each element by 5:
Q3. Find the value of \(k\) if \(k\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 8 & 4 \\ 0 & 12 \end{pmatrix}\).
Answer: Comparing elements: \(k \times 2 = 8\), so \(k = 4\).
Let us verify: \(k \times 1 = 4\) ✓, \(k \times 0 = 0\) ✓, \(k \times 3 = 12\) ✓.
Therefore, \(k = 4\).
6. Matrix Multiplication
Matrix multiplication is one of the most important operations. It is NOT the same as multiplying element by element!
6.1 Condition for Multiplication
Two matrices \(A\) (of order \(m \times p\)) and \(B\) (of order \(p \times n\)) can be multiplied as \(AB\) only if the number of columns of \(A\) equals the number of rows of \(B\). The resulting matrix \(AB\) has order \(m \times n\).
6.2 How to Multiply
To find the element in row \(i\), column \(j\) of the product \(AB\): take the \(i\)-th row of \(A\) and the \(j\)-th column of \(B\), multiply corresponding elements, and add them up.
Worked Example 1:
Find \(AB\).
Solution:
\(AB\) has order \(2 \times 2\).
Worked Example 2: Now find \(BA\) and compare with \(AB\).
Worked Example 3: Multiplying a \(2 \times 3\) matrix by a \(3 \times 2\) matrix.
Solution: \(AB\) will be \(2 \times 2\).
6.3 Properties of Matrix Multiplication
- NOT commutative: \(AB \neq BA\) in general
- Associative: \((AB)C = A(BC)\)
- Distributive over addition: \(A(B + C) = AB + AC\)
- \(AI = IA = A\) (identity matrix acts like 1)
- \(AO = OA = O\) (zero matrix gives zero product)
- Inner dimensions must match: \(m \times \boxed{p}\) times \(\boxed{p} \times n\) gives \(m \times n\).
- Multiplication is NOT commutative: \(AB \neq BA\).
- Each element = sum of products of row of \(A\) and column of \(B\).
- \(AB = O\) does NOT mean \(A = O\) or \(B = O\).
- \(AI = IA = A\).
- \((AB)^T = B^T A^T\) (reverse order).
Q1. If \(A = \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix}\) and \(B = \begin{pmatrix} 2 & 1 \\ 0 & -2 \end{pmatrix}\), find \(AB\) and \(BA\). Is \(AB = BA\)?
Answer:
\(AB \neq BA\). Matrix multiplication is not commutative.
Q2. Can we multiply a \(3 \times 4\) matrix by a \(2 \times 3\) matrix? Explain.
Answer: No. For \(AB\), \(A\) is \(3 \times \boxed{4}\) and \(B\) is \(\boxed{2} \times 3\). The inner dimensions (4 and 2) do not match, so multiplication \(AB\) is not defined.
However, we could multiply \(BA\): \(B\) is \(2 \times \boxed{3}\) and \(A\) is \(\boxed{3} \times 4\). Inner dimensions match (3 = 3), so \(BA\) is defined and would be \(2 \times 4\).
Q3. If \(A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) and \(B = \begin{pmatrix} 3 & 5 \\ 2 & 7 \end{pmatrix}\), find \(AB\) and \(BA\). What do you notice?
Answer:
Both \(AB = B\) and \(BA = B\). This is because \(A = I_2\) (the identity matrix). The identity matrix leaves any matrix unchanged when multiplied — just like multiplying a number by 1!
7. Determinant of a Matrix
The determinant is a special number that we can calculate from a square matrix only. It tells us important things about the matrix — for example, whether the matrix has an inverse.
7.1 Determinant of a \(2 \times 2\) Matrix
For a \(2 \times 2\) matrix:
The determinant is:
Worked Example 1:
\(|A| = (3)(7) – (5)(2) = 21 – 10 = 11\)
Worked Example 2:
\(|B| = (4)(-3) – (-2)(6) = -12 + 12 = 0\)
When the determinant is zero, we say the matrix is singular (it has no inverse). When the determinant is non-zero, the matrix is non-singular (it has an inverse).
7.2 Determinant of a \(3 \times 3\) Matrix
For a \(3 \times 3\) matrix:
We expand along the first row:
Worked Example:
Solution:
Remember the sign pattern: \(\begin{pmatrix} + & – & + \\ – & + & – \\ + & – & + \end{pmatrix}\)
This alternation of signs is crucial when expanding along any row or column.
7.3 Properties of Determinants
- \(|A^T| = |A|\)
- If two rows (or columns) are swapped, the determinant changes sign.
- If two rows (or columns) are identical, \(|A| = 0\).
- \(|kA| = k^n |A|\) where \(n\) is the order of the square matrix.
- \(|AB| = |A| \cdot |B|\)
- \(|A^{-1}| = \dfrac{1}{|A|}\) (when \(|A| \neq 0\))
- Determinant is defined only for square matrices.
- For \(2 \times 2\): \(|A| = ad – bc\).
- For \(3 \times 3\): expand along a row/column using cofactors.
- \(|A| = 0\): singular (no inverse); \(|A| \neq 0\): non-singular (has inverse).
- \(|AB| = |A||B|\).
- \(|kA| = k^n|A|\) for an \(n \times n\) matrix — NOT just \(k|A|\)!
Q1. Find the determinant of \(A = \begin{pmatrix} 5 & 3 \\ -2 & 4 \end{pmatrix}\) and state whether \(A\) is singular or non-singular.
Answer:
\(|A| = (5)(4) – (3)(-2) = 20 + 6 = 26\)
Since \(|A| = 26 \neq 0\), the matrix \(A\) is non-singular (it has an inverse).
Q2. Find \(|A|\) where \(A = \begin{pmatrix} 2 & -3 & 1 \\ 0 & 1 & 4 \\ 5 & 0 & -1 \end{pmatrix}\).
Answer: Expanding along the first row:
So \(|A| = -67\). The matrix is non-singular.
Q3. If \(A\) is a \(3 \times 3\) matrix with \(|A| = 5\), find \(|2A|\) and \(|A^T|\).
Answer:
\(|2A| = 2^3 |A| = 8 \times 5 = 40\) (since \(A\) is \(3 \times 3\), we use \(n = 3\))
\(|A^T| = |A| = 5\) (determinant of transpose equals determinant of original)
8. Inverse of a Matrix
The inverse of a square matrix \(A\), denoted \(A^{-1}\), is a matrix such that:
An inverse exists only if \(|A| \neq 0\) (i.e., \(A\) is non-singular).
8.1 Inverse of a \(2 \times 2\) Matrix
For \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\):
Worked Example 1: Find the inverse of \(A = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}\).
Solution:
Step 1: \(|A| = (3)(4) – (1)(2) = 12 – 2 = 10 \neq 0\) ✓
Step 2: Swap diagonal: \(\begin{pmatrix} 4 & ? \\ ? & 3 \end{pmatrix}\)
Step 3: Negate off-diagonal: \(\begin{pmatrix} 4 & -1 \\ -2 & 3 \end{pmatrix}\)
Step 4: Divide by \(|A|\):
Verification:
8.2 Inverse of a \(3 \times 3\) Matrix
For a \(3 \times 3\) matrix, the process involves:
- Find the determinant \(|A|\). If \(|A| = 0\), stop — no inverse exists.
- Find the matrix of cofactors (each element replaced by its cofactor).
- Take the transpose of the cofactor matrix to get the adjoint (also called adjugate), written as \(\text{adj}(A)\).
- Apply the formula: \(A^{-1} = \dfrac{1}{|A|} \cdot \text{adj}(A)\)
Worked Example: Find the inverse of \(A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 2 & 3 & 4 \end{pmatrix}\).
Solution:
Step 1: Find \(|A|\):
Step 2: Find cofactors for each element:
Cofactor matrix: \(\begin{pmatrix} 1 & 2 & -2 \\ 1 & -2 & 1 \\ -1 & -1 & 1 \end{pmatrix}\)
Step 3: Transpose to get adjoint:
Step 4: Apply formula:
- Inverse exists only when \(|A| \neq 0\) (non-singular matrix).
- For \(2 \times 2\): swap diagonal, negate off-diagonal, divide by determinant.
- For \(3 \times 3\): \(A^{-1} = \dfrac{1}{|A|} \text{adj}(A)\)
- \(\text{adj}(A) = (\text{Cofactor matrix})^T\) — do not forget the transpose!
- \((A^{-1})^{-1} = A\)
- \((AB)^{-1} = B^{-1}A^{-1}\) (reverse order, like transpose)
Q1. Find the inverse of \(A = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\) and verify that \(AA^{-1} = I\).
Answer:
\(|A| = (2)(3) – (5)(1) = 6 – 5 = 1\)
Verification:
Q2. Does the matrix \(B = \begin{pmatrix} 4 & 2 \\ 6 & 3 \end{pmatrix}\) have an inverse? Explain.
Answer: \(|B| = (4)(3) – (2)(6) = 12 – 12 = 0\).
Since \(|B| = 0\), the matrix \(B\) is singular and does NOT have an inverse.
Notice that the second row is \(\frac{3}{2}\) times the first row — the rows are proportional, which always gives determinant = 0.
Q3. If \(A^{-1} = \begin{pmatrix} 3 & -1 \\ 2 & 1 \end{pmatrix}\), find matrix \(A\).
Answer: Since \((A^{-1})^{-1} = A\), we need to find the inverse of \(A^{-1}\).
\(|A^{-1}| = (3)(1) – (-1)(2) = 3 + 2 = 5\)
9. Solving Systems of Linear Equations Using Matrices
One of the most important applications of matrices is solving systems of linear equations. Let us see how!
9.1 Matrix Form of a System
A system of linear equations can be written as \(AX = B\), where:
- \(A\) = coefficient matrix
- \(X\) = column matrix of variables
- \(B\) = column matrix of constants
Example: Consider the system:
In matrix form:
So \(AX = B\), where \(A = \begin{pmatrix} 2 & 3 \\ 1 & -1 \end{pmatrix}\), \(X = \begin{pmatrix} x \\ y \end{pmatrix}\), \(B = \begin{pmatrix} 8 \\ 1 \end{pmatrix}\).
9.2 Solving Using the Inverse Method
If \(A\) is non-singular (\(|A| \neq 0\)), then:
Worked Example: Solve the system above using the inverse method.
Solution:
Step 1: \(|A| = (2)(-1) – (3)(1) = -2 – 3 = -5 \neq 0\) ✓
Step 2: Find \(A^{-1}\):
Step 3: \(X = A^{-1}B\):
So \(x = \dfrac{11}{5}\) and \(y = \dfrac{6}{5}\).
9.3 What if \(|A| = 0\)?
If \(|A| = 0\), the system either has:
- No solution (inconsistent system), or
- Infinitely many solutions (dependent system)
You cannot use the inverse method in this case. You would need other methods (like row reduction) to determine which case applies.
9.4 Cramer’s Rule (Alternative Method)
For a \(2 \times 2\) system \(AX = B\):
where \(A_x\) is matrix \(A\) with the first column replaced by \(B\), and \(A_y\) is \(A\) with the second column replaced by \(B\).
Example: Solve \(2x + 3y = 8\), \(x – y = 1\) using Cramer’s Rule.
Same answer! Both methods give \(x = \frac{11}{5}\), \(y = \frac{6}{5}\). ✓
- Write the system as \(AX = B\).
- If \(|A| \neq 0\): unique solution \(X = A^{-1}B\).
- If \(|A| = 0\): either no solution or infinitely many solutions.
- Cramer’s Rule: \(x = \dfrac{|A_x|}{|A|}\), \(y = \dfrac{|A_y|}{|A|}\) (for \(2 \times 2\)).
- For \(3 \times 3\) systems, Cramer’s Rule extends similarly.
Q1. Solve using the inverse method: \(3x + y = 7\), \(x + 2y = 5\).
Answer:
\(A = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}\), \(|A| = 6 – 1 = 5\)
So \(x = \dfrac{9}{5}\), \(y = \dfrac{8}{5}\).
Verification: \(3(\frac{9}{5}) + \frac{8}{5} = \frac{27+8}{5} = \frac{35}{5} = 7\) ✓ \(\frac{9}{5} + 2(\frac{8}{5}) = \frac{9+16}{5} = \frac{25}{5} = 5\) ✓
Q2. Solve using Cramer’s Rule: \(x + y = 5\), \(2x – y = 1\).
Answer:
So \(x = 2\), \(y = 3\).
Q3. Can the system \(x + 2y = 3\), \(2x + 4y = 7\) be solved using the inverse method? Explain.
Answer: No.
\(A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}\), \(|A| = 4 – 4 = 0\).
Since \(|A| = 0\), \(A\) is singular and has no inverse. The inverse method cannot be used.
Notice: the second equation is \(2 \times\) the first equation’s left side, but \(2 \times 3 = 6 \neq 7\). So the equations are inconsistent — this system has no solution.
Matrices — Quick Revision Notes
A rectangular arrangement of numbers in rows and columns. Order: \(m \times n\) (rows × columns). Element \(a_{ij}\) = element in row \(i\), column \(j\).
Key Definitions
| Type | Definition | Example |
|---|---|---|
| Row Matrix | Exactly 1 row; \(1 \times n\) | \(\begin{pmatrix} 2 & 5 & 1 \end{pmatrix}\) |
| Column Matrix | Exactly 1 column; \(m \times 1\) | \(\begin{pmatrix} 3 \\ 7 \end{pmatrix}\) |
| Square Matrix | Rows = Columns; \(n \times n\) | \(\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\) |
| Diagonal Matrix | Square + off-diagonal = 0 | \(\begin{pmatrix} 5 & 0 \\ 0 & 3 \end{pmatrix}\) |
| Scalar Matrix | Diagonal + all diagonal elements equal | \(\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}\) |
| Identity Matrix \(I\) | Scalar with diagonal = 1 | \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) |
| Zero Matrix \(O\) | All elements = 0 | \(\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\) |
| Symmetric | \(A = A^T\) (i.e., \(a_{ij} = a_{ji}\)) | \(\begin{pmatrix} 1 & 3 \\ 3 & 5 \end{pmatrix}\) |
| Skew-Symmetric | \(A = -A^T\) (diagonal = 0) | \(\begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}\) |
Key Formulas
\((A^T)^T = A\) | \((A + B)^T = A^T + B^T\) | \((AB)^T = B^T A^T\)
\(2 \times 2\): \(\left|\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right| = ad – bc\)
\(3 \times 3\): Expand along any row/column using cofactors with sign pattern:
\(\begin{pmatrix} + & – & + \\ – & + & – \\ + & – & + \end{pmatrix}\)
\(|A^T| = |A|\) | \(|AB| = |A||B|\) | \(|kA| = k^n|A|\) for \(n \times n\) matrix
\(2 \times 2\): If \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), then \(A^{-1} = \dfrac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\)
\(3 \times 3\): \(A^{-1} = \dfrac{1}{|A|} \cdot \text{adj}(A)\) where \(\text{adj}(A) = (\text{Cofactor matrix})^T\)
\((AB)^{-1} = B^{-1}A^{-1}\) | \((A^{-1})^{-1} = A\) | \(|A^{-1}| = \dfrac{1}{|A|}\)
Write as \(AX = B\). If \(|A| \neq 0\): \(X = A^{-1}B\)
Cramer’s Rule (\(2 \times 2\)): \(x = \dfrac{|A_x|}{|A|}\), \(y = \dfrac{|A_y|}{|A|}\)
Common Mistakes to Avoid
Fix: Always check that both matrices have the same \(m \times n\) order before adding or subtracting.
Fix: Matrix multiplication is NOT commutative. Always compute \(AB\) and \(BA\) separately if both are asked.
Fix: \(\text{adj}(A) = (\text{Cofactor matrix})^T\) — the transpose step is essential!
Fix: For an \(n \times n\) matrix, multiplying every element by \(k\) means \(k\) is factored out \(n\) times from each row.
Fix: Unlike regular numbers, two non-zero matrices can multiply to give a zero matrix. Do not make this assumption.
Fix: Memorize the pattern: \(\begin{pmatrix} + & – & + \\ – & + & – \\ + & – & + \end{pmatrix}\). The sign at position \((i,j)\) is \((-1)^{i+j}\).
Fix: The correct formula is \((AB)^T = B^T A^T\) — the order reverses!
Quick Examples for Revision
Ex 1: If \(A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}\), find \(|A|\) and \(A^{-1}\).
Solution: \(|A| = 8 – (-3) = 11\). \(A^{-1} = \frac{1}{11}\begin{pmatrix} 4 & 1 \\ -3 & 2 \end{pmatrix}\).
Ex 2: If \(|A| = 3\) for a \(2 \times 2\) matrix, find \(|4A|\).
Solution: \(|4A| = 4^2 |A| = 16 \times 3 = 48\).
Ex 3: Is \(\begin{pmatrix} 0 & 5 \\ -5 & 0 \end{pmatrix}\) symmetric or skew-symmetric?
Solution: \(A^T = \begin{pmatrix} 0 & -5 \\ 5 & 0 \end{pmatrix} = -A\). So it is skew-symmetric.
Challenge Exam Questions — Matrices
Test yourself with these exam-style questions. Try each one before clicking “Show Answer”!
Multiple Choice Questions
Q1. If \(A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\), then \(A^T\) is:
Answer: (A)
\(A^T\) is obtained by interchanging rows and columns. The first row \((1, 2)\) becomes the first column, and the second row \((3, 4)\) becomes the second column.
Q2. The determinant of \(\begin{pmatrix} 6 & 3 \\ 4 & 2 \end{pmatrix}\) is:
Answer: (C)
\(|A| = (6)(2) – (3)(4) = 12 – 12 = 0\).
The second column is \(\frac{1}{2}\) times the first column, so the columns are proportional and the determinant is zero. The matrix is singular.
Q3. If \(A\) is a \(3 \times 4\) matrix and \(B\) is a \(4 \times 2\) matrix, then the order of \(AB\) is:
Answer: (A)
\(A\) is \(3 \times \boxed{4}\) and \(B\) is \(\boxed{4} \times 2\). Inner dimensions match (4 = 4). Product \(AB\) has order \(3 \times 2\) (outer dimensions).
Q4. Which of the following is a scalar matrix?
Answer: (D)
A scalar matrix is a diagonal matrix where all diagonal elements are equal. Option (A) has diagonal elements both equal to 1 (it is also an identity matrix). Option (B) has diagonal elements both equal to 3. Both are scalar matrices. Option (C) has different diagonal elements, so it is diagonal but not scalar.
Q5. If \(|A| = 5\) and \(A\) is \(3 \times 3\), then \(|3A|\) equals:
Answer: (C)
\(|3A| = 3^3 |A| = 27 \times 5 = 135\). For an \(n \times n\) matrix, \(|kA| = k^n |A|\). Here \(n = 3\).
Fill in the Blank
Q6. A square matrix \(A\) is symmetric if \(A =\) ________.
Answer: \(A^T\)
A symmetric matrix satisfies \(A = A^T\), meaning \(a_{ij} = a_{ji}\) for all \(i, j\).
Q7. The inverse of a matrix exists only when its determinant is ________.
Answer: non-zero (or not equal to zero)
If \(|A| = 0\), the matrix is singular and has no inverse. If \(|A| \neq 0\), the matrix is non-singular and \(A^{-1}\) exists.
Q8. For matrices, \((AB)^T =\) ________.
Answer: \(B^T A^T\)
The order reverses when taking the transpose of a product. This is a very commonly tested property!
Q9. The adjoint of a matrix is the ________ of its cofactor matrix.
Answer: transpose
\(\text{adj}(A) = (\text{Cofactor matrix of } A)^T\). Do not forget this transpose step!
Q10. In a skew-symmetric matrix, all diagonal elements are ________.
Answer: zero
Because \(a_{ii} = -a_{ii}\) for a skew-symmetric matrix, which gives \(2a_{ii} = 0\), so \(a_{ii} = 0\).
Short Answer Questions
Q11. If \(A = \begin{pmatrix} 1 & -2 \\ 3 & 0 \end{pmatrix}\) and \(B = \begin{pmatrix} 4 & 1 \\ -1 & 5 \end{pmatrix}\), find \(2A + B\).
Answer:
Q12. Find \(x\) if \(\begin{pmatrix} x & 1 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 10 \\ 8 \end{pmatrix}\).
Answer:
Equating: \(3x + 4 = 10 \Rightarrow 3x = 6 \Rightarrow x = 2\).
(The second equation \(8 = 8\) is already satisfied — it serves as a check.)
Q13. Show that \(A = \begin{pmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{pmatrix}\) is skew-symmetric.
Answer:
Since \(A = -A^T\), the matrix is skew-symmetric. ✓
Also note: all diagonal elements are 0, which is required for skew-symmetry. ✓
Step-by-Step Calculation Questions
Q14. Find the inverse of \(A = \begin{pmatrix} 2 & 1 & 3 \\ 0 & 1 & 2 \\ 1 & 0 & 2 \end{pmatrix}\).
Answer:
Step 1: Find \(|A|\) by expanding along row 1:
Step 2: Find all 9 cofactors:
Cofactor matrix: \(\begin{pmatrix} 2 & 2 & -1 \\ -2 & 1 & 1 \\ -1 & -4 & 2 \end{pmatrix}\)
Step 3: Transpose to get adjoint:
Step 4: \(A^{-1} = \dfrac{1}{3}\begin{pmatrix} 2 & -2 & -1 \\ 2 & 1 & -4 \\ -1 & 1 & 2 \end{pmatrix}\)
Q15. Solve the system using the matrix inverse method:
\(x + y + z = 6\)
\(x + 2y + 3z = 14\)
\(2x + y + z = 5\)
Answer:
\(A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{pmatrix}\), \(X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\), \(B = \begin{pmatrix} 6 \\ 14 \\ 5 \end{pmatrix}\)
Step 1: Find \(|A|\):
Step 2: Find cofactors:
\(\text{adj}(A) = \begin{pmatrix} -1 & 0 & 1 \\ 5 & -1 & -2 \\ -3 & 1 & 1 \end{pmatrix}\)
Step 3: \(A^{-1} = \dfrac{1}{1}\text{adj}(A) = \begin{pmatrix} -1 & 0 & 1 \\ 5 & -1 & -2 \\ -3 & 1 & 1 \end{pmatrix}\)
Step 4: \(X = A^{-1}B\):
So \(x = -1\), \(y = 6\), \(z = 1\).
Verification: \(-1 + 6 + 1 = 6\) ✓ \(-1 + 12 + 3 = 14\) ✓ \(-2 + 6 + 1 = 5\) ✓
Q16. If \(A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\) and \(B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\), verify that \((AB)^T = B^T A^T\).
Answer:
Left side: \(AB = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\)
Right side: \(A^T = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}\), \(B^T = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\)
Since \((AB)^T = B^T A^T = \begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix}\), the property is verified. ✓
Q17. Solve using Cramer’s Rule:
\(2x – y = 3\)
\(4x + 3y = 5\)
Answer:
So \(x = \dfrac{7}{5}\), \(y = -\dfrac{1}{5}\).
Q18. Prove that every square matrix can be expressed as the sum of a symmetric matrix and a skew-symmetric matrix. (Hint: Use \(A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A – A^T)\))
Answer:
Let \(A\) be any square matrix. Define:
Then \(A = P + Q\).
Check if \(P\) is symmetric:
Check if \(Q\) is skew-symmetric:
Therefore, \(A = P + Q\) where \(P\) is symmetric and \(Q\) is skew-symmetric. This proves the statement.
Example: If \(A = \begin{pmatrix} 1 & 5 \\ 2 & 3 \end{pmatrix}\), then:
\(P = \frac{1}{2}\begin{pmatrix} 2 & 7 \\ 7 & 6 \end{pmatrix} = \begin{pmatrix} 1 & 3.5 \\ 3.5 & 3 \end{pmatrix}\) (symmetric) ✓
\(Q = \frac{1}{2}\begin{pmatrix} 0 & 3 \\ -3 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1.5 \\ -1.5 & 0 \end{pmatrix}\) (skew-symmetric) ✓
Q19. Find the value of \(x\) if the matrix \(A = \begin{pmatrix} x+1 & 2 \\ 3 & 4 \end{pmatrix}\) is singular.
Answer:
A matrix is singular when \(|A| = 0\).
Setting \(|A| = 0\): \(4x – 2 = 0 \Rightarrow x = \dfrac{1}{2}\)
Q20. If \(A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\), show that \(A \cdot A^T = I\). (This is called an orthogonal matrix.)
Answer:
Since \(\cos^2\theta + \sin^2\theta = 1\) and \(\cos\theta\sin\theta – \sin\theta\cos\theta = 0\).
This shows \(A\) is an orthogonal matrix. For orthogonal matrices, \(A^{-1} = A^T\).