Operations on Matrices

Introduction

Matrices are not just a way to store data; they can be added, subtracted, and multiplied, much like numbers. This section covers the fundamental operations of matrix algebra, which form the basis for solving complex systems and performing transformations.

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1. Addition of Matrices

Matrix addition is the simplest operation. It's just like adding corresponding elements together.

The Golden Rule: Two matrices can be added only if they have the exact same order (same number of rows and same number of columns). You can't add a 2x3 matrix to a 2x2 matrix because their elements wouldn't line up.

• Rule: If $A = [a_{ij}]$ and $B = [b_{ij}]$ are two matrices, both of order m x n, then their sum $A+B$ is a new m x n matrix $C = [c_{ij}]$, where each new element is the sum of the corresponding old elements: $$c_{ij} = a_{ij} + b_{ij}$$

Example: If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$, then: $$A+B = \begin{bmatrix} 1+5 & 2+6 \\ 3+7 & 4+8 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}$$

Properties of Matrix Addition

Matrix addition behaves just like regular addition of numbers.

• Commutative: $A + B = B + A$. (The order you add them in doesn't matter).

• Associative: $(A + B) + C = A + (B + C)$. (The way you group them doesn't matter).

• Existence of Additive Identity: $A + O = O + A = A$. The Zero Matrix (O), which is a matrix of all zeros, acts like the number '0' in regular addition.

• Existence of Additive Inverse: For any matrix A, there is a matrix $-A$ (its negative) such that $A + (-A) = O$. The matrix $-A$ is found by negating every single element of A.

  • If $A = \begin{bmatrix} 1 & -2 \\ 3 & 0 \end{bmatrix}$, then $-A = \begin{bmatrix} -1 & 2 \\ -3 & 0 \end{bmatrix}$.

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2. Multiplication of a Matrix by a Scalar

A scalar is just a regular number (like 3, -1, or 0.5). To multiply a matrix by a scalar, you simply multiply every element in the matrix by that number. Think of it as scaling the entire matrix up or down.

• Rule: If $A = [a_{ij}]$ is an m x n matrix and k is a scalar, then $kA$ is the m x n matrix where each element is $k \times a_{ij}$.

• Example: If $A = \begin{bmatrix} 1 & 10 \\ 3 & -4 \end{bmatrix}$, then $5A$ is: $$5A = 5 \begin{bmatrix} 1 & 10 \\ 3 & -4 \end{bmatrix} = \begin{bmatrix} 5(1) & 5(10) \\ 5(3) & 5(-4) \end{bmatrix} = \begin{bmatrix} 5 & 50 \\ 15 & -20 \end{bmatrix}$$

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3. Matrix Multiplication

This is the most complex operation. It is not element-by-element multiplication. It's a row-by-column operation.

The Golden Rule: Two matrices A and B can be multiplied to find the product $AB$ only if the number of columns in the first matrix (A) is equal to the number of rows in the second matrix (B).

• Rule for Order: If A is an m x n matrix and B is an n x p matrix, the inner dimensions ($n$) must match. The resulting matrix, $AB$, will have the outer dimensions: m x p.

  • (2 x 3) $\times$ (3 x 4) $\implies$ Result is a 2 x 4 matrix. (OK)
  • (2 x 3) $\times$ (2 x 2) $\implies$ Result is undefined. (Mismatch)

• How to Calculate: To find the element in the i-th row and j-th column of $AB$, you take the i-th row of A and the j-th column of B. You multiply their corresponding elements, and then add up the results.

Example: Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}_{2 \times 2}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}_{2 \times 2}$. Find $AB$.

1. Order Check: A is 2x2, B is 2x2. The inner dimensions match. The result $AB$ will be a 2x2 matrix: $\begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix}$.

2. Calculate $c_{11}$ (Row 1 of A, Column 1 of B): $$\begin{bmatrix} \mathbf{1} & \mathbf{2} \\ 3 & 4 \end{bmatrix} \begin{bmatrix} \mathbf{5} & 6 \\ \mathbf{7} & 8 \end{bmatrix}$$ $\implies c_{11} = (1)(5) + (2)(7) = 5 + 14 = 19$.

3. Calculate $c_{12}$ (Row 1 of A, Column 2 of B): $$\begin{bmatrix} \mathbf{1} & \mathbf{2} \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & \mathbf{6} \\ 7 & \mathbf{8} \end{bmatrix}$$ $\implies c_{12} = (1)(6) + (2)(8) = 6 + 16 = 22$.

4. Calculate $c_{21}$ (Row 2 of A, Column 1 of B): $$\begin{bmatrix} 1 & 2 \\ \mathbf{3} & \mathbf{4} \end{bmatrix} \begin{bmatrix} \mathbf{5} & 6 \\ \mathbf{7} & 8 \end{bmatrix}$$ $\implies c_{21} = (3)(5) + (4)(7) = 15 + 28 = 43$.

5. Calculate $c_{22}$ (Row 2 of A, Column 2 of B): $$\begin{bmatrix} 1 & 2 \\ \mathbf{3} & \mathbf{4} \end{bmatrix} \begin{bmatrix} 5 & \mathbf{6} \\ 7 & \mathbf{8} \end{bmatrix}$$ $\implies c_{22} = (3)(6) + (4)(8) = 18 + 32 = 50$.

6. Final Result: $AB = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}$.

Properties of Matrix Multiplication

• Not Commutative (Crucial!): In general, $AB \ne BA$. Order matters!

  • In our example, $BA = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 23 & 34 \\ 31 & 46 \end{bmatrix}$, which is not equal to $AB$.

• Associative: $(AB)C = A(BC)$. (As long as the orders are compatible).

• Distributive: $A(B+C) = AB+AC$. (The distributive law holds).

• Existence of Multiplicative Identity: For any square matrix A, $AI = IA = A$, where I is the identity matrix of the same order. $I$ acts like the number '1' in matrix multiplication.

Example 1: Matrix Addition

Question: If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$, find $A+B$.

Explanation:

1. Check Order: Both matrices are 2x2. Since they have the same order, they can be added.
2. Add Corresponding Elements: We add the element in the i-th row and j-th column of A to the element in the i-th row and j-th column of B.
   • $a_{11} + b_{11} = 1 + 5 = 6$
   • $a_{12} + b_{12} = 2 + 6 = 8$
   • $a_{21} + b_{21} = 3 + 7 = 10$
   • $a_{22} + b_{22} = 4 + 8 = 12$
3. Construct the Result Matrix: $A+B = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}$.

Answer: $\begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}$

Example 2: Scalar Multiplication and Subtraction

Question: If $A = \begin{bmatrix} 8 & 0 \\ 4 & -2 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -2 \\ 4 & 2 \end{bmatrix}$, find the matrix X such that $2A+3X=5B$.

Explanation:

1. Isolate X: Start by algebraically solving for the matrix X. $$2A+3X=5B \implies 3X = 5B - 2A \implies X = \frac{1}{3}(5B - 2A)$$
2. Calculate Scalar Multiples:
   • $$5B = 5\begin{bmatrix} 2 & -2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 10 & -10 \\ 20 & 10 \end{bmatrix}$$
   • $$2A = 2\begin{bmatrix} 8 & 0 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 16 & 0 \\ 8 & -4 \end{bmatrix}$$
3. Perform Matrix Subtraction (5B - 2A): $$5B - 2A = \begin{bmatrix} 10 & -10 \\ 20 & 10 \end{bmatrix} - \begin{bmatrix} 16 & 0 \\ 8 & -4 \end{bmatrix}$$ $$= \begin{bmatrix} 10-16 & -10-0 \\ 20-8 & 10-(-4) \end{bmatrix} = \begin{bmatrix} -6 & -10 \\ 12 & 14 \end{bmatrix}$$
4. Solve for X: $$X = \frac{1}{3} \begin{bmatrix} -6 & -10 \\ 12 & 14 \end{bmatrix} = \begin{bmatrix} -6/3 & -10/3 \\ 12/3 & 14/3 \end{bmatrix} = \begin{bmatrix} -2 & -10/3 \\ 4 & 14/3 \end{bmatrix}$$

Answer: $X = \begin{bmatrix} -2 & -10/3 \\ 4 & 14/3 \end{bmatrix}$.

Example 3: Basic Matrix Multiplication

Question: Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 \\ 6 \end{bmatrix}$. Find AB.

Explanation:

1. Check Order: Matrix A is 2x2. Matrix B is 2x1. The inner dimensions match (2 and 2). The resulting matrix AB will have the outer dimensions: 2x1.

2. Calculate Element (1,1): Take the 1st row of A and the 1st column of B. Multiply corresponding elements and add. $$(1)(5) + (2)(6) = 5+12 = 17$$

3. Calculate Element (2,1): Take the 2nd row of A and the 1st column of B. $$(3)(5) + (4)(6) = 15+24 = 39$$

4. Construct the Result Matrix:

Answer: $$AB = \begin{bmatrix} 17 \\ 39 \end{bmatrix}$$

Example 4: 2x2 Matrix Multiplication

Question: If $A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}$, find AB.

Explanation:

1. Check Order: A is 2x2, B is 2x2. The inner dimensions match. The result AB will be a 2x2 matrix.

2. Calculate Element (1,1): (Row 1 of A) $\times$ (Col 1 of B) $$(2)(1) + (1)(-1) = 2 - 1 = 1$$

3. Calculate Element (1,2): (Row 1 of A) $\times$ (Col 2 of B) $$(2)(0) + (1)(2) = 0 + 2 = 2$$

4. Calculate Element (2,1): (Row 2 of A) $\times$ (Col 1 of B) $$(3)(1) + (2)(-1) = 3 - 2 = 1$$

5. Calculate Element (2,2): (Row 2 of A) $\times$ (Col 2 of B) $$(3)(0) + (2)(2) = 0 + 4 = 4$$

6. Construct the Result Matrix:

Answer: $$AB = \begin{bmatrix} 1 & 2 \\ 1 & 4 \end{bmatrix}$$

Example 5: Non-Commutativity

Question: Using the matrices from the previous example, find BA and show that $AB \ne BA$.

Explanation:

1. Matrices: $A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}$. We need to find $BA$.
2. Check Order: B is 2x2, A is 2x2. The inner dimensions match. The result BA will be a 2x2 matrix.
3. Calculate Element (1,1): (Row 1 of B) $\times$ (Col 1 of A) $$(1)(2) + (0)(3) = 2 + 0 = 2$$
4. Calculate Element (1,2): (Row 1 of B) $\times$ (Col 2 of A) $$(1)(1) + (0)(2) = 1 + 0 = 1$$
5. Calculate Element (2,1): (Row 2 of B) $\times$ (Col 1 of A) $$(-1)(2) + (2)(3) = -2 + 6 = 4$$
6. Calculate Element (2,2): (Row 2 of B) $\times$ (Col 2 of A) $$(-1)(1) + (2)(2) = -1 + 4 = 3$$
7. Result for BA: $$BA = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}$$
8. Compare: From Example 4, $AB = \begin{bmatrix} 1 & 2 \\ 1 & 4 \end{bmatrix}$. Since the matrices are not identical, $AB \ne BA$. This demonstrates that matrix multiplication is not commutative.

Example 6: Solving a Matrix Equation

Question: Find x if $\begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = O$.

Explanation:

1. Multiply the first two matrices: (Order 1x3) $\times$ (Order 3x3) $\to$ (Order 1x3).

   Let the result be $$C = \begin{bmatrix} c_{11} & c_{12} & c_{13} \end{bmatrix}$$

   • $$c_{11} = (1)(1) + (x)(2) + (1)(15) = 1+2x+15 = 2x+16$$
   • $$c_{12} = (1)(3) + (x)(5) + (1)(3) = 3+5x+3 = 5x+6$$
   • $$c_{13} = (1)(2) + (x)(1) + (1)(2) = 2+x+2 = x+4$$ So, $$\begin{bmatrix} 2x+16 & 5x+6 & x+4 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = O$$

2. Multiply the resulting matrices: (Order 1x3) $\times$ (Order 3x1) $\to$ (Order 1x1).

   The zero matrix $O$ is the 1x1 matrix $[0]$. $$[(2x+16)(1) + (5x+6)(2) + (x+4)(x)] = [0]$$

3. Form the Equation: $$(2x+16) + (10x+12) + (x^2+4x) = 0$$ $$x^2 + (2x+10x+4x) + (16+12) = 0$$ $$x^2 + 16x + 28 = 0$$

4. Solve the Quadratic Equation: Factoring the quadratic: $$(x+2)(x+14) = 0$$

Answer: The solutions are x = -2 and x = -14.

Example 7: Powers of a Matrix

Question: If $A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$, find $A^2$ and $A^3$.

Explanation:

1. Calculate $A^2$: $A^2 = A \cdot A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$.

   • (Row 1, Col 1): $(1)(1) + (0)(1) = 1$.
   • (Row 1, Col 2): $(1)(0) + (0)(1) = 0$.
   • (Row 2, Col 1): $(1)(1) + (1)(1) = 2$.
   • (Row 2, Col 2): $(1)(0) + (1)(1) = 1$. So, $A^2 = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$.

2. Calculate $A^3$: $A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$.

   • (Row 1, Col 1): $(1)(1) + (0)(1) = 1$.
   • (Row 1, Col 2): $(1)(0) + (0)(1) = 0$.
   • (Row 2, Col 1): $(2)(1) + (1)(1) = 3$.
   • (Row 2, Col 2): $(2)(0) + (1)(1) = 1$. So, $A^3 = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}$.

Answer: $A^2 = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$ and $A^3 = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}$.

Example 8: Associative Property

Question: Let $A = \begin{bmatrix} 1 & 2 \end{bmatrix}, B = \begin{bmatrix} 3 \\ 4 \end{bmatrix}, C = \begin{bmatrix} 5 & 6 \end{bmatrix}$. Verify that $(AB)C = A(BC)$.

Explanation: 1. Calculate LHS: (AB)C

• First, find $AB$: (Order 1x2) $\times$ (2x1) $\to$ (Order 1x1). $$AB = \begin{bmatrix} (1)(3) + (2)(4) \end{bmatrix} = \begin{bmatrix} 3+8 \end{bmatrix} = \begin{bmatrix} 11 \end{bmatrix}$$
• Then, find $(AB)C$: (Order 1x1) $\times$ (1x2) $\to$ (Order 1x2). $$(AB)C = \begin{bmatrix} 11 \end{bmatrix} \begin{bmatrix} 5 & 6 \end{bmatrix} = \begin{bmatrix} (11)(5) & (11)(6) \end{bmatrix} = \begin{bmatrix} 55 & 66 \end{bmatrix}$$

2. Calculate RHS: A(BC)

• First, find $BC$: (Order 2x1) $\times$ (Order 1x2) $\to$ (Order 2x2). $$BC = \begin{bmatrix} 3 \\ 4 \end{bmatrix} \begin{bmatrix} 5 & 6 \end{bmatrix} = \begin{bmatrix} (3)(5) & (3)(6) \\ (4)(5) & (4)(6) \end{bmatrix} = \begin{bmatrix} 15 & 18 \\ 20 & 24 \end{bmatrix}$$
• Then, find $A(BC)$: (Order 1x2) $\times$ (2x2) $\to$ (Order 1x2). $$A(BC) = \begin{bmatrix} 1 & 2 \end{bmatrix} \begin{bmatrix} 15 & 18 \\ 20 & 24 \end{bmatrix}$$ $$= \begin{bmatrix} (1)(15)+(2)(20) & (1)(18)+(2)(24) \end{bmatrix}$$ $$= \begin{bmatrix} 15+40 & 18+48 \end{bmatrix} = \begin{bmatrix} 55 & 66 \end{bmatrix}$$

3. Conclusion: Since LHS = RHS, the associative property is verified.

Answer: LHS = $\begin{bmatrix} 55 & 66 \end{bmatrix}$ and RHS = $\begin{bmatrix} 55 & 66 \end{bmatrix}$.

Example 9: Distributive Property

Question: If $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}, B = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, C = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, verify $A(B+C)=AB+AC$.

Explanation: 1. Calculate LHS: A(B+C)

• First, find $B+C$: $$B+C = \begin{bmatrix} 1+0 & 0+1 \\ 0+1 & -1+0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$$
• Now, find $A(B+C)$: $$A(B+C) = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$$ $$= \begin{bmatrix} (1)(1)+(1)(1) & (1)(1)+(1)(-1) \\ (1)(1)+(1)(1) & (1)(1)+(1)(-1) \end{bmatrix}$$ $$= \begin{bmatrix} 2 & 0 \\ 2 & 0 \end{bmatrix}$$

2. Calculate RHS: AB+AC

• First, find $AB$: $$AB = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$$ $$= \begin{bmatrix} (1)(1)+(1)(0) & (1)(0)+(1)(-1) \\ (1)(1)+(1)(0) & (1)(0)+(1)(-1) \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}$$
• Next, find $AC$: $$AC = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$ $$= \begin{bmatrix} (1)(0)+(1)(1) & (1)(1)+(1)(0) \\ (1)(0)+(1)(1) & (1)(1)+(1)(0) \end{bmatrix}$$ $$= \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$
• Now, add the results: $$AB+AC = \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$ $$= \begin{bmatrix} 1+1 & -1+1 \\ 1+1 & -1+1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 2 & 0 \end{bmatrix}$$

3. Conclusion: Since LHS = RHS, the distributive property is verified.

Answer: LHS = $\begin{bmatrix} 2 & 0 \\ 2 & 0 \end{bmatrix}$ and RHS = $\begin{bmatrix} 2 & 0 \\ 2 & 0 \end{bmatrix}$.

Example 10: Special Matrix Product

Question: If $A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$, show that $A^2 = \begin{bmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{bmatrix}$.

Explanation: This matrix is a rotation matrix by an angle $\theta$. Squaring it should be equivalent to applying the rotation twice, resulting in a rotation by $2\theta$.

1. Set up the multiplication $A^2 = A \cdot A$: $$A^2 = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$$

2. Calculate each element:

   • Element (1,1): $$(\cos\theta)(\cos\theta) + (\sin\theta)(-\sin\theta) = \cos^2\theta - \sin^2\theta$$
   • Element (1,2): $$(\cos\theta)(\sin\theta) + (\sin\theta)(\cos\theta) = 2\sin\theta\cos\theta$$
   • Element (2,1): $$(-\sin\theta)(\cos\theta) + (\cos\theta)(-\sin\theta) = -2\sin\theta\cos\theta$$
   • Element (2,2): $$(-\sin\theta)(\sin\theta) + (\cos\theta)(\cos\theta) = \cos^2\theta - \sin^2\theta$$

3. Form the resulting matrix: $$A^2 = \begin{bmatrix} \cos^2\theta - \sin^2\theta & 2\sin\theta\cos\theta \\ -2\sin\theta\cos\theta & \cos^2\theta - \sin^2\theta \end{bmatrix}$$

4. Apply Double Angle Identities: Using the identities $$\cos 2\theta = \cos^2\theta - \sin^2\theta$$ and $$\sin 2\theta = 2\sin\theta\cos\theta$$, we get: $$A^2 = \begin{bmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{bmatrix}$$

Answer: The statement is verified.