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📐 Adding and subtracting matrices

Algebra - Section 1 - Matrix Operations

Mathematics Lessons | Sec 1 math Egypt

Algebra: Adding and subtracting matrices

Sec 1.

Welcome to Math Egypt by Mr. Ayman Hassan. This matrices lesson for Sec 1 students focuses on adding and subtracting matrices. Master how to perform operations on matrices of the same order and explore the properties of addition. Before starting, ensure you have reviewed Matrix Basics & Organizing Data and tested your skills with our Math Quiz.

Condition for Addition and Subtraction

If $A$ and $B$ are two matrices of the same order, then the addition operation is possible. The result of addition is a matrix of the same order and each of its elements is the sum of the two corresponding elements in $A$ and $B$.

If $A = (a_{ij})$ and $B = (b_{ij})$ are two matrices of order $m \times n$, then:

$A \pm B = (a_{ij} \pm b_{ij})$

The result is a matrix of the same order $m \times n$.

Example & Solution (1)

The Matrices:

If $A = \begin{pmatrix} 1 & -2 \\ 3 & 5 \\ 4 & 2 \end{pmatrix}$, $B = \begin{pmatrix} -1 & -2 \\ 5 & 1 \\ 3 & 2 \end{pmatrix}$ and $C = \begin{pmatrix} 2 & 0 & -1 \\ 3 & 4 & 6 \end{pmatrix}$

Find each of the following if it is possible:

(1) $2A + C^t$ (2) $B + C$

Solution Steps:

(1) Finding $2A + C^t$ :

$2A + C^t = 2 \begin{pmatrix} 1 & -2 \\ 3 & 5 \\ 4 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 0 & -1 \\ 3 & 4 & 6 \end{pmatrix}^t$

$= \begin{pmatrix} 2 & -4 \\ 6 & 10 \\ 8 & 4 \end{pmatrix} + \begin{pmatrix} 2 & 3 \\ 0 & 4 \\ -1 & 6 \end{pmatrix} = \begin{pmatrix} 4 & -1 \\ 6 & 14 \\ 7 & 10 \end{pmatrix}$

(2) Finding $B + C$ :

It is impossible to add $B$ and $C$, because they don't have the same order. $B$ is $(3 \times 2)$ while $C$ is $(2 \times 3)$.

Properties of Matrix Addition

1. Closure property: The sum of any two matrices of the same order is a matrix of the same order.

2. Commutative property: $A + B = B + A$

3. Associative property: $(A + B) + C = A + (B + C)$

4. Additive Identity: The zero matrix $\mathcal{O}$ is the additive identity where $A + \mathcal{O} = \mathcal{O} + A = A$.

5. Additive Inverse: For any matrix $A$, there exists an additive inverse $-A$ such that $A + (-A) = \mathcal{O}$.

Remark:

Subtracting matrices operation is not commutative and not associative.

The Transpose Property

Rule of Transpose in Addition

$(A \pm B)^T = A^T \pm B^T$

Example & Solution (Transpose Property)

If $A = \begin{pmatrix} 2 & 3 \\ -1 & 5 \\ 6 & 7 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 4 \\ 2 & -3 \\ 4 & -8 \end{pmatrix}$, check that: $(A + B)^t = A^t + B^t$

Solution:

$\because A + B = \begin{pmatrix} 2 & 3 \\ -1 & 5 \\ 6 & 7 \end{pmatrix} + \begin{pmatrix} 0 & 4 \\ 2 & -3 \\ 4 & -8 \end{pmatrix} = \begin{pmatrix} 2 & 7 \\ 1 & 2 \\ 10 & -1 \end{pmatrix}$

$\therefore (A + B)^t = \begin{pmatrix} 2 & 1 & 10 \\ 7 & 2 & -1 \end{pmatrix} \quad \text{--- (1)}$

$\because A^t = \begin{pmatrix} 2 & -1 & 6 \\ 3 & 5 & 7 \end{pmatrix} \quad, \quad B^t = \begin{pmatrix} 0 & 2 & 4 \\ 4 & -3 & -8 \end{pmatrix}$

$\therefore A^t + B^t = \begin{pmatrix} 2 & -1 & 6 \\ 3 & 5 & 7 \end{pmatrix} + \begin{pmatrix} 0 & 2 & 4 \\ 4 & -3 & -8 \end{pmatrix} = \begin{pmatrix} 2 & 1 & 10 \\ 7 & 2 & -1 \end{pmatrix} \quad \text{--- (2)}$

From (1) and (2), we deduce that: $(A + B)^t = A^t + B^t$
Example & Solution (Finding unknown values)

Find the values of $a, b$ and $c$ that satisfy the equation:

$3 \begin{pmatrix} a & b \\ c & 3 \end{pmatrix} = 2 \begin{pmatrix} a & 6 \\ -1 & 3 \end{pmatrix} + \begin{pmatrix} 4 & b+4 \\ c+3 & 3 \end{pmatrix}$

Solution:

$\begin{pmatrix} 3a & 3b \\ 3c & 9 \end{pmatrix} = \begin{pmatrix} 2a & 12 \\ -2 & 6 \end{pmatrix} + \begin{pmatrix} 4 & b+4 \\ c+3 & 3 \end{pmatrix}$

$\therefore \begin{pmatrix} 3a & 3b \\ 3c & 9 \end{pmatrix} = \begin{pmatrix} 2a+4 & b+16 \\ c+1 & 9 \end{pmatrix}$

From equality of matrices:

$\therefore 3a = 2a + 4 \implies a = 4$
$\therefore 3b = b + 16 \implies 2b = 16 \implies b = 8$
$\therefore 3c = c + 1 \implies 2c = 1 \implies c = \frac{1}{2}$
Example & Solution (Matrix Equations)

If $A = \begin{pmatrix} 2 & 3 & -2 \\ -1 & 4 & 5 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & -1 & 3 \\ 5 & 2 & -4 \end{pmatrix}$, find the matrix $X$ that satisfies: $2[X^t - A] = 3B$

Solution:

$\because 2[X^t - A] = 3B \implies 2X^t - 2A = 3B$

$\therefore 2X^t = 3B + 2A$

$$ 2X^t = 3\begin{pmatrix} 0 & -1 & 3 \\ 5 & 2 & -4 \end{pmatrix} + 2\begin{pmatrix} 2 & 3 & -2 \\ -1 & 4 & 5 \end{pmatrix} $$

$$ 2X^t = \begin{pmatrix} 0 & -3 & 9 \\ 15 & 6 & -12 \end{pmatrix} + \begin{pmatrix} 4 & 6 & -4 \\ -2 & 8 & 10 \end{pmatrix} = \begin{pmatrix} 4 & 3 & 5 \\ 13 & 14 & -2 \end{pmatrix} $$

$\therefore X^t = \frac{1}{2}\begin{pmatrix} 4 & 3 & 5 \\ 13 & 14 & -2 \end{pmatrix} = \begin{pmatrix} 2 & \frac{3}{2} & \frac{5}{2} \\ \frac{13}{2} & 7 & -1 \end{pmatrix}$

Final Answer:

$$ X = (X^t)^t = \begin{pmatrix} 2 & \frac{13}{2} \\ \frac{3}{2} & 7 \\ \frac{5}{2} & -1 \end{pmatrix} $$

Frequently Asked Questions (FAQ)

Q1: Can we add two matrices with different orders?

No, the fundamental condition for matrix addition or subtraction is that both matrices must have the same order (same number of rows and columns).

Q2: Is matrix subtraction commutative?

No, matrix subtraction is NOT commutative ($A - B \neq B - A$). It is also NOT associative.

Q3: What happens when we add the zero matrix $\mathcal{O}$ to any matrix $A$?

The zero matrix $\mathcal{O}$ is the additive identity, so $A + \mathcal{O} = A$. The matrix remains unchanged.

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