1. Matrix Foundations & Types

Square: \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\)
Row: \(\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}\)
Column: \(\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\)
Identity: \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

2. Equality of Matrices (Detailed Solutions)

Two matrices are equal if they have the same order and their corresponding elements are identical.

Problem: Find \(x\) and \(y\) if \(\begin{bmatrix} x+y & 3 \\ 5 & x-y \end{bmatrix} = \begin{bmatrix} 7 & 3 \\ 5 & 1 \end{bmatrix}\)

Method 1: Elimination

1. Equate elements: \(x+y=7\) (i) and \(x-y=1\) (ii).

2. Add (i) and (ii): \(2x = 8 \implies \mathbf{x = 4}\).

3. Substitute \(x=4\) in (i): \(4+y=7 \implies \mathbf{y=3}\).

Method 2: Substitution

1. From (ii), \(x = 1+y\) (iii).

2. Substitute (iii) into (i): \((1+y) + y = 7 \implies 1+2y=7 \implies 2y=6 \implies \mathbf{y=3}\).

3. Put \(y=3\) into (iii): \(x = 1+3 \implies \mathbf{x=4}\).

3. Matrix Operations

Addition: Add corresponding positions.

\(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}\)

Scalar Multiplication: Multiply every entry by \(k\).

\(3 \times \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 6 \\ 9 & 12 \end{bmatrix}\)

Matrix Multiplication: Row-by-Column.

\(\begin{bmatrix} a & b \\ c & d \end{bmatrix} \times \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} ax+by \\ cx+dy \end{bmatrix}\)