Matrices (Pt. 1): Intro

Matrices

Matrix represents the bulk of our discussion as all possible discussed entities are arguably matrices with varying sizes.

Size of Matrix

Matrices are expressed through rows by cols, with

• m x n (read as m by n)

• 3 x 2 (read as 3 by 2)

Type	Size	Example
Matrix	m x n	[1, 2, 3] [4, 5, 6]
Square Matrix	n x n	[1, 2, 3] [4, 5, 6] [7, 8, 9]

It may though be superficial, but what distinguishes between a vector or a matrix is that both m and n variables has to be greater than 1.

Notation of Elements

An element (or item) of a matrix is conventionally referred to by its location with

• i being its row number, while

• j being its column number

An element of matrix a(i, j)

Hence, for Matrix "a" is particular, we could reference an element by calling for example "a one-two".

Matrix Arithmetic

Matrices can undergo addition, subtraction, and mostly multiplication, but never division (lol) 💀.

The following operations can be read at face-value. For example, Matrix-"some-entity" "some-operation" .

Matrix-Scalar Add/Subtract/Multiply Operation

Scalar-value is applied to all elements of the matrix.

$$\left[\begin{array}{cc} 4 & 5\\ 6 & 7 \end{array}\right] +2 = \left[\begin{array}{cc} (4+2) & (5+2)\\ (6+2) & (7+2) \end{array}\right] = \left[\begin{array}{cc} 6 & 7\\ 8 & 9 \end{array}\right]$$

$$\left[\begin{array}{cc} 6 & 7\\ 8 & 9 \end{array}\right] - 3 = \left[\begin{array}{cc} (6-3) & (7-3)\\ (8-3) & (9-3) \end{array}\right]= \left[\begin{array}{cc} 3 & 4\\ 5 & 6 \end{array}\right]$$

$$\left[\begin{array}{cc} 1 & 2\\ 3 & 4 \end{array}\right] \times 2 = \left[\begin{array}{cc} (1\times 2) & (2\times 4)\\ (3 \times 2) & (4\times 2) \end{array}\right]= \left[\begin{array}{cc} 2 & 4\\ 6 & 8 \end{array}\right]$$

Matrix-Matrix Add/Subtract Operation

Scalar-value of each elements are applied to respective elements of matrices.

• Matrices must be of the same size

$$\left[\begin{array}{cc} 1 & 2\\ 3 & 4 \end{array}\right] + \left[\begin{array}{cc} 5 & 6\\ 7 & 8 \end{array}\right] = \left[\begin{array}{cc} (1+5) & (2+6)\\ (3+7) & (4+8) \end{array}\right]= \left[\begin{array}{cc} 6 & 8\\ 10 & 12 \end{array}\right]$$

$$\left[\begin{array}{cc} 7 & 8 & 9 \end{array}\right] - \left[\begin{array}{cc} 3 & 2 & 1 \end{array}\right] = \left[\begin{array}{cc} (7-3) & (8-2) & (9-1) \end{array}\right] = \left[\begin{array}{cc} 4 & 6 & 8 \end{array}\right]$$

More on the next page ⏭