Revise Matrices for Cambridge 9231 (1.4)

A matrix is a rectangular array of numbers, called elements, arranged in rows and columns. The 'order' or 'dimension' of a matrix is given as $m \times n$, where $m$ is the number of rows and $n$ is the number of columns. For example, a $2 \times 3$ matrix has 2 rows and 3 columns.

Matrices of the same order can be added or subtracted by adding or subtracting their corresponding elements. Any matrix can be multiplied by a scalar (a single number) by multiplying every element in the matrix by that scalar.

Multiplying two matrices together is more complex. To find the product $\mathbf{AB}$, the number of columns in matrix $\mathbf{A}$ must equal the number of rows in matrix $\mathbf{B}$. If $\mathbf{A}$ is of order $m \times p$ and $\mathbf{B}$ is of order $p \times n$, their product $\mathbf{AB}$ will be of order $m \times n$. To find the element in row $i$ and column $j$ of the product matrix, you multiply each element of row $i$ of $\mathbf{A}$ by the corresponding element of column $j$ of $\mathbf{B}$, and sum the results. Crucially, matrix multiplication is not commutative; in general, $\mathbf{AB} \neq \mathbf{BA}$.

For any square matrix (where rows = columns), we can calculate a single number called its determinant. This value is incredibly useful. If the determinant is non-zero, the matrix is 'non-singular' and has an inverse. If the determinant is zero, the matrix is 'singular' and has no inverse. The inverse of a matrix $\mathbf{A}$, denoted $\mathbf{A}^{-1}$, is a matrix such that $\mathbf{A}\mathbf{A}^{-1} = \mathbf{A}^{-1}\mathbf{A} = \mathbf{I}$, where $\mathbf{I}$ is the identity matrix.

For a 2x2 matrix $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$: Determinant: $\det(\mathbf{A}) = ad - bc$ Inverse: $\mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

For a 3x3 matrix, the determinant is found by cofactor expansion. The inverse is found using a four-step process: calculate the determinant, find the matrix of cofactors, transpose it to get the adjugate matrix, and finally multiply the adjugate by $1/\det(\mathbf{A})$.

A key application of matrices is solving systems of linear equations. A system like: $2x + y - z = 8$ $x + 3y + 2z = 5$ $-x + 2y + z = 4$ can be written in the matrix form $\mathbf{A}\mathbf{x} = \mathbf{b}$, where $\mathbf{A}$ is the matrix of coefficients, $\mathbf{x}$ is the column vector of variables, and $\mathbf{b}$ is the column vector of constants.

\begin{pmatrix} 2 & 1 & -1 \ 1 & 3 & 2 \ -1 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 8 \ 5 \ 4 \end{pmatrix}

If the coefficient matrix $\mathbf{A}$ is non-singular (i.e., $\det(\mathbf{A}) \neq 0$), we can find its inverse $\mathbf{A}^{-1}$. By pre-multiplying both sides of the equation by $\mathbf{A}^{-1}$, we get $\mathbf{A}^{-1}\mathbf{A}\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}$, which simplifies to $\mathbf{I}\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}$, giving the unique solution $\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}$.