Linear algebra

Matrix operations

Matrix products are computed "row times column". Once products are routine, learn how to find inverses and solve matrix equations.

Basic, Standard, Advanced: University Year 1 · 1st semester

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Sums, scalar multiples and products

Sums, differences and scalar multiples are computed entry by entry. The $(i,\ j)$ entry of $AB$ is row $i$ of $A$ times column $j$ of $B$ :

\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} p & q \\ r & s \end{pmatrix} = \begin{pmatrix} ap + br & aq + bs \\ cp + dr & cq + ds \end{pmatrix}

Caution

In general $AB \neq BA$ . An $m \times n$ matrix times an $n \times l$ matrix is $m \times l$ ; if the inner sizes differ, the product is not defined.

The transpose $\,{}^tA$ swaps rows and columns, and $\,{}^t(AB) = {}^tB\,{}^tA$ .

Finding inverses

2×2 inverse

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix},\quad ad - bc \neq 0:\quad A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

For larger matrices use row reduction: reduce $(A \mid E)$ until the left half is the identity $E$ ; the right half is then $A^{-1}$ . You can also use the adjugate: $A^{-1} = \frac{1}{|A|}\tilde{A}$ .

Solve $AX = B$ by multiplying on the left by $A^{-1}$ : $X = A^{-1}B$ . For $XA = B$ , multiply on the right: $X = BA^{-1}$ .

Cayley–Hamilton and $A^n$

Cayley–Hamilton theorem

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}:\quad A^2 - (a + d)A + (ad - bc)E = O

This writes $A^2$ as a combination of $A$ and $E$ , so $A^3$ and $A^4$ can be reduced too. For $A^n$ , compute $A^2$ and $A^3$ to find a pattern, or diagonalize $A$ using its eigenvalues.

Q1BasicUniv Y1 · Sem 1Matrix operations

For the matrices $A$ and $B$ below, compute $-A + B$ .

A = \begin{pmatrix}1 & 0 \\ 2 & 3\end{pmatrix},\quad B = \begin{pmatrix}-2 & 2 \\ 0 & 0\end{pmatrix}

Answer

\begin{pmatrix}-3 & 2 \\ -2 & -3\end{pmatrix}

Solution

Entry by entry,
$-A + B = \begin{pmatrix}-1 & 0 \\ -2 & -3\end{pmatrix} + \begin{pmatrix}-2 & 2 \\ 0 & 0\end{pmatrix} = \begin{pmatrix}-3 & 2 \\ -2 & -3\end{pmatrix}$

Q2StandardUniv Y1 · Sem 1Matrix operations

Find the inverse of the matrix.

A = \begin{pmatrix}2 & 1 & 1 \\ -1 & 0 & 2 \\ 2 & 1 & 0\end{pmatrix}

Answer

A^{-1} = \begin{pmatrix}2 & -1 & -2 \\ -4 & 2 & 5 \\ 1 & 0 & -1\end{pmatrix}

Solution

Form $(A \mid E)$ :
$\left(\begin{array}{ccc|ccc}2 & 1 & 1 & 1 & 0 & 0 \\ -1 & 0 & 2 & 0 & 1 & 0 \\ 2 & 1 & 0 & 0 & 0 & 1\end{array}\right)$
Row-reduce until the left half is the identity; the right half is the inverse.
$A^{-1} = \begin{pmatrix}2 & -1 & -2 \\ -4 & 2 & 5 \\ 1 & 0 & -1\end{pmatrix}$
(Check: $\det A = -1$ and $AA^{-1} = E$ .)

Q3AdvancedUniv Y1 · Sem 1Matrix operations

For the matrices $A$ and $B$ below, find the matrix $X$ with $XA = B$ .

A = \begin{pmatrix}-3 & 1 \\ -1 & 1\end{pmatrix},\quad B = \begin{pmatrix}10 & -4 \\ 1 & -1\end{pmatrix}

Answer

X = \begin{pmatrix}-3 & -1 \\ 0 & -1\end{pmatrix}

Solution

The inverse of $A$ is
$A^{-1} = \begin{pmatrix}-\frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{3}{2}\end{pmatrix}$
Multiplying both sides on the right by $A^{-1}$ ,
$X = BA^{-1} = \begin{pmatrix}10 & -4 \\ 1 & -1\end{pmatrix}\begin{pmatrix}-\frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{3}{2}\end{pmatrix} = \begin{pmatrix}-3 & -1 \\ 0 & -1\end{pmatrix}$

Q1BasicUniv Y1 · Sem 1Matrix operations

For the matrices below, compute $A\,{}^tA$ (where $\,{}^tX$ is the transpose of $X$ ).

A = \begin{pmatrix}1 & -1 & 3 \\ -1 & -3 & -2\end{pmatrix}

Q2StandardUniv Y1 · Sem 1Matrix operations

For the matrix $A$ below, find $A^{7}$ .

A = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}

Q3AdvancedUniv Y1 · Sem 1Matrix operations

Find the values of the constant $k$ for which the matrix has no inverse.

\begin{pmatrix}k & 4 \\ 1 & k - 3\end{pmatrix}

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