Matrices and Cramer's rule

The idea

A circuit with three loops gives you three simultaneous equations. Writing them as a matrix keeps them tidy, and the determinant gives a mechanical way to solve them. You’ll use this in Mesh analysis, and it also turns up in WM1AC for vectors and robot kinematics.

Matrices

An $m\times n$ matrix has $m$ rows and $n$ columns; $a_{ij}$ is the element in row $i$, column $j$.

• Add or subtract: same size only, element by element.
• Multiply $AB$: the columns of $A$ must equal the rows of $B$. $(m\times n)(n\times p)\to (m\times p)$. Each element is a row of $A$ dotted with a column of $B$. $\left[\begin{array}{cc}2& 3\\ 1& 4\end{array}\right]\left[\begin{array}{cc}5& 1\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}10& 8\\ 5& 9\end{array}\right]$
• Order matters: in general $AB\ne BA$.
• There’s no division. "$A÷B$" means $A{B}^{-1}$, which only exists if $\det B\ne 0$.

Determinants

A single number from a square matrix. If it’s zero, the matrix has no inverse and the equations have no unique solution (the rows aren’t independent).

2×2: $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$

3×3: expand along any row or column. Each element times its minor (the 2×2 left when you cover its row and column), with signs from the chequerboard: $\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}$ Choose the row or column with the most zeros to save work.

2×2 inverse: swap the main diagonal, negate the other two, divide by the determinant: ${\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

Cramer’s rule

For equations $A\mathbf{x}=\mathbf{b}$:

1. Find $\Delta =\det A$.
2. For each unknown, replace its column of $A$ with $\mathbf{b}$ and take that determinant, ${\Delta }_k$.
3. $x_k={\Delta }_k/\Delta$.
4. Substitute back to check.

Worked example

Find $\det \left[\begin{array}{ccc}3& 1& 2\\ 0& 4& 1\\ 2& -1& 3\end{array}\right]$, then solve $2x+3y=13$, $x-y=-1$ with Cramer’s rule.

Answer

Expand along column 1 (it has a zero): $3\left|\begin{array}{cc}4& 1\\ -1& 3\end{array}\right|-0+2\left|\begin{array}{cc}1& 2\\ 4& 1\end{array}\right|=3(13)+2(1-8)=39-14=$ 25 $\Delta =\left|\begin{array}{cc}2& 3\\ 1& -1\end{array}\right|=-2-3=-5$ ${\Delta }_x=\left|\begin{array}{cc}13& 3\\ -1& -1\end{array}\right|=-13+3=-10$ → $x=2$ ${\Delta }_y=\left|\begin{array}{cc}2& 13\\ 1& -1\end{array}\right|=-2-13=-15$ → $y=3$ Check: $2(2)+3(3)=13$ ✓, $2-3=-1$ ✓

On the calculator and in MATLAB

• Scientific calculators (e.g. Casio fx-991) have a Matrix mode for determinants and an Equation / Simul mode that solves 3×3 systems directly. Use Cramer when you need to show working.
• MATLAB: A = [2, 3; 1, -1] (commas separate columns, semicolons rows), det(A), and x = A\b with b = [13; -1]. b must be a column, so use ;.

Common mistakes

• Forgetting the minus sign in the middle of the $+-+$ pattern.
• Multiplying matrices element by element.
• Putting the constants in the wrong column for ${\Delta }_k$.

Check yourself

1. Inverse of $\left[\begin{array}{cc}4& 2\\ 3& 5\end{array}\right]$?

$\Delta =20-6=14$, so $\frac{1}{14}\left[\begin{array}{cc}5& -2\\ -3& 4\end{array}\right]$.

2. Why does $\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$ have no inverse?

Its determinant is $4-4=0$: the second row is just twice the first.