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Cramer's Rule

Today we'll use matrices to derive Cramer's Rule, a rule by which we can solve systems of equations by just plugging in numbers. (Cramer's Rule works for $n \times n$ matrices but we'll use the $2 \times 2$ case only.)

Let's start with a generic system of two linear equations in two unknowns:

\begin{array}{lll} a x + b y & = & e \\ c x + d y & = & f \end{array}

(1)

Now let $A = (\begin{array}{cc} a & b \\ c & d \end{array})$ ; $X = (\begin{array}{c} x \\ y \end{array})$ ; $E = (\begin{array}{c} e \\ f \end{array})$ .

We can see that the equation $A X = E$ is equivalent to the system (1):

\begin{array}{rcl} A X & = & E \\ (\begin{array}{cc} a & b \\ c & d \end{array}) (\begin{array}{c} x \\ y \end{array}) & = & (\begin{array}{c} e \\ f \end{array}) \\ (\begin{array}{c} a x + b y \\ c x + d y \end{array}) & = & (\begin{array}{c} e \\ f \end{array}) \end{array}

Finding a solution means finding values for $x$ and $y$ that make it true, or in the language of matrices, finding a value for $X$ that makes it true. Well, we already have a single (matrix) equation involving $X$ :

\begin{array}{lll} A X & = & E \end{array}

How do we solve it? If we were solving a simple equation about numbers, we could divide both sides by $A$ , or equivalently, multiply both sides by $\frac{1}{A}$ . With matrices, we can do almost the same thing: multiply both sides by $A^{- 1}$ :

\begin{array}{lll} A^{- 1} A X & = & A^{- 1} E \end{array}

(2)

And, just like, say, $\frac{1}{3} \cdot 3 = 1$ , it is also true that $A^{- 1} A = I$ so $A^{- 1} A X = I X = X$ . Thus we can simplify equation (2) thus:

X = A^{- 1} E

(3)

This tells us our solution! Because we learned recently that ${(\begin{array}{cc} a & b \\ c & d \end{array})}^{- 1} = \frac{1}{| \begin{array}{c} A \end{array} |} (\begin{array}{cc} d & - b \\ - c & a \end{array})$ , we can work out the matrix multiplication:

\begin{array}{rcl} X & = & A^{- 1} E \\ = & \frac{1}{| \begin{array}{c} A \end{array} |} (\begin{array}{cc} d & - b \\ - c & a \end{array}) (\begin{array}{c} e \\ f \end{array}) \\ = & \frac{1}{| \begin{array}{c} A \end{array} |} (\begin{array}{c} d e - b f \\ a f - c e \end{array}) (4) \end{array}

This (4) is Cramer's Rule.

Example

Solve the system

\begin{array}{rcl} 8 x + 5 y & = & 2 \\ 2 x - 4 y & = & - 10 \end{array}

Solution (using Cramer's Rule):

\begin{array}{rcl} X & = & \frac{1}{| \begin{array}{cc} 8 & 5 \\ 2 & - 4 \end{array} |} (\begin{array}{c} 2 (- 4) - 5 (- 10) \\ 8 (- 10) - 2 (2) \end{array}) \\ = & \frac{1}{8 (- 4) - 2 \cdot 5} (\begin{array}{c} 42 \\ - 84 \end{array}) \\ = & \frac{1}{- 42} (\begin{array}{c} 42 \\ - 84 \end{array}) \\ = & (\begin{array}{c} \frac{42}{- 42} \\ \frac{- 84}{- 42} \end{array}) \\ = & (\begin{array}{c} - 1 \\ 2 \end{array}) \end{array}

Now we should check that we have a solution. We can either plug into the system or into the matrix form. Let's try the latter.

\begin{array}{rcl} (\begin{array}{cc} 8 & 5 \\ 2 & - 4 \end{array}) (\begin{array}{c} - 1 \\ 2 \end{array}) & = & (\begin{array}{c} 8 (- 1) + 5 \cdot 2 \\ 2 (- 1) + (- 4) 2 \end{array}) \\ = & (\begin{array}{c} 2 \\ - 10 \end{array}) \end{array}

Whew. Let's try one more: Solve the system

\begin{array}{rcl} 2 x + y & = & 3 \\ 5 x + 6 y & = & 4 \end{array}

So let $A = (\begin{array}{cc} 2 & 1 \\ 5 & 6 \end{array})$ ; $X = (\begin{array}{c} x \\ y \end{array})$ ; $E = (\begin{array}{c} 3 \\ 4 \end{array})$ , and we wish to solve $A X = E$ . Cramer's Rule says

\begin{array}{rcl} X & = & \frac{1}{2 \cdot 6 - 5 \cdot 1} (\begin{array}{c} 6 \cdot 3 - 1 \cdot 4 \\ 2 \cdot 4 - 5 \cdot 3 \end{array}) \\ = & \frac{1}{7} (\begin{array}{c} 14 \\ - 7 \end{array}) \\ = & (\begin{array}{c} 2 \\ - 1 \end{array}) \end{array}

Let's check: $(\begin{array}{cc} 2 & 1 \\ 5 & 6 \end{array}) (\begin{array}{c} 2 \\ - 1 \end{array}) = (\begin{array}{c} 2 \cdot 2 + 1 (- 1) \\ 5 \cdot 2 + 6 (- 1) \end{array}) = (\begin{array}{c} 3 \\ 4 \end{array})$ . Whew again.