11.3: Determinants and Cramer's Rule for 2 X 2 Matrices

$2 \times 2$ Determinants

Let $A$ be an $n\times n$ matrix. That is, let $A$ be a square matrix. The determinant of $A$, denoted by $\det \left( A\right)$ is a very important number which we will explore throughout this section.

If $A$ is a 2$\times 2$ matrix, the determinant is given by the following formula.

Definition $\PageIndex{1}$ Determinant of a Two By Two Matrix

Let $A=\left[ \begin{array}{rr} a & b \\ c & d \end{array} \right] .$ Then $$\det \left( A\right) = ad-bc\nonumber$$

The determinant is also often denoted by enclosing the matrix with two vertical lines. Thus $$\det \left[ \begin{array}{rr} a & b \\ c & d \end{array} \right] =\left| \begin{array}{rr} a & b \\ c & d \end{array} \right| =ad - bc\nonumber$$

The following is an example of finding the determinant of a $2 \times 2$ matrix.

Example $\PageIndex{1}$

Find $\det\left(A\right)$ for the matrix $A = \left[ \begin{array}{rr} 2 & 4 \\ -1 & 6 \end{array} \right] .$

Solution

From Definition $\PageIndex{1}$, $$\det \left( A\right) = \left( 2\right) \left( 6\right) -\left( 4\right) \left( -1\right) = 12 + 4 = 16\nonumber$$

Cramer’s Rule for a $2\times 2$ Matrix

Cramer's Rule gives a way to solve systems of equations using determinants. Although it may seem as though it is a more difficult way to solve systems of equations when the coefficients are constants, it is a much more efficient way to solve systems with functions as coefficients, which is predominantly what we see in differential equations. We omit the proof as it requires a significant amount of linear algebra.

Theorem $\PageIndex{1}$ Cramer's Rule for a $2\times 2$ Matrix

Consider the following system of equations: $$\begin{aligned}a_{11}x_1+a_{12}x_2&=b_1 \\ a_{21}x_1+a_{22}x_2&=b_2\end{aligned}\nonumber$$

Let $A=\left[ \begin{array}{rr} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]$ and b$= \left[ \begin{array}{r} b_1 \\ b_2 \end{array} \right].$

Let $A_1$ be the matrix where the first column of $A$ has been replaced with b and $A_2$ the matrix where the second column of $A$ has been replaced with b.

$$x_1= \frac{\det \left(A_{1}\right)}{\det \left(A\right)} = \frac{\left| \begin{array}{rrr} b_1 & a_{12} \\ b_2 & a_{22} \end{array} \right| }{\left| \begin{array}{rrr} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| }\nonumber$$ and $$x_2= \frac{\det \left(A_{2}\right)}{\det \left(A\right)} = \frac{\left| \begin{array}{rrr} a_{11} & b_1 \\ a_{21} & b_2 \end{array} \right| }{\left| \begin{array}{rrr} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| }\nonumber$$

Example $\PageIndex{2}$

Use Cramer's Rule to solve

$$\begin{aligned}3x_1+5x_2&=19 \\ 7x_1-4x_2&=13\end{aligned}\nonumber$$

Solution

We have Let $A=\left[ \begin{array}{rr} 3 & 5 \\ 7 & -4 \end{array} \right]$ and b$= \left[ \begin{array}{r} 19 \\ 13 \end{array} \right].\nonumber$

$$x_1= \frac{\det \left(A_{1}\right)}{\det \left(A\right)} = \frac{\left| \begin{array}{rrr} 19 & 5 \\ 13 & -4 \end{array} \right| }{\left| \begin{array}{rrr} 3 & 5 \\ 7 & -4 \end{array} \right| }={-141\over -47}=3\nonumber$$ and $$x_2= \frac{\det \left(A_{2}\right)}{\det \left(A\right)}=\frac{\left| \begin{array}{rrr} 3 & 19 \\ 7 & 13 \end{array} \right| }{ \left| \begin{array}{rrr} 3 & 5 \\ 7 & -4 \end{array} \right|}={-94\over -47}=2\nonumber$$