11.3: Determinants and Cramer's Rule for 2 X 2 Matrices
- Page ID
- 108334
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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.
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.
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.
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 \[x_2= \frac{\det \left(A_{2}\right)}{\det \left(A\right)} = \frac{\left| \begin{array}{rrr} a_{11} & b_1 \\ a_{21} & b_2 \end{array} \right| }{\left| \begin{array}{rrr} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| }\nonumber\]
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 \[x_2= \frac{\det \left(A_{2}\right)}{\det \left(A\right)}=\frac{\left| \begin{array}{rrr} 3 & 19 \\ 7 & 13 \end{array} \right| }{ \left| \begin{array}{rrr} 3 & 5 \\ 7 & -4 \end{array} \right|}={-94\over -47}=2\nonumber\]