# Determinants

### Evaluating Determinants

The determinant of a square matrix is a value related to properties of the matrix.

The determinant of a square matrix is a value calculated from the entries (or numbers) of the matrix. This value provides information about the inverse of the matrix and about the system of linear equations associated with the matrix. The determinant of matrix $A$ is denoted $\left | A\right |$. If the value of the determinant of a matrix is zero, then the inverse of the matrix does not exist.

The determinant of a $2\times 2$ matrix $A$ can be found using a formula.
$A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\hspace{40pt}\left | A\right |=ad-bc$
Notice that the determinant for a $2\times 2$ matrix is the same as the denominator in the formula for the inverse of a $2\times 2$ matrix.
\begin{aligned}A^{-1}&=\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\\&=\frac{1}{\left | A\right | }\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\end{aligned}
Matrix Determinant Does the inverse exist?
$A=\begin{bmatrix} 3 & -1 \\ 5 & \phantom{-}2 \end{bmatrix}$ $\left | A\right |=(3\cdot2)-(-1\cdot5)=11$ $A^{-1}$ exists.
$B=\begin{bmatrix} \phantom{-}4 & \phantom{-}2 \\ -6 & -3 \end{bmatrix}$ $\left | B\right |=(4\cdot-3)-(2\cdot-6)=0$ $B^{-1}$ does not exist.

The determinant of a $3\times 3$ matrix $A$ can also be found using a formula.
$A= \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\hspace{40pt}\left | A \right |= a(ei - fh) - b(di - fg) + c(dh - eg)$
Step-By-Step Example
Finding the Determinant of a $3\times 3$ Matrix with an Inverse
Calculate the determinant of $C$. Determine whether the inverse exists.
$C= \begin{bmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 3 & 1 & 2 \end{bmatrix}$
Step 1
The determinant of $C$ is found using the formula:
$C= \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\hspace{40pt}\left | C \right |= a(ei - fh) - b(di - fg) + c(dh - eg)$
Calculate:
${\color{#c42126}a}({\color{#0f6a21}ei} - {\color{#0047af}fh})$
$\begin{bmatrix} {\color{#c42126}2} & 1 & 0 \\ 1 & {\color{#0f6a21}3} & {\color{#0047af}1} \\ 3 & {\color{#0047af}1} & {\color{#0f6a21}2} \end{bmatrix}$
${\color{#c42126}2}({\color{#0f6a21}3} \cdot {\color{#0f6a21}2} - {\color{#0047af}1} \cdot {\color{#0047af}1}) = 10$
Step 2
Calculate:
${\color{#c42126}b}({\color{#0f6a21}di} - {\color{#0047af}fg})$
$\begin{bmatrix} 2 & {\color{#c42126}1} & 0 \\ {\color{#0f6a21}1} & 3 & {\color{#0047af}1} \\ {\color{#0047af}3} & 1 & {\color{#0f6a21}2} \end{bmatrix}$
${\color{#c42126}1}({\color{#0f6a21}1} \cdot {\color{#0f6a21}2} - {\color{#0047af}1} \cdot {\color{#0047af}3}) = -1$
Step 3
Calculate:
${\color{#c42126}c}({\color{#0f6a21}dh} - {\color{#0047af}eg})$
$\begin{bmatrix} 2 & 1 & {\color{#c42126}0} \\ {\color{#0f6a21}1} & {\color{#0047af}3} & 1 \\ {\color{#0047af}3} & {\color{#0f6a21}1} & 2 \end{bmatrix}$
${\color{#c42126}0}({\color{#0f6a21}1} \cdot {\color{#0f6a21}1} - {\color{#0047af}3} \cdot {\color{#0047af}3}) = 0$
Step 4
Calculate:
$a(ei - fh) - b(di - fg) + c(dh - eg)$
$10 - (-1) + 0 = 11$
Solution
$\left | C \right |= 11$
Since the determinant of $C$ is not equal to zero, $C^{-1}$ exists.
Step-By-Step Example
Finding the Determinant of a $3\times 3$ Matrix without an Inverse
Calculate the determinant of $D$. Tell whether the inverse exists.
$D= \begin{bmatrix} 1 & 3 & 2 \\ 5 & 0 & 1 \\ 2 & 6 & 4 \end{bmatrix}$
Step 1
The determinant of $D$ is found using the formula:
$D= \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\hspace{40pt}\left | D \right |= a(ei - fh) - b(di - fg) + c(dh - eg)$
Calculate:
${\color{#c42126}a}({\color{#0f6a21}ei} - {\color{#0047af}fh})$
$\begin{bmatrix} {\color{#c42126}1} & 3 & 2 \\ 5 & {\color{#0f6a21}0} & {\color{#0047af}1} \\ 2 & {\color{#0047af}6} & {\color{#0f6a21}4} \end{bmatrix}$
${\color{#c42126}1}({\color{#0f6a21}0} \cdot {\color{#0f6a21}4} - {\color{#0047af}1} \cdot {\color{#0047af}6}) = -6$
Step 2
Calculate:
${\color{#c42126}b}({\color{#0f6a21}di} - {\color{#0047af}fg})$
$\begin{bmatrix} 1 & {\color{#c42126}3} & 2 \\ {\color{#0f6a21}5} & 0 & {\color{#0047af}1} \\ {\color{#0047af}2} & 6 & {\color{#0f6a21}4} \end{bmatrix}$
${\color{#c42126}3}({\color{#0f6a21}5} \cdot {\color{#0f6a21}4} - {\color{#0047af}1} \cdot {\color{#0047af}2}) = 54$
Step 3
Calculate:
${\color{#c42126}c}({\color{#0f6a21}dh} - {\color{#0047af}eg})$
$\begin{bmatrix} 1 & 3 & {\color{#c42126}2} \\ {\color{#0f6a21}5} & {\color{#0047af}0} & 1 \\ {\color{#0047af}2} & {\color{#0f6a21}6} & 4 \end{bmatrix}$
${\color{#c42126}2}({\color{#0f6a21}5} \cdot {\color{#0f6a21}6} - {\color{#0047af}0} \cdot {\color{#0047af}2}) = 60$
Step 4
Calculate:
$a(ei - fh) - b(di - fg) + c(dh - eg)$
$-6 - 54 + 60 = 0$
Solution
$\left | D \right |= 0$
Since the determinant of $D$ is equal to zero, $D^{-1}$ does not exist.
The procedure for finding the determinant of larger matrices follows a similar pattern, but technology can also be used.

### Using Determinants to Solve Systems of Equations

Cramer's rule is a method of using determinants to solve systems of linear equations.

Another method of solving systems of equations by using matrices is called Cramer's rule. To apply Cramer's rule, the system of linear equations should be in standard form.

1. Let $D$ represent the matrix of coefficients of the system and $C$ represent the matrix of constants. Note that $C$ has only one column.
2. Write a new matrix that replaces one column of $D$ with the constant matrix $C$. Write $D_x$ by replacing the coefficients of $x$ with the constant matrix $C$. Write $D_y$ by replacing the coefficients of $y$ with the constant matrix $C$. If there are three variables, then write $D_z$ by replacing the coefficients of $z$ with the constant matrix $C$.
3. Evaluate the determinants of $D$ and the new matrices. Divide the determinants to solve for each variable.
\begin{aligned}x=\frac{\mid D_x\mid}{\mid D\mid}\\y=\frac{\mid D_y\mid}{\mid D\mid}\\z=\frac{\mid D_z\mid}{\mid D\mid}\end{aligned}
Consider the system of equations in two variables:
\begin{aligned} ax+by&={\color{#c42126} c} \\ dx+ey&={\color{#c42126} f} \end{aligned}
The coefficient matrix is $D$. The constant matrix is $C$.
$D= \begin{bmatrix} a & b \\ d & e \end{bmatrix}\hspace{30pt}C= \begin{bmatrix} {\color{#c42126} c} \\ {\color{#c42126} f} \end{bmatrix}$
Replace the first column of $D$ with $C$ to write $D_x$. Replace the second column of $D$ with $C$ to write $D_y$.
$D_x= \begin{bmatrix} {\color{#c42126} c} & b \\ {\color{#c42126} f} & e \end{bmatrix}\hspace{30pt} D_y= \begin{bmatrix} a & {\color{#c42126} c} \\ d & {\color{#c42126} f} \end{bmatrix}$
Step-By-Step Example
Solving a System of Equations in Two Variables Using Cramer's Rule
Use Cramer's rule to solve the system of equations in two variables.
\begin{aligned} x+y &=3 \\ 2x-y &=0 \ \end{aligned}
Step 1
Write the system of equations as a coefficient matrix $D$ and a constant matrix.
\begin{aligned} x+y&=3 \\ 2x-y&=0 \end{aligned}
Coefficient matrix:
$D= \begin{bmatrix}1&\phantom{-}1\\2&-1 \end{bmatrix}$
Constant matrix:
$\begin{bmatrix} {\color{#c42126} 3} \\ {\color{#c42126} 0} \end{bmatrix}$
Step 2
Write two new matrices by replacing each column of the coefficient matrix with the constant matrix.
$D_x= \begin{bmatrix} {\color{#c42126} 3} & \phantom{-}1 \\ {\color{#c42126} 0} & -1 \end{bmatrix}$
$D_y =\begin{bmatrix} 1 & {\color{#c42126} 3} \\ 2 & {\color{#c42126} 0} \end{bmatrix}$
Step 3
Calculate the determinant of each matrix.
$\left | D \right |=\begin{vmatrix} 1&\phantom{-}1 \\ 2&-1 \end{vmatrix} = -3$
$\left | D_x \right |=\begin{vmatrix} 3&\phantom{-}1 \\ 0&-1 \end{vmatrix} = -3$
$\left | D_y \right |=\begin{vmatrix} 1&3 \\ 2&0 \end{vmatrix} = -6$
Step 4
Divide the determinants to solve for each variable.
$x=\frac{\mid D_x \mid}{\mid D \mid} = \frac {-3}{-3} = 1$
$y=\frac{\mid D_y \mid}{\mid D \mid} = \frac {-6}{-3} = 2$
Solution
The solution to the system of equations is $(1,2)$.
Consider a system of equations in three variables:
\begin{aligned} ax+by+cz&={\color{#c42126} d} \\ ex+fy+gx&={\color{#c42126} h} \\ jx+ky+mx&={\color{#c42126} n}\end{aligned}
The coefficient matrix is $D$. The constant matrix is $C$.
$D= \begin{bmatrix} a & b & c \\ e & f & g \\ j & k & m\end{bmatrix}\hspace{30pt}C= \begin{bmatrix} {\color{#c42126} d} \\ {\color{#c42126} h} \\ {\color{#c42126} n} \end{bmatrix}$
Replace the first column of $D$ with $C$ to write $D_x$. Replace the second column of $D$ with $C$ to write $D_y$. Replace the third column of $D$ with $C$ to write $D_z$.
$D_x= \begin{bmatrix} {\color{#c42126} d} & b & c \\ {\color{#c42126} h} & f & g \\ {\color{#c42126} n} & k & m\end{bmatrix}\hspace{30pt} D_y= \begin{bmatrix} a & {\color{#c42126} d} & c \\ e & {\color{#c42126} h} & g \\ j & {\color{#c42126} n} & m\end{bmatrix}\hspace{30pt} D_z= \begin{bmatrix} a & b & {\color{#c42126} d} \\ e & f & {\color{#c42126} h} \\ j & k & {\color{#c42126} n}\end{bmatrix}$
Step-By-Step Example
Solving a System of Equations in Three Variables Using Cramer's Rule
Use Cramer's rule to solve the system of equations in three variables.
\begin{aligned} 3x-2y+z &=11 \\ x+2y+2z &=1 \\ 2x+y-z &=5 \ \end{aligned}
Step 1
Write the system of equations as a coefficient matrix $D$ and a constant matrix.
\begin{aligned} 3x-2y+z &=11 \\ x+2y+2z &=1 \\ 2x+y-z &=5 \ \end{aligned}
Coefficient matrix:
$D= \begin{bmatrix} 3 & -2 & \phantom{-}1 \\1 & \phantom{-}2 & \phantom{-}2 \\2 & \phantom{-}1 & -1 \end{bmatrix}$
Constant matrix:
$\begin{bmatrix} {\color{#c42126} 11} \\ {\color{#c42126} 1} \\ {\color{#c42126} 5} \end{bmatrix}$
Step 2
Write three new matrices by replacing each column of the coefficient matrix with the constant matrix.
$D_x= \begin{bmatrix} {\color{#c42126} 11} & -2& \phantom{-}1 \\ {\color{#c42126} 1} & \phantom{-}2 & \phantom{-}2 \\ {\color{#c42126} 5} & \phantom{-}1 & -1 \end{bmatrix}$
$D_y = \begin{bmatrix} 3 & {\color{#c42126} 11} & \phantom{-}1 \\ 1 & {\color{#c42126} 1} & \phantom{-}2 \\ 2 & {\color{#c42126} 5} & -1 \end{bmatrix}$
$D_z = \begin{bmatrix} 3 & -2 & {\color{#c42126} 11} \\ 1 & \phantom{-}2 & {\color{#c42126} 1} \\ 2 & \phantom{-}1 & {\color{#c42126} 5} \end{bmatrix}$
Step 3
Identify the determinant of each matrix by using the formula or technology.
$\left | D \right |= \begin{vmatrix} 3 & -2 & \phantom{-}1 \\ 1 & \phantom{-}2 & \phantom{-}2 \\ 2 & \phantom{-}1 & -1\end{vmatrix} = -25$
$\left | D_x \right |=\begin{vmatrix} 11&-2&\phantom{-}1 \\ 1&\phantom{-}2&\phantom{-}2 \\5 & \phantom{-}1 & -1 \end{vmatrix} = -75$
$\left | D_y \right |=\begin{vmatrix} 3&11&\phantom{-}1 \\ 1&1&\phantom{-}2 \\ 2&5&-1 \end{vmatrix} = 25$
$\left | D_z \right |=\begin{vmatrix} 3&-2&11 \\ 1&\phantom{-}2&1 \\ 2&\phantom{-}1&5 \end{vmatrix} = 0$
Step 4
Divide the determinants to solve for each variable.
$x=\frac{\mid D_x \mid}{\mid D \mid} = \frac {-75}{-25} = 3$
$y=\frac{\mid D_y \mid}{\mid D \mid} = \frac {25}{-25} = -1$
$z=\frac{\mid D_z \mid}{\mid D \mid} = \frac {0}{-25} = 0$
Solution
The solution to the system of equations is $(3,-1,0)$.