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College Algebra/Matrices/Determinants

Matrices

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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 AA is denoted ∣A∣\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 2×22\times 2 matrix AA can be found using a formula.
A=[abcd]∣A∣=ad−bcA=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\hspace{40pt}\left | A\right |=ad-bc
Notice that the determinant for a 2×22\times 2 matrix is the same as the denominator in the formula for the inverse of a 2×22\times 2 matrix.
A−1=1ad−bc[d−b−ca]=1∣A∣[d−b−ca]\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=[3−15−2]A=\begin{bmatrix} 3 & -1 \\ 5 & \phantom{-}2 \end{bmatrix} ∣A∣=(3⋅2)−(−1⋅5)=11\left | A\right |=(3\cdot2)-(-1\cdot5)=11 A−1A^{-1} exists.
B=[−4−2−6−3]B=\begin{bmatrix} \phantom{-}4 & \phantom{-}2 \\ -6 & -3 \end{bmatrix} ∣B∣=(4⋅−3)−(2⋅−6)=0\left | B\right |=(4\cdot-3)-(2\cdot-6)=0 B−1B^{-1} does not exist.

The determinant of a 3×33\times 3 matrix AA can also be found using a formula.
A=[abcdefghi]∣A∣=a(ei−fh)−b(di−fg)+c(dh−eg)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 3×33\times 3 Matrix with an Inverse
Calculate the determinant of CC. Determine whether the inverse exists.
C=[210131312]C= \begin{bmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 3 & 1 & 2 \end{bmatrix}
Step 1
The determinant of CC is found using the formula:
C=[abcdefghi]∣C∣=a(ei−fh)−b(di−fg)+c(dh−eg)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:
a(ei−fh){\color{#c42126}a}({\color{#0f6a21}ei} - {\color{#0047af}fh})
[210131312]\begin{bmatrix} {\color{#c42126}2} & 1 & 0 \\ 1 & {\color{#0f6a21}3} & {\color{#0047af}1} \\ 3 & {\color{#0047af}1} & {\color{#0f6a21}2} \end{bmatrix}
2(3⋅2−1⋅1)=10{\color{#c42126}2}({\color{#0f6a21}3} \cdot {\color{#0f6a21}2} - {\color{#0047af}1} \cdot {\color{#0047af}1}) = 10
Step 2
Calculate:
b(di−fg){\color{#c42126}b}({\color{#0f6a21}di} - {\color{#0047af}fg})
[210131312]\begin{bmatrix} 2 & {\color{#c42126}1} & 0 \\ {\color{#0f6a21}1} & 3 & {\color{#0047af}1} \\ {\color{#0047af}3} & 1 & {\color{#0f6a21}2} \end{bmatrix}
1(1⋅2−1⋅3)=−1{\color{#c42126}1}({\color{#0f6a21}1} \cdot {\color{#0f6a21}2} - {\color{#0047af}1} \cdot {\color{#0047af}3}) = -1
Step 3
Calculate:
c(dh−eg){\color{#c42126}c}({\color{#0f6a21}dh} - {\color{#0047af}eg})
[210131312]\begin{bmatrix} 2 & 1 & {\color{#c42126}0} \\ {\color{#0f6a21}1} & {\color{#0047af}3} & 1 \\ {\color{#0047af}3} & {\color{#0f6a21}1} & 2 \end{bmatrix}
0(1⋅1−3⋅3)=0{\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)a(ei - fh) - b(di - fg) + c(dh - eg)
10−(−1)+0=1110 - (-1) + 0 = 11
Solution
∣C∣=11\left | C \right |= 11
Since the determinant of CC is not equal to zero, C−1C^{-1} exists.
Step-By-Step Example
Finding the Determinant of a 3×33\times 3 Matrix without an Inverse
Calculate the determinant of DD. Tell whether the inverse exists.
D=[132501264]D= \begin{bmatrix} 1 & 3 & 2 \\ 5 & 0 & 1 \\ 2 & 6 & 4 \end{bmatrix}
Step 1
The determinant of DD is found using the formula:
D=[abcdefghi]∣D∣=a(ei−fh)−b(di−fg)+c(dh−eg)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:
a(ei−fh){\color{#c42126}a}({\color{#0f6a21}ei} - {\color{#0047af}fh})
[132501264]\begin{bmatrix} {\color{#c42126}1} & 3 & 2 \\ 5 & {\color{#0f6a21}0} & {\color{#0047af}1} \\ 2 & {\color{#0047af}6} & {\color{#0f6a21}4} \end{bmatrix}
1(0⋅4−1⋅6)=−6{\color{#c42126}1}({\color{#0f6a21}0} \cdot {\color{#0f6a21}4} - {\color{#0047af}1} \cdot {\color{#0047af}6}) = -6
Step 2
Calculate:
b(di−fg){\color{#c42126}b}({\color{#0f6a21}di} - {\color{#0047af}fg})
[132501264]\begin{bmatrix} 1 & {\color{#c42126}3} & 2 \\ {\color{#0f6a21}5} & 0 & {\color{#0047af}1} \\ {\color{#0047af}2} & 6 & {\color{#0f6a21}4} \end{bmatrix}
3(5⋅4−1⋅2)=54{\color{#c42126}3}({\color{#0f6a21}5} \cdot {\color{#0f6a21}4} - {\color{#0047af}1} \cdot {\color{#0047af}2}) = 54
Step 3
Calculate:
c(dh−eg){\color{#c42126}c}({\color{#0f6a21}dh} - {\color{#0047af}eg})
[132501264]\begin{bmatrix} 1 & 3 & {\color{#c42126}2} \\ {\color{#0f6a21}5} & {\color{#0047af}0} & 1 \\ {\color{#0047af}2} & {\color{#0f6a21}6} & 4 \end{bmatrix}
2(5⋅6−0⋅2)=60{\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)a(ei - fh) - b(di - fg) + c(dh - eg)
−6−54+60=0-6 - 54 + 60 = 0
Solution
∣D∣=0\left | D \right |= 0
Since the determinant of DD is equal to zero, D−1D^{-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 DD represent the matrix of coefficients of the system and CC represent the matrix of constants. Note that CC has only one column.
2. Write a new matrix that replaces one column of DD with the constant matrix CC. Write DxD_x by replacing the coefficients of xx with the constant matrix CC. Write DyD_y by replacing the coefficients of yy with the constant matrix CC. If there are three variables, then write DzD_z by replacing the coefficients of zz with the constant matrix CC.
3. Evaluate the determinants of DD and the new matrices. Divide the determinants to solve for each variable.
x=∣Dx∣∣D∣y=∣Dy∣∣D∣z=∣Dz∣∣D∣\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:
ax+by=cdx+ey=f\begin{aligned} ax+by&={\color{#c42126} c} \\ dx+ey&={\color{#c42126} f} \end{aligned}
The coefficient matrix is DD. The constant matrix is CC.
D=[abde]C=[cf]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 DD with CC to write DxD_x. Replace the second column of DD with CC to write DyD_y.
Dx=[cbfe]Dy=[acdf]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.
x+y=32x−y=0 \begin{aligned} x+y &=3 \\ 2x-y &=0 \ \end{aligned}
Step 1
Write the system of equations as a coefficient matrix DD and a constant matrix.
x+y=32x−y=0\begin{aligned} x+y&=3 \\ 2x-y&=0 \end{aligned}
Coefficient matrix:
D=[1−12−1]D= \begin{bmatrix}1&\phantom{-}1\\2&-1 \end{bmatrix}
Constant matrix:
[30]\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.
Dx=[3−10−1]D_x= \begin{bmatrix} {\color{#c42126} 3} & \phantom{-}1 \\ {\color{#c42126} 0} & -1 \end{bmatrix}
Dy=[1320]D_y =\begin{bmatrix} 1 & {\color{#c42126} 3} \\ 2 & {\color{#c42126} 0} \end{bmatrix}
Step 3
Calculate the determinant of each matrix.
∣D∣=∣1−12−1∣=−3\left | D \right |=\begin{vmatrix} 1&\phantom{-}1 \\ 2&-1 \end{vmatrix} = -3
∣Dx∣=∣3−10−1∣=−3\left | D_x \right |=\begin{vmatrix} 3&\phantom{-}1 \\ 0&-1 \end{vmatrix} = -3
∣Dy∣=∣1320∣=−6\left | D_y \right |=\begin{vmatrix} 1&3 \\ 2&0 \end{vmatrix} = -6
Step 4
Divide the determinants to solve for each variable.
x=∣Dx∣∣D∣=−3−3=1x=\frac{\mid D_x \mid}{\mid D \mid} = \frac {-3}{-3} = 1
y=∣Dy∣∣D∣=−6−3=2y=\frac{\mid D_y \mid}{\mid D \mid} = \frac {-6}{-3} = 2
Solution
The solution to the system of equations is (1,2)(1,2).

Consider a system of equations in three variables:
ax+by+cz=dex+fy+gx=hjx+ky+mx=n\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 DD. The constant matrix is CC.
D=[abcefgjkm]C=[dhn]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 DD with CC to write DxD_x. Replace the second column of DD with CC to write DyD_y. Replace the third column of DD with CC to write DzD_z.
Dx=[dbchfgnkm]Dy=[adcehgjnm]Dz=[abdefhjkn]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.
3x−2y+z=11x+2y+2z=12x+y−z=5 \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 DD and a constant matrix.
3x−2y+z=11x+2y+2z=12x+y−z=5 \begin{aligned} 3x-2y+z &=11 \\ x+2y+2z &=1 \\ 2x+y-z &=5 \ \end{aligned}
Coefficient matrix:
D=[3−2−11−2−22−1−1]D= \begin{bmatrix} 3 & -2 & \phantom{-}1 \\1 & \phantom{-}2 & \phantom{-}2 \\2 & \phantom{-}1 & -1 \end{bmatrix}
Constant matrix:
[1115]\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.
Dx=[11−2−11−2−25−1−1]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}
Dy=[311−111−225−1]D_y = \begin{bmatrix} 3 & {\color{#c42126} 11} & \phantom{-}1 \\ 1 & {\color{#c42126} 1} & \phantom{-}2 \\ 2 & {\color{#c42126} 5} & -1 \end{bmatrix}
Dz=[3−2111−212−15]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.
∣D∣=∣3−2−11−2−22−1−1∣=−25\left | D \right |= \begin{vmatrix} 3 & -2 & \phantom{-}1 \\ 1 & \phantom{-}2 & \phantom{-}2 \\ 2 & \phantom{-}1 & -1\end{vmatrix} = -25
∣Dx∣=∣11−2−11−2−25−1−1∣=−75\left | D_x \right |=\begin{vmatrix} 11&-2&\phantom{-}1 \\ 1&\phantom{-}2&\phantom{-}2 \\5 & \phantom{-}1 & -1 \end{vmatrix} = -75
∣Dy∣=∣311−111−225−1∣=25\left | D_y \right |=\begin{vmatrix} 3&11&\phantom{-}1 \\ 1&1&\phantom{-}2 \\ 2&5&-1 \end{vmatrix} = 25
∣Dz∣=∣3−2111−212−15∣=0\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=∣Dx∣∣D∣=−75−25=3x=\frac{\mid D_x \mid}{\mid D \mid} = \frac {-75}{-25} = 3
y=∣Dy∣∣D∣=25−25=−1y=\frac{\mid D_y \mid}{\mid D \mid} = \frac {25}{-25} = -1
z=∣Dz∣∣D∣=0−25=0z=\frac{\mid D_z \mid}{\mid D \mid} = \frac {0}{-25} = 0
Solution
The solution to the system of equations is (3,−1,0)(3,-1,0).
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