9-05 Determinants of Matrices

Determinant of 2×2 Matrix

Evaluate \(\left|\begin{matrix} 2 & 4 \\ -1 & 3 \end{matrix}\right|\)

Solution

Multiply the elements on the downward diagonal, then subtract the product of the upward diagonal.

$$ \require{cancel} \left|\begin{matrix} 2 & 4 \\ -1 & 3 \end{matrix}\right| $$

$$ = \left|\color{blue}{\bcancel{\color{red}{\cancel{\color{black}{\begin{matrix} 2 & 4 \\ -1 & 3 \end{matrix}}}}}}\right| $$

=(2·3) − (−1·4)

= 10

Determinant of 3×3 Matrix

Evaluate \(\left|\begin{matrix} 1 & 0 & 2 \\ -2 & 3 & 1 \\ -1 & 2 & 0 \end{matrix}\right|\)

Solution

Copy the first two columns after the matrix.

$$ \left|\begin{matrix} 1 & 0 & 2 \\ -2 & 3 & 1 \\ -1 & 2 & 0 \end{matrix}\right| \begin{matrix} 1 & 0 \\ -2 & 3 \\ -1 & 2 \end{matrix} $$

Add the products of the downward diagonals and subtract the products of the upward diagonals.

$$ \require{cancel} \left|\begin{matrix} \color{blue}{\bcancel{\color{black}{1}}} & \color{indigo}{\bcancel{\color{black}{0}}} & \color{purple}{\bcancel{\color{red}{\cancel{\color{black}{2}}}}} \\ -2 & \color{blue}{\bcancel{\color{red}{\cancel{\color{black}{3}}}}} & \color{indigo}{\bcancel{\color{orange}{\cancel{\color{black}{1}}}}} \\ \color{red}{\cancel{\color{black}{-1}}} & \color{orange}{\cancel{\color{black}{2}}} & \color{blue}{\bcancel{\color{green}{\cancel{\color{black}{0}}}}} \end{matrix}\right| \begin{matrix} \color{orange}{\cancel{\color{black}{1}}} & \color{green}{\cancel{\color{black}{0}}} \\ \color{purple}{\bcancel{\color{green}{\cancel{\color{black}{-2}}}}} & 3 \\ \color{indigo}{\bcancel{\color{black}{-1}}} & \color{purple}{\bcancel{\color{black}{2}}} \end{matrix} $$

= (1·3·0) + (0·1·−1) + (2·−2·2) − (−1·3·2) − (2·1·1) − (0·−2·0)

= 0 + 0 + (−8) − (−6) − 2 − 0

= −4

Minors and Cofactors

Given the matrix \(\left[\begin{matrix} 1 & 0 & 2 \\ -2 & 1 & 0 \\ 3 & -2 & -1 \end{matrix}\right]\), find

1. the minor M₂₁
2. cofactor C₂₁

Solution

1. The first number in the subscript is the row number and the second number is the column number. The minor M₂₁ indicates to cross out the 2nd row and 1st column.

   $$ \require{enclose} \left[\begin{matrix} \enclose{verticalstrike}{1} & 0 & 2 \\ \enclose{circle}{-2} & \enclose{horizontalstrike}{1} & \enclose{horizontalstrike}{0} \\ \enclose{verticalstrike}{3} & -2 & -1 \end{matrix}\right] $$

   The minor is the determinant of a matrix made out of all the noncrossed out numbers

   $$ \left|\begin{matrix} 0 & 2 \\ -2 & -1 \end{matrix}\right| $$

   $$ \require{cancel} \left|\color{blue}{\bcancel{\color{red}{\cancel{\color{black}{\begin{matrix} 0 & 2 \\ -2 & -1 \end{matrix}}}}}}\right| $$

   = 0·(−1) − (−2)·2

   M₂₁ = 4

2. The cofactor C₂₁ means to multiply the sign from the sign pattern and the minor M₂₁. To find the sign, look at the sign pattern and determine the sign for the element in the 2nd row, 1st column.

   $$ \require{enclose} \left[\begin{matrix} + & - & + & - & \cdots \\ \enclose{circle}{-} & + & - & + & \cdots \\ + & - & + & - & \cdots \\ - & + & - & + & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix}\right] $$

   Multiply the − from the sign pattern with the minor from part a.

   C₂₁ = −4

Expansion by Cofactors

Evaluate using expansion by cofactors \(\left[\begin{matrix} 1 & 0 & 2 \\ -2 & 1 & 0 \\ 3 & -2 & -1 \end{matrix}\right]\)

Solution

Choose a row or column that has a lot of zeros in it. For this problem, that is either the 1st row, 2nd row, 2nd column, or 3rd column which each have one zero. I will choose the 1st row. The determinant is the elements of the 1st row multiplied by their respective cofactors.

det = 1C₁₁ + 0C₁₂ + 2C₁₃

Start by crossing out the 1st row and 1st column and find the cofactor C₁₁.

$$ \require{enclose} \left[\begin{matrix} \color{blue}{\enclose{circle}{\color{black}{1}}} & \color{blue}{\enclose{horizontalstrike}{\color{black}{0}}} & \color{blue}{\enclose{horizontalstrike}{\color{black}{2}}} \\ \color{blue}{\enclose{verticalstrike}{\color{black}{-2}}} & 1 & 0 \\ \color{blue}{\enclose{verticalstrike}{\color{black}{3}}} & -2 & -1 \end{matrix}\right]\ $$

The sign pattern has a "+" in position 1,1.

$$ C_{11} = \left|\color{blue}{\bcancel{\color{red}{\cancel{\color{black}{\begin{matrix} 1 & 0 \\ -2 & -1 \end{matrix}}}}}}\right| $$

C₁₁ = (1·−1) − (−2·0)

C₁₁ = −1

Next move on to the next entry in the 1st row. Cross out the 1st row and 2nd column to find the cofactor C₁₂.

$$ \require{enclose} \left[\begin{matrix} \color{blue}{\enclose{horizontalstrike}{\color{black}{1}}} & \color{blue}{\enclose{circle}{\color{black}{0}}} & \color{blue}{\enclose{horizontalstrike}{\color{black}{2}}} \\ -2 & \color{blue}{\enclose{verticalstrike}{\color{black}{1}}} & 0 \\ 3 & \color{blue}{\enclose{verticalstrike}{\color{black}{-2}}} & -1 \end{matrix}\right]\ $$

The sign pattern has a "−" in position 1,2.

$$ C_{12} = -\left|\color{blue}{\bcancel{\color{red}{\cancel{\color{black}{\begin{matrix} -2 & 0 \\ 3 & -1 \end{matrix}}}}}}\right| $$

C₁₂ = −((−2·−1) − (3·0))

C₁₂ = −2

Next move on to the next entry in the 1st row. Cross out the 1st row and 3rd column to find the cofactor C₁₃.

$$ \require{enclose} \left[\begin{matrix} \color{blue}{\enclose{horizontalstrike}{\color{black}{1}}} & \color{blue}{\enclose{horizontalstrike}{\color{black}{0}}} & \color{blue}{\enclose{circle}{\color{black}{2}}} \\ -2 & 1 & \color{blue}{\enclose{verticalstrike}{\color{black}{0}}} \\ 3 & -2 & \color{blue}{\enclose{verticalstrike}{\color{black}{-1}}} \end{matrix}\right]\ $$

The sign pattern has a "+" in position 1,3.

$$ C_{13} = \left|\color{blue}{\bcancel{\color{red}{\cancel{\color{black}{\begin{matrix} -2 & 1 \\ 3 & -2 \end{matrix}}}}}}\right| $$

C₁₃ = (−2·−2) − (3·1)

C₁₃ = 1

The determinant is the elements from the 1st row multiplied by their cofactors.

det = 1C₁₁ + 0C₁₂ + 2C₁₃

det = 1(−1) + 0(−2) + 2(1)

det = 1

Notice that C₁₂ is multiplied by 0, so we did not have to calculate it.

Expansion by Cofactors

Evaluate \(\left[\begin{matrix} 1 & 0 & -1 & 2 \\ -2 & 1 & 1 & 0 \\ 4 & 2 & -1 & 0 \\ 0 & -3 & 1 & 2 \end{matrix}\right]\)

Solution

Choose a row or column that has a lot of zeros in it. For this problem, the 4th column has the most zeros. The determinant is the elements of the 4th column multiplied by their respective cofactors.

det = 2C₁₄ + 0C₂₄ + 0C₃₄ + 2C₄₄

Notice that C₂₄ and C₃₄ are multiplied by 0, so it does not matter what those two cofactors equal since their terms will equal 0.

Start by crossing out the 1st row and 4th column and find the cofactor C₁₄.

$$ \require{enclose} \left[\begin{matrix} \color{blue}{\enclose{horizontalstrike}{\color{black}{1}}} & \color{blue}{\enclose{horizontalstrike}{\color{black}{0}}} & \color{blue}{\enclose{horizontalstrike}{\color{black}{-1}}} & \color{blue}{\enclose{circle}{\color{black}{2}}} \\ -2 & 1 & 1 & \color{blue}{\enclose{verticalstrike}{\color{black}{0}}} \\ 4 & 2 & -1 & \color{blue}{\enclose{verticalstrike}{\color{black}{0}}} \\ 0 & -3 & 1 & \color{blue}{\enclose{verticalstrike}{\color{black}{2}}} \end{matrix}\right]\ $$

The sign pattern has a "−" in position 1,4.

$$ C_{14} = -\left|\begin{matrix} -2 & 1 & 1 \\ 4 & 2 & -1 \\ 0 & -3 & 1 \end{matrix}\right| $$

$$ -\require{cancel} \left|\begin{matrix} \color{blue}{\bcancel{\color{black}{-2}}} & \color{indigo}{\bcancel{\color{black}{1}}} & \color{purple}{\bcancel{\color{red}{\cancel{\color{black}{1}}}}} \\ 4 & \color{blue}{\bcancel{\color{red}{\cancel{\color{black}{2}}}}} & \color{indigo}{\bcancel{\color{orange}{\cancel{\color{black}{-1}}}}} \\ \color{red}{\cancel{\color{black}{0}}} & \color{orange}{\cancel{\color{black}{-3}}} & \color{blue}{\bcancel{\color{green}{\cancel{\color{black}{1}}}}} \end{matrix}\right| \begin{matrix} \color{orange}{\cancel{\color{black}{-2}}} & \color{green}{\cancel{\color{black}{1}}} \\ \color{purple}{\bcancel{\color{green}{\cancel{\color{black}{4}}}}} & 2 \\ \color{indigo}{\bcancel{\color{black}{0}}} & \color{purple}{\bcancel{\color{black}{-3}}} \end{matrix} $$

C₁₄ = −((−2·2·1) + (1·−1·0) + (1·4·−3) − (0·2·1) − (−3·−1·−2) − (1·4·1))

C₁₄ = −(−4 + 0 + (−12) − 0 − (−6) − 4)

C₁₄ = 14

Next move on to the next entry in the 4th column. Cross out the 2nd row and 4th column to find the cofactor C₂₄, but wait! This will be multiplied by 0, so it does not matter what C₂₄ equals. We will therefore skip calculating it.

Next move on to the next entry in the 4th column. Cross out the 3rd row and 4th column to find the cofactor C₃₄, but wait! This will be multiplied by 0, so it does not matter what C₃₄ equals. We will therefore skip calculating it.

Next move on to the next entry in the 4th column. Cross out the 4th row and 4th column to find the cofactor C₄₄.

$$ \require{enclose} \left[\begin{matrix} 1 & 0 & -1 & \color{blue}{\enclose{verticalstrike}{\color{black}{2}}} \\ -2 & 1 & 1 & \color{blue}{\enclose{verticalstrike}{\color{black}{0}}} \\ 4 & 2 & -1 & \color{blue}{\enclose{verticalstrike}{\color{black}{0}}} \\ \color{blue}{\enclose{horizontalstrike}{\color{black}{0}}} & \color{blue}{\enclose{horizontalstrike}{\color{black}{-3}}} & \color{blue}{\enclose{horizontalstrike}{\color{black}{1}}} & \color{blue}{\enclose{circle}{\color{black}{2}}} \end{matrix}\right]\ $$

The sign pattern has a "+" in position 4,4.

$$ C_{44} = \left|\begin{matrix} 1 & 0 & -1 \\ -2 & 1 & 1 \\ 4 & 2 & -1 \end{matrix}\right| $$

$$ \require{cancel} \left|\begin{matrix} \color{blue}{\bcancel{\color{black}{1}}} & \color{indigo}{\bcancel{\color{black}{0}}} & \color{purple}{\bcancel{\color{red}{\cancel{\color{black}{-1}}}}} \\ -2 & \color{blue}{\bcancel{\color{red}{\cancel{\color{black}{1}}}}} & \color{indigo}{\bcancel{\color{orange}{\cancel{\color{black}{1}}}}} \\ \color{red}{\cancel{\color{black}{4}}} & \color{orange}{\cancel{\color{black}{2}}} & \color{blue}{\bcancel{\color{green}{\cancel{\color{black}{-1}}}}} \end{matrix}\right| \begin{matrix} \color{orange}{\cancel{\color{black}{1}}} & \color{green}{\cancel{\color{black}{0}}} \\ \color{purple}{\bcancel{\color{green}{\cancel{\color{black}{-2}}}}} & 1 \\ \color{indigo}{\bcancel{\color{black}{4}}} & \color{purple}{\bcancel{\color{black}{2}}} \end{matrix} $$

C₄₄ = (1·1·−1) + (0·1·4) + (−1·−2·2) − (4·1·−1) − (2·1·1) − (−1·−2·0)

C₄₄ = −1 + 0 + 4 − (−4) − 2 − 0

C₄₄ = 5

The determinant is the elements from the 1st row multiplied by their cofactors.

det = 2C₁₄ + 0C₂₄ + 0C₃₄ + 2C₄₄

det = 2(14) + 0 + 0 + 2(5)

det = 38