Precalculus by Richard Wright

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Jesus said, “My kingdom is not of this world. If it were, my servants would fight to prevent my arrest by the Jewish leaders. But now my kingdom is from another place.” John‬ ‭18‬:‭36‬ ‭NIV‬‬‬‬‬‬‬‬‬

9-05 Determinants of Matrices

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.6.4

La Roche-Guyon castle's vegetable garden
Figure 1: La Roche-Guyon castle's vegetable garden. credit (wikimedia/DocteurCosmos)

The La Roche-Guyon castle's vegetable garden in France is open to the public. The shapes of these individual gardens are triangular. Let's say the gardener needs to put three inches of fertilizer in one of the gardens. If he were given just a map with coordinates, how could he calculate the amount of fertilizer needed?

He could find the lengths of the sides using the distance formula, or he could use a determinant. A determinant is a number associated with a square matrix. It is symbolized by vertical lines around a matrix instead of brackets, |A|. It is not unique, as different matrices can have the same determinant, but it is useful in diverse applications such as finding areas of triangles, finding equations of lines, and solving systems of linear equations.

Shortcuts for Finding Determinants

There are shortcuts for finding determinants of 2×2 and 3×3 matrices. For both of these, remember to add downward products and subtract upward products.

Determinants of 2×2 and 3×3 Matrices

2×2 Determinant

Product of down diagonal − product of up diagonal

$$ \require{cancel} \left|\begin{matrix} a & b \\ c & d \end{matrix}\right| = \left|\color{blue}{\bcancel{\color{red}{\cancel{\color{black}{\begin{matrix} a & b \\ c & d \end{matrix}}}}}}\right| = \color{blue}{a·d} - \color{red}{c·b} $$

3×3 Determinant

    $$ \left|\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right| $$

  1. Copy the first two columns after the matrix.
  2. $$ \left|\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right| \begin{matrix} a & b \\ d & e \\ g & h \end{matrix} $$

  3. Add the products of the downward diagonals and subtract the products of the upward diagonals.
  4. $$ \require{cancel} \left|\begin{matrix} \color{blue}{\bcancel{\color{black}{a}}} & \color{indigo}{\bcancel{\color{black}{b}}} & \color{purple}{\bcancel{\color{red}{\cancel{\color{black}{c}}}}} \\ d & \color{blue}{\bcancel{\color{red}{\cancel{\color{black}{e}}}}} & \color{indigo}{\bcancel{\color{orange}{\cancel{\color{black}{f}}}}} \\ \color{red}{\cancel{\color{black}{g}}} & \color{orange}{\cancel{\color{black}{h}}} & \color{blue}{\bcancel{\color{green}{\cancel{\color{black}{i}}}}} \end{matrix}\right| \begin{matrix} \color{orange}{\cancel{\color{black}{a}}} & \color{green}{\cancel{\color{black}{b}}} \\ \color{purple}{\bcancel{\color{green}{\cancel{\color{black}{d}}}}} & e \\ \color{indigo}{\bcancel{\color{black}{g}}} & \color{purple}{\bcancel{\color{black}{h}}} \end{matrix} $$

    $$ = \color{blue}{(a·e·i)} + \color{indigo}{(b·f·g)} + \color{purple}{(e·d·h)} - \color{red}{(g·e·c)} - \color{orange}{(h·f·a)} - \color{green}{(i·d·b)} $$

Example 1: 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.

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

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

$$ = \color{blue}{(2·3)} - \color{red}{(-1·4)} $$

$$ = 10 $$

Try It 1

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

Answer

−2

Example 2: 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.

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

$$ = \color{blue}{(1·3·0)} + \color{indigo}{(0·1·-1)} + \color{purple}{(2·-2·2)} - \color{red}{(-1·3·2)} - \color{orange}{(2·1·1)} - \color{green}{(0·-2·0)} $$

$$ = \color{blue}{0} + \color{indigo}{0} + \color{purple}{(-8)} - \color{red}{(-6)} - \color{orange}{2} - \color{green}{0} $$

$$ = -4 $$

Try It 2

Evaluate \(\left|\begin{matrix} -2 & 3 & 0 \\ 1 & -2 & 3 \\ 5 & 0 & -1 \end{matrix}\right|\)

Answer

44

Determinant by Expansion by Cofactors

The diagonals only work to find determinants of 2×2 or 3×3. To find the determinant of any larger matrix, another method must be used.

The first item to know is the minor which is the determinant of a matrix created by crossing out a row and column of a larger matrix. A cofactor is a minor with the sign from the sign pattern below.

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

Example 3: 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 \(M_{21}\)
  2. cofactor \(C_{21}\)
Solution
  1. The first number in the subscript is the row number and the second number is the column number. The minor \(M_{21}\) 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

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

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

    $$ = \color{blue}{0·(-1)} - \color{red}{(-2)·2} $$

    $$ M_{21} = 4 $$

  2. The cofactor \(C_{21}\) means to multiply the sign from the sign pattern and the minor \(M_{21}\). 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.

    $$ C_{21} = -4 $$

Expansion by Cofactors

To find a determinant by expansion by cofactors

  1. Choose a row or column. A row or column with the most zeros possible is preferred.
  2. Starting with the top of the column or left of the row, write the entry and multiply it by the cofactor. Repeat for the entire row or column and add them together.

Example 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.

$$ det = 1C_{11} + 0C_{12} + 2C_{13} $$

Start by crossing out the 1st row and 1st column and find the cofactor \(C_{11}\).

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

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

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

$$ C_{11} = \color{blue}{(1·-1)} - \color{red}{(-2·0)} $$

$$ C_{11} = -1 $$

Next move on to the next entry in the 1st row. Cross out the 1st row and 2nd column to find the cofactor \(C_{12}\).

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

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

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

$$ C_{12} = -(\color{blue}{(-2·-1)} - \color{red}{(3·0)}) $$

$$ C_{12} = -2 $$

Next move on to the next entry in the 1st row. Cross out the 1st row and 3rd column to find the cofactor \(C_{13}\).

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

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

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

$$ C_{13} = \color{blue}{(-2·-2)} - \color{red}{(3·1)} $$

$$ C_{13} = 1 $$

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

$$ det = 1C_{11} + 0C_{12} + 2C_{13} $$

$$ det = 1(-1) + 0(-2) + 2(1) $$

$$ det = 1 $$

Notice that \(C_{12}\) is multiplied by 0, so we did not have to calculate it.

Example 5: 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.

$$ det = 2C_{14} + \color{blue}{0C_{24} + 0C_{34}} + 2C_{44} $$

Notice that \(C_{24}\) and \(C_{34}\) 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 \(C_{14}\).

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

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

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

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

$$ C_{14} = -(\color{blue}{(-2·2·1)} + \color{indigo}{(1·-1·0)} + \color{purple}{(1·4·-3)} - \color{red}{(0·2·1)} - \color{orange}{(-3·-1·-2)} - \color{green}{(1·4·1)}) $$

$$ C_{14} = -(\color{blue}{-4} + \color{indigo}{0} + \color{purple}{(-12)} - \color{red}{0} - \color{orange}{(-6)} - \color{green}{4}) $$

$$ C_{14} = 14 $$

Next move on to the next entry in the 4th column. Cross out the 2nd row and 4th column to find the cofactor \(C_{24}\), but wait! This will be multiplied by 0, so it does not matter what \(C_{24}\) 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 \(C_{34}\), but wait! This will be multiplied by 0, so it does not matter what \(C_{34}\) 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 \(C_{44}\).

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

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

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

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

$$ C_{44} = \color{blue}{(1·1·-1)} + \color{indigo}{(0·1·4)} + \color{purple}{(-1·-2·2)} - \color{red}{(4·1·-1)} - \color{orange}{(2·1·1)} - \color{green}{(-1·-2·0)} $$

$$ C_{44} = \color{blue}{-1} + \color{indigo}{0} + \color{purple}{4} - \color{red}{(-4)} - \color{orange}{2} - \color{green}{0} $$

$$ C_{44} = 5 $$

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

$$ det = 2C_{14} + 0C_{24} + 0C_{34} + 2C_{44} $$

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

$$ det = 38 $$

Try It 3

Evaluate using expansion by cofactors \(\left[\begin{matrix} 0 & 3 & -1 \\ 4 & 1 & -3 \\ 0 & 1 & -1 \end{matrix}\right]\)

Answer

8

Lesson Summary

Determinants of 2×2 and 3×3 Matrices

2×2 Determinant

Product of down diagonal − product of up diagonal

$$ \require{cancel} \left|\begin{matrix} a & b \\ c & d \end{matrix}\right| = \left|\color{blue}{\bcancel{\color{red}{\cancel{\color{black}{\begin{matrix} a & b \\ c & d \end{matrix}}}}}}\right| = \color{blue}{a·d} - \color{red}{c·b} $$

3×3 Determinant

    $$ \left|\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right| $$

  1. Copy the first two columns after the matrix.
  2. $$ \left|\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right| \begin{matrix} a & b \\ d & e \\ g & h \end{matrix} $$

  3. Add the products of the downward diagonals and subtract the products of the upward diagonals.
  4. $$ \require{cancel} \left|\begin{matrix} \color{blue}{\bcancel{\color{black}{a}}} & \color{indigo}{\bcancel{\color{black}{b}}} & \color{purple}{\bcancel{\color{red}{\cancel{\color{black}{c}}}}} \\ d & \color{blue}{\bcancel{\color{red}{\cancel{\color{black}{e}}}}} & \color{indigo}{\bcancel{\color{orange}{\cancel{\color{black}{f}}}}} \\ \color{red}{\cancel{\color{black}{g}}} & \color{orange}{\cancel{\color{black}{h}}} & \color{blue}{\bcancel{\color{green}{\cancel{\color{black}{i}}}}} \end{matrix}\right| \begin{matrix} \color{orange}{\cancel{\color{black}{a}}} & \color{green}{\cancel{\color{black}{b}}} \\ \color{purple}{\bcancel{\color{green}{\cancel{\color{black}{d}}}}} & e \\ \color{indigo}{\bcancel{\color{black}{g}}} & \color{purple}{\bcancel{\color{black}{h}}} \end{matrix} $$

    $$ = \color{blue}{(a·e·i)} + \color{indigo}{(b·f·g)} + \color{purple}{(e·d·h)} - \color{red}{(g·e·c)} - \color{orange}{(h·f·a)} - \color{green}{(i·d·b)} $$


Expansion by Cofactors

To find a determinant by expansion by cofactors

  1. Choose a row or column. A row or column with the most zeros possible is preferred.
  2. Starting with the top of the column or left of the row, write the entry and multiply it by the cofactor. Repeat for the entire row or column and add them together.

Helpful videos about this lesson.

Practice Exercises

  1. Describe what the subscripts on the minors and cofactors mean, i.e. \(M_{35}\) or \(C_{42}\).
  2. Evaluate the determinant using the shortcuts.

  3. \(\left|\begin{matrix} 2 & 0 \\ 5 & 1 \end{matrix}\right|\)
  4. \(\left|\begin{matrix} 3 & -1 \\ 6 & 7 \end{matrix}\right|\)
  5. \(\left|\begin{matrix} -2 & 6 \\ -5 & -4 \end{matrix}\right|\)
  6. \(\left|\begin{matrix} 1 & 0 & 2 \\ 3 & 7 & -5 \\ -2 & 4 & 0 \end{matrix}\right|\)
  7. \(\left|\begin{matrix} -1 & -2 & -3 \\ 3 & 2 & 1 \\ 4 & -4 & 5 \end{matrix}\right|\)
  8. \(\left|\begin{matrix} 1 & 2 & 1 \\ 0 & 5 & -2 \\ -4 & -1 & 3 \end{matrix}\right|\)
  9. Find the (a) minor and (b) cofactor of the matrix
    \(\left|\begin{matrix} 2 & 1 & 4 \\ -1 & 5 & -2 \\ 0 & -1 & 3 \end{matrix}\right|\)

  10. a. \(M_{23}\) b. \(C_{23}\)
  11. a. \(M_{13}\) b. \(C_{13}\)
  12. a. \(M_{32}\) b. \(C_{32}\)
  13. a. \(M_{31}\) b. \(C_{31}\)
  14. Evaluate the determinant using expansion by cofactors.

  15. \(\left|\begin{matrix} 1 & 3 & -2 \\ 0 & 0 & 2 \\ -5 & 1 & 4 \end{matrix}\right|\)
  16. \(\left|\begin{matrix} -1 & 4 & -2 \\ 1 & 2 & 5 \\ 0 & -3 & 2 \end{matrix}\right|\)
  17. \(\left|\begin{matrix} 1 & 0 & -2 & 1 \\ 0 & -1 & 2 & -1 \\ 3 & 2 & 0 & -2 \\ 0 & 5 & 1 & 1 \end{matrix}\right|\)
  18. \(\left|\begin{matrix} -1 & 3 & 5 & 1 & 2 \\ 0 & 2 & 4 & 0 & 0 \\ 0 & 4 & -3 & 0 & 3 \\ 0 & 0 & 2 & 0 & 0 \\ 1 & -4 & -5 & 4 & -2 \end{matrix}\right|\)
  19. Mixed Review

  20. (9-04) Find the inverse of \(\left[\begin{matrix} 2 & 3 \\ -4 & 1 \end{matrix}\right]\).
  21. (9-04) Use an inverse matrix to solve the system \(\left\{\begin{align} 2x + 3y &= 5 \\ -4x + y &= 11 \end{align}\right.\).
  22. (9-03) Simplify \(\left[\begin{matrix} 2 & 1 \\ -3 & 1 \end{matrix}\right] \left[\begin{matrix} 5 & -3 \\ 2 & 10 \end{matrix}\right]\).
  23. (9-01) Use Gaussian Elimination to solve \(\left\{\begin{align} x + y + z &= -1 \\ y - z &= 9 \\ 2y + z &= 0 \end{align}\right.\).
  24. (8-05) Solve by graphing \(\left\{\begin{align} x - y &≥ -2 \\ x + y &≤ 6 \\ y &≥ x^2 - 6x + 9 \end{align}\right.\).

Answers

  1. The first number is the row and the second number is the column which are crossed out to create the new matrix. Take the determinant of that matrix.
  2. 2
  3. 27
  4. 38
  5. 72
  6. 68
  7. 49
  8. −2; 2
  9. 1; 1
  10. 0; 0
  11. −22; −22
  12. −32
  13. −21
  14. −37
  15. −60
  16. \(\left[\begin{matrix} \frac{1}{14} & -\frac{3}{14} \\ \frac{2}{7} & \frac{1}{7} \end{matrix}\right]\)
  17. (−2, 3)
  18. \(\left[\begin{matrix} 12 & 4 \\ -13 & 19 \end{matrix}\right]\)
  19. (2, 3, −6)
  20. answer