Linear Algebra for Team-Based Inquiry Learning

2023 Edition

Steven Clontz Drew Lewis
University of South Alabama

August 24, 2023

Section 5.2: Computing Determinants (GT2)

Remark 5.2.1

We’ve seen that row reducing all the way into RREF gives us a method of computing determinants.

However, we learned in Chapter 1 Systems of Linear Equations (LE) that this can be tedious for large matrices. Thus, we will try to figure out how to turn the determinant of a larger matrix into the determinant of a smaller matrix.

Activity 5.2.1 (~5 min)

The following image illustrates the transformation of the unit cube by the matrix \(\left[\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 0 & 1\end{array}\right]\text{.}\)

Figure 1. Transformation of the unit cube by the linear transformation.

Recall that for this solid \(V=Bh\text{,}\) where \(h\) is the height of the solid and \(B\) is the area of its parallelogram base. So what must its volume be?

  1. \(\displaystyle \det \left[\begin{array}{cc} 1 & 1 \\ 1 & 3 \end{array}\right]\)

  2. \(\displaystyle \det \left[\begin{array}{cc} 1 & 0 \\ 3 & 1 \end{array}\right]\)

  3. \(\displaystyle \det \left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right]\)

  4. \(\displaystyle \det \left[\begin{array}{cc} 1 & 3 \\ 0 & 0 \end{array}\right]\)

Fact 5.2.2

If row \(i\) contains all zeros except for a \(1\) on the main (upper-left to lower-right) diagonal, then both column and row \(i\) may be removed without changing the value of the determinant.

\begin{equation*} \det \left[\begin{array}{cccc} 3 & {\color{red} 2} & -1 & 3 \\ {\color{red} 0} & {\color{red} 1} & {\color{red} 0} & {\color{red} 0} \\ -1 & {\color{red} 4} & 1 & 0 \\ 5 & {\color{red} 0} & 11 & 1 \end{array}\right] = \det \left[\begin{array}{ccc} 3 & -1 & 3 \\ -1 & 1 & 0 \\ 5 & 11 & 1 \end{array}\right] \end{equation*}

Since row and column operations affect the determinants in the same way, the same technique works for a column of all zeros except for a \(1\) on the main diagonal.

\begin{equation*} \det \left[\begin{array}{cccc} 3 & {\color{red} 0} & -1 & 5 \\ {\color{red} 2} & {\color{red} 1} & {\color{red} 4} & {\color{red} 0} \\ -1 & {\color{red} 0} & 1 & 11 \\ 3 & {\color{red} 0} & 0 & 1 \end{array}\right] = \det \left[\begin{array}{ccc} 3 & -1 & 5 \\ -1 & 1 & 11 \\ 3 & 0 & 1 \end{array}\right] \end{equation*}

Put another way, if you have either a column or row from the identity matrix, you can cancel both the column and row containing the \(1\text{.}\)

Activity 5.2.2 (~5 min)

Remove an appropriate row and column of \(\det \left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 5 & 12 \\ 3 & 2 & -1 \end{array}\right]\) to simplify the determinant to a \(2\times 2\) determinant.

Activity 5.2.3 (~5 min)

Simplify \(\det \left[\begin{array}{ccc} 0 & 3 & -2 \\ 2 & 5 & 12 \\ 0 & 2 & -1 \end{array}\right]\) to a multiple of a \(2\times 2\) determinant by first doing the following:

Part 1.

Factor out a \(2\) from a column.

Activity 5.2.3 (~5 min)

Simplify \(\det \left[\begin{array}{ccc} 0 & 3 & -2 \\ 2 & 5 & 12 \\ 0 & 2 & -1 \end{array}\right]\) to a multiple of a \(2\times 2\) determinant by first doing the following:

Part 2.

Swap rows or columns to put a \(1\) on the main diagonal.

Activity 5.2.4 (~5 min)

Simplify \(\det \left[\begin{array}{ccc} 4 & -2 & 2 \\ 3 & 1 & 4 \\ 1 & -1 & 3\end{array}\right]\) to a multiple of a \(2\times 2\) determinant by first doing the following:

Part 1.

Use row/column operations to create two zeroes in the same row or column.

Activity 5.2.4 (~5 min)

Simplify \(\det \left[\begin{array}{ccc} 4 & -2 & 2 \\ 3 & 1 & 4 \\ 1 & -1 & 3\end{array}\right]\) to a multiple of a \(2\times 2\) determinant by first doing the following:

Part 2.

Factor/swap as needed to get a row/column of all zeroes except a \(1\) on the main diagonal.

Observation 5.2.4

Using row/column operations, you can introduce zeros and reduce dimension to whittle down the determinant of a large matrix to a determinant of a smaller matrix.

\begin{align*} \det\left[\begin{array}{cccc} 4 & 3 & 0 & 1 \\ 2 & -2 & 4 & 0 \\ -1 & 4 & 1 & 5 \\ 2 & 8 & 0 & 3 \end{array}\right] &= \det\left[\begin{array}{cccc} 4 & 3 & {\color{red} 0} & 1 \\ 6 & -18 & {\color{red} 0} & -20 \\ {\color{red} {-1}} & {\color{red} 4} & {\color{red} 1} & {\color{red} 5} \\ 2 & 8 & {\color{red} 0} & 3 \end{array}\right] = \det\left[\begin{array}{ccc} 4 & 3 & 1 \\ 6 & -18 & -20 \\ 2 & 8 & 3 \end{array}\right]\\ &=\dots= -2\det\left[\begin{array}{ccc} {\color{red} 1} & {\color{red} 3} & {\color{red} 4} \\ {\color{red} 0} & 21 & 43 \\ {\color{red} 0} & -1 & -10 \end{array}\right] = -2\det\left[\begin{array}{cc} 21 & 43 \\ -1 & -10 \end{array}\right]\\ &= \dots= -2\det\left[\begin{array}{cc} -167 & {\color{red}{21}} \\ {\color{red} 0} & {\color{red} 1} \end{array}\right] = -2\det[-167]\\ &=-2(-167)\det(I)= 334 \end{align*}

Activity 5.2.5 (~10 min)

Rewrite

\begin{equation*} \det \left[\begin{array}{cccc} 2 & 1 & -2 & 1 \\ 3 & 0 & 1 & 4 \\ -2 & 2 & 3 & 0 \\ -2 & 0 & -3 & -3 \end{array}\right] \end{equation*}
as a multiple of a determinant of a \(3\times3\) matrix.

Activity 5.2.6 (~20 min)

Compute \(\det\left[\begin{array}{cccc} 2 & 3 & 5 & 0 \\ 0 & 3 & 2 & 0 \\ 1 & 2 & 0 & 3 \\ -1 & -1 & 2 & 2 \end{array}\right]\) by using any combination of row/column operations.

Observation 5.2.5

Another option is to take advantage of the fact that the determinant is linear in each row or column. This approach is called Laplace expansion or cofactor expansion.

For example, since \(\color{blue}{ \left[\begin{array}{ccc} 1 & 2 & 4 \end{array}\right] = 1\left[\begin{array}{ccc} 1 & 0 & 0 \end{array}\right] + 2\left[\begin{array}{ccc} 0 & 1 & 0 \end{array}\right] + 4\left[\begin{array}{ccc} 0 & 0 & 1 \end{array}\right]} \text{,}\)

\begin{align*} \det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ {\color{blue} 1} & {\color{blue} 2} & {\color{blue} 4} \end{array}\right] &= {\color{blue} 1}\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ {\color{blue} 1} & {\color{blue} 0} & {\color{blue} 0} \end{array}\right] + {\color{blue} 2}\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ {\color{blue} 0} & {\color{blue} 1} & {\color{blue} 0} \end{array}\right] + {\color{blue} 4}\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ {\color{blue} 0} & {\color{blue} 0} & {\color{blue} 1} \end{array}\right]\\ &= -1\det \left[\begin{array}{ccc} 5 & 3 & 2 \\ 5 & 3 & -1 \\ 0 & 0 & 1 \end{array}\right] -2\det \left[\begin{array}{ccc} 2 & 5 & 3 \\ -1 & 5 & 3 \\ 0 & 0 & 1 \end{array}\right] + 4\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ 0 & 0 & 1 \end{array}\right]\\ &= -\det \left[\begin{array}{cc} 5 & 3 \\ 5 & 3 \end{array}\right] -2 \det \left[\begin{array}{cc} 2 & 5 \\ -1 & 5 \end{array}\right] +4 \det \left[\begin{array}{cc} 2 & 3 \\ -1 & 3 \end{array}\right] \end{align*}

Observation 5.2.6

Recall the formula for a \(2\times 2\) determinant found in Observation 5.1.18 :

\begin{equation*} \det \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] = ad-bc\text{.} \end{equation*}

There are formulas and algorithms for the determinants of larger matrices, but they can be pretty tedious to use. For example, writing out a formula for a \(4\times 4\) determinant would require 24 different terms!

\begin{equation*} \det\left[\begin{array}{cccc} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{array}\right] = a_{11}(a_{22}(a_{33}a_{44}-a_{43}a_{34})-a_{23}(a_{32}a_{44}-a_{42}a_{34})+\dots)+\dots \end{equation*}

Activity 5.2.7 (~5 min)

Based on the previous activities, which technique is easier for computing determinants?

  1. Memorizing formulas.

  2. Using row/column operations.

  3. Laplace expansion.

  4. Some other technique.

Activity 5.2.8 (~10 min)

Use your preferred technique to compute \(\det\left[\begin{array}{cccc} 4 & -3 & 0 & 0 \\ 1 & -3 & 2 & -1 \\ 3 & 2 & 0 & 3 \\ 0 & -3 & 2 & -2 \end{array}\right] \text{.}\)