## Section5.2Computing Determinants (GT2)

### Subsection5.2.1Class Activities

#### Remark5.2.1.

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

However, we learned in Chapter 1 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.

#### Activity5.2.2.

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

#### Activity5.2.4.

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.

#### Activity5.2.5.

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:

##### (a)

Factor out a $$2$$ from a column.

##### (b)

Swap rows or columns to put a $$1$$ on the main diagonal.

#### Activity5.2.6.

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:

##### (a)

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

##### (b)

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

#### Observation5.2.7.

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*}

#### Activity5.2.8.

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.

#### Activity5.2.9.

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.

#### Observation5.2.10.

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*}

#### Observation5.2.11.

Applying Laplace expansion to a $$2 \times 2$$ matrix yields a short formula you may have seen:

\begin{equation*} \det \left[\begin{array}{cc} {\color{blue} a} & {\color{blue} b} \\ c & d \end{array}\right] = {\color{blue} a}\det \left[\begin{array}{cc} {\color{blue} 1} & {\color{blue} 0} \\ c & d \end{array}\right] + {\color{blue} b} \det \left[\begin{array}{cc} {\color{blue} 0} & {\color{blue} 1} \\ c & d \end{array}\right] = a\det \left[\begin{array}{cc} {\color{red} 1} & {\color{red} 0} \\ {\color{red} c} & d \end{array}\right] - b \det \left[\begin{array}{cc} {\color{red} 1} & {\color{red} 0} \\ {\color{red} d} & c \end{array}\right] = ad-bc\text{.} \end{equation*}

There are formulas 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*}

So this is why we either use Laplace expansion or row/column operations directly.

#### Activity5.2.12.

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 (be prepared to describe it).

#### Activity5.2.13.

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

#### Insight5.2.14.

det([4,-3,0,0; 1,-3,2,-1; 3,2,0,3; 0,-3,2,-2])


### Subsection5.2.3Slideshow

Slideshow of activities available at https://teambasedinquirylearning.github.io/linear-algebra/2022/GT2.slides.html.

### Subsection5.2.5Mathematical Writing Explorations

#### Exploration5.2.15.

Prove that the equation of a line in the plane, through points $$(x_1,y_1), (x_2,y_2)\text{,}$$ when $$x_1 \neq x_2$$ is given by the equation $$\mbox{det}\left(\begin{array}{ccc}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{array}\right) = 0.$$

#### Exploration5.2.16.

Prove that the determinant of any diagonal matrix, upper triangular matrix, or lower triangular matrix, is the product of it's diagonal entries.

#### Exploration5.2.17.

Show that, if an $$n \times n$$ matrix $$M$$ has a non-zero determinant, then any $$\vec{v} \in \mathbb{R}^n$$ can be represented as a linear combination of the columns of $$M\text{.}$$

#### Exploration5.2.18.

What is the smallest number of zeros necessary to place in a $$4 \times 4$$ matrix, and the placement of those zeros, such that the matrix has a zero determinant?

### Subsection5.2.6Sample Problem and Solution

Sample problem Example B.1.22.