## Section5.1Row Operations and Determinants (GT1)

### Subsection5.1.1Class Activities

#### Activity5.1.1.

The image in Figure 46 illustrates how the linear transformation $$T : \IR^2 \rightarrow \IR^2$$ given by the standard matrix $$A = \left[\begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array}\right]$$ transforms the unit square.

##### (a)

What are the lengths of $$A\vec e_1$$ and $$A\vec e_2\text{?}$$

##### (b)

What is the area of the transformed unit square?

#### Activity5.1.2.

The image below illustrates how the linear transformation $$S : \IR^2 \rightarrow \IR^2$$ given by the standard matrix $$B = \left[\begin{array}{cc} 2 & 3 \\ 0 & 4 \end{array}\right]$$ transforms the unit square.

##### (a)

What are the lengths of $$B\vec e_1$$ and $$B\vec e_2\text{?}$$

##### (b)

What is the area of the transformed unit square?

#### Observation5.1.3.

It is possible to find two nonparallel vectors that are scaled but not rotated by the linear map given by $$B\text{.}$$

\begin{equation*} B\vec e_1=\left[\begin{array}{cc} 2 & 3 \\ 0 & 4 \end{array}\right]\left[\begin{array}{c}1\\0\end{array}\right] =\left[\begin{array}{c}2\\0\end{array}\right]=2\vec e_1 \end{equation*}
\begin{equation*} B\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right] = \left[\begin{array}{cc} 2 & 3 \\ 0 & 4 \end{array}\right]\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right] = \left[\begin{array}{c}3\\2\end{array}\right] = 4\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right] \end{equation*}

The process for finding such vectors will be covered later in this chapter.

#### Observation5.1.4.

Notice that while a linear map can transform vectors in various ways, linear maps always transform parallelograms into parallelograms, and these areas are always transformed by the same factor: in the case of $$B=\left[\begin{array}{cc} 2 & 3 \\ 0 & 4 \end{array}\right]\text{,}$$ this factor is $$8\text{.}$$

Since this change in area is always the same for a given linear map, it will be equal to the value of the transformed unit square (which begins with area $$1$$).

#### Remark5.1.5.

We will define the determinant of a square matrix $$B\text{,}$$ or $$\det(B)$$ for short, to be the factor by which $$B$$ scales areas. In order to figure out how to compute it, we first figure out the properties it must satisfy.

#### Activity5.1.6.

The transformation of the unit square by the standard matrix $$[\vec{e}_1\hspace{0.5em} \vec{e}_2]=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]=I$$ is illustrated below. If $$\det([\vec{e}_1\hspace{0.5em} \vec{e}_2])=\det(I)$$ is the area of resulting parallelogram, what is the value of $$\det([\vec{e}_1\hspace{0.5em} \vec{e}_2])=\det(I)\text{?}$$

The value for $$\det([\vec{e}_1\hspace{0.5em} \vec{e}_2])=\det(I)$$ is:

1. 0

2. 1

3. 2

4. 4

#### Activity5.1.7.

The transformation of the unit square by the standard matrix $$[\vec{v}\hspace{0.5em} \vec{v}]$$ is illustrated below: both $$T(\vec{e}_1)=T(\vec{e}_2)=\vec{v}\text{.}$$ If $$\det([\vec{v}\hspace{0.5em} \vec{v}])$$ is the area of the generated parallelogram, what is the value of $$\det([\vec{v}\hspace{0.5em} \vec{v}])\text{?}$$

The value of $$\det([\vec{v}\hspace{0.5em} \vec{v}])$$ is:

1. 0

2. 1

3. 2

4. 4

#### Activity5.1.8.

The transformations of the unit square by the standard matrices $$[\vec{v}\hspace{0.5em} \vec{w}]$$ and $$[c\vec{v}\hspace{0.5em} \vec{w}]$$ are illustrated below. Describe the value of $$\det([c\vec{v}\hspace{0.5em} \vec{w}])\text{.}$$

Describe the value of $$\det([c\vec{v}\hspace{0.5em} \vec{w}])\text{:}$$

1. $$\displaystyle \det([\vec{v}\hspace{0.5em} \vec{w}])$$

2. $$\displaystyle c\det([\vec{v}\hspace{0.5em} \vec{w}])$$

3. $$\displaystyle c^2\det([\vec{v}\hspace{0.5em} \vec{w}])$$

4. Cannot be determined from this information.

Consider the vectors $$\vec{u}\text{,}$$ $$\vec{v}\text{,}$$ $$\vec{u}+\vec{v}\text{,}$$ and $$\vec{w}$$ displayed below. Each pair of vectors generates a parallelogram, and the area of each parallelogram can be described in terms of determinants.

For example, $$\det([\vec{u}\hspace{0.5em} \vec{w}])$$ represents the shaded area shown below.

Similarly, $$\det([\vec{v}\hspace{0.5em} \vec{w}])$$ represents the shaded area shown below.

#### Activity5.1.9.

The paralellograms generated by the standard matrices $$[\vec{u}\hspace{0.5em} \vec{w}]\text{,}$$ $$[\vec{v}\hspace{0.5em} \vec{w}]$$ and $$[\vec{u}+\vec{v}\hspace{0.5em} \vec{w}]$$ are illustrated below.

Describe the value of $$\det([\vec{u}+\vec{v}\hspace{0.5em} \vec{w}])\text{.}$$

1. $$\displaystyle \det([\vec{u}\hspace{0.5em} \vec{w}])=\det([\vec{v}\hspace{0.5em} \vec{w}])$$

2. $$\displaystyle \det([\vec{u}\hspace{0.5em} \vec{w}])+\det([\vec{v}\hspace{0.5em} \vec{w}])$$

3. $$\displaystyle \det([\vec{u}\hspace{0.5em} \vec{w}])\det([\vec{v}\hspace{0.5em} \vec{w}])$$

4. Cannot be determined from this information.

#### Definition5.1.10.

The determinant is the unique function $$\det:M_{n,n}\to\IR$$ satisfying these properties:

1. $$\displaystyle \det(I)=1$$

2. $$\det(A)=0$$ whenever two columns of the matrix are identical.

3. $$\det[\cdots\hspace{0.5em}c\vec{v}\hspace{0.5em}\cdots]= c\det[\cdots\hspace{0.5em}\vec{v}\hspace{0.5em}\cdots]\text{,}$$ assuming no other columns change.

4. $$\det[\cdots\hspace{0.5em}\vec{v}+\vec{w}\hspace{0.5em}\cdots]= \det[\cdots\hspace{0.5em}\vec{v}\hspace{0.5em}\cdots]+ \det[\cdots\hspace{0.5em}\vec{w}\hspace{0.5em}\cdots]\text{,}$$ assuming no other columns change.

Note that these last two properties together can be phrased as “The determinant is linear in each column.”

#### Observation5.1.11.

The determinant must also satisfy other properties. Consider $$\det([\vec v \hspace{1em}\vec w+c \vec{v}])$$ and $$\det([\vec v\hspace{1em}\vec w])\text{.}$$

The base of both parallelograms is $$\vec{v}\text{,}$$ while the height has not changed, so the determinant does not change either. This can also be proven using the other properties of the determinant:

\begin{align*} \det([\vec{v}+c\vec{w}\hspace{1em}\vec{w}]) &= \det([\vec{v}\hspace{1em}\vec{w}])+ \det([c\vec{w}\hspace{1em}\vec{w}])\\ &= \det([\vec{v}\hspace{1em}\vec{w}])+ c\det([\vec{w}\hspace{1em}\vec{w}])\\ &= \det([\vec{v}\hspace{1em}\vec{w}])+ c\cdot 0\\ &= \det([\vec{v}\hspace{1em}\vec{w}]) \end{align*}

#### Remark5.1.12.

Swapping columns may be thought of as a reflection, which is represented by a negative determinant. For example, the following matrices transform the unit square into the same parallelogram, but the second matrix reflects its orientation.

\begin{equation*} A=\left[\begin{array}{cc}2&3\\0&4\end{array}\right]\hspace{1em}\det A=8\hspace{3em} B=\left[\begin{array}{cc}3&2\\4&0\end{array}\right]\hspace{1em}\det B=-8 \end{equation*}

#### Observation5.1.13.

The fact that swapping columns multiplies determinants by a negative may be verified by adding and subtracting columns.

\begin{align*} \det([\vec{v}\hspace{1em}\vec{w}]) &= \det([\vec{v}+\vec{w}\hspace{1em}\vec{w}])\\ &= \det([\vec{v}+\vec{w}\hspace{1em}\vec{w}-(\vec{v}+\vec{w})])\\ &= \det([\vec{v}+\vec{w}\hspace{1em}-\vec{v}])\\ &= \det([\vec{v}+\vec{w}-\vec{v}\hspace{1em}-\vec{v}])\\ &= \det([\vec{w}\hspace{1em}-\vec{v}])\\ &= -\det([\vec{w}\hspace{1em}\vec{v}]) \end{align*}

#### Activity5.1.15.

The transformation given by the standard matrix $$A$$ scales areas by $$4\text{,}$$ and the transformation given by the standard matrix $$B$$ scales areas by $$3\text{.}$$ By what factor does the transformation given by the standard matrix $$AB$$ scale areas?

1. $$\displaystyle 1$$

2. $$\displaystyle 7$$

3. $$\displaystyle 12$$

4. Cannot be determined

#### Remark5.1.17.

Recall that row operations may be produced by matrix multiplication.

• Multiply the first row of $$A$$ by $$c\text{:}$$ $$\left[\begin{array}{cccc} c & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]A$$

• Swap the first and second row of $$A\text{:}$$ $$\left[\begin{array}{cccc} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]A$$

• Add $$c$$ times the third row to the first row of $$A\text{:}$$ $$\left[\begin{array}{cccc} 1 & 0 & c & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]A$$

#### Activity5.1.19.

Consider the row operation $$R_1+4R_3\to R_1$$ applied as follows to show $$A\sim B\text{:}$$

\begin{equation*} A=\left[\begin{array}{cccc}1&2&3 & 4\\5&6 & 7 & 8\\9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16\end{array}\right] \sim \left[\begin{array}{cccc}1+4(9)&2+4(10)&3+4(11) & 4+4(12) \\5&6 & 7 & 8\\9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16\end{array}\right]=B \end{equation*}
##### (a)

Find a matrix $$R$$ such that $$B=RA\text{,}$$ by applying the same row operation to $$I=\left[\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right]\text{.}$$

##### (b)

Find $$\det R$$ by comparing with the previous slide.

##### (c)

If $$C \in M_{4,4}$$ is a matrix with $$\det(C)= -3\text{,}$$ find

\begin{equation*} \det(RC)=\det(R)\det(C). \end{equation*}

#### Activity5.1.20.

Consider the row operation $$R_1\leftrightarrow R_3$$ applied as follows to show $$A\sim B\text{:}$$

\begin{equation*} A=\left[\begin{array}{cccc}1&2&3&4\\5&6&7&8\\9&10&11&12 \\ 13 & 14 & 15 & 16\end{array}\right] \sim \left[\begin{array}{cccc}9&10&11&12\\5&6&7&8\\1&2&3&4 \\ 13 & 14 & 15 & 16\end{array}\right]=B \end{equation*}
##### (a)

Find a matrix $$R$$ such that $$B=RA\text{,}$$ by applying the same row operation to $$I\text{.}$$

##### (b)

If $$C \in M_{4,4}$$ is a matrix with $$\det(C)= 5\text{,}$$ find $$\det(RC)\text{.}$$

#### Activity5.1.21.

Consider the row operation $$3R_2\to R_2$$ applied as follows to show $$A\sim B\text{:}$$

\begin{equation*} A=\left[\begin{array}{cccc}1&2&3&4\\5&6&7&8\\9&10&11&12 \\ 13 & 14 & 15 & 16\end{array}\right] \sim \left[\begin{array}{cccc}1&2&3&4\\3(5)&3(6)&3(7)&3(8)\\9&10&11&12 \\ 13 & 14 & 15 & 16\end{array}\right]=B \end{equation*}
##### (a)

Find a matrix $$R$$ such that $$B=RA\text{.}$$

##### (b)

If $$C \in M_{4,4}$$ is a matrix with $$\det(C)= -7\text{,}$$ find $$\det(RC)\text{.}$$

#### Remark5.1.22.

Recall that the column versions of the three row-reducing operations a matrix may be used to simplify a determinant:

1. Multiplying columns by scalars:

\begin{equation*} \det([\cdots\hspace{0.5em}c\vec{v}\hspace{0.5em} \cdots])= c\det([\cdots\hspace{0.5em}\vec{v}\hspace{0.5em} \cdots]) \end{equation*}
2. Swapping two columns:

\begin{equation*} \det([\cdots\hspace{0.5em}\vec{v}\hspace{0.5em} \cdots\hspace{1em}\vec{w}\hspace{0.5em} \cdots])= -\det([\cdots\hspace{0.5em}\vec{w}\hspace{0.5em} \cdots\hspace{1em}\vec{v}\hspace{0.5em} \cdots]) \end{equation*}
3. Adding a multiple of a column to another column:

\begin{equation*} \det([\cdots\hspace{0.5em}\vec{v}\hspace{0.5em} \cdots\hspace{1em}\vec{w}\hspace{0.5em} \cdots])= \det([\cdots\hspace{0.5em}\vec{v}+c\vec{w}\hspace{0.5em} \cdots\hspace{1em}\vec{w}\hspace{0.5em} \cdots]) \end{equation*}

#### Remark5.1.23.

The determinants of row operation matrices may be computed by manipulating columns to reduce each matrix to the identity:

• Scaling a row: $$\left[\begin{array}{cccc} 1 & 0 & 0 &0 \\ 0 & c & 0 &0\\ 0 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 \end{array}\right]$$

• Swapping rows: $$\left[\begin{array}{cccc} 0 & 1 & 0 &0 \\ 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$$

• Adding a row multiple to another row: $$\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & c & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$$

#### Observation5.1.25.

So we may compute the determinant of $$\left[\begin{array}{cc} 2 & 4 \\ 2 & 3 \end{array}\right]$$ by manipulating its rows/columns to reduce the matrix to $$I\text{:}$$

\begin{align*} \det\left[\begin{array}{cc} 2 & 4 \\ 2 & 3 \end{array}\right] &= 2 \det \left[\begin{array}{cc} 1 & 2 \\ 2 & 3 \end{array}\right]\\ &= %2 \det \left[\begin{array}{cc} 1 & 2 \\ 2-2(1) & 3-2(2)\end{array}\right]= 2 \det \left[\begin{array}{cc} 1 & 2 \\ 0 & -1 \end{array}\right]\\ &= %2(-1) \det \left[\begin{array}{cc} 1 & -2 \\ 0 & +1 \end{array}\right]= -2 \det \left[\begin{array}{cc} 1 & -2 \\ 0 & 1 \end{array}\right]\\ &= %-2 \det \left[\begin{array}{cc} 1+2(0) & -2+2(1) \\ 0 & 1\end{array}\right] = -2 \det \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\\ &= %-2\det I = %-2(1) = -2 \end{align*}

### Subsection5.1.3Slideshow

Slideshow of activities available at GT1.slides.html.

### Exercises5.1.4Exercises

Exercises available at https://stevenclontz.github.io/checkit-tbil-la-2021-dev/#/bank/GT1/.

### Subsection5.1.5Sample Problem and Solution

Sample problem Example B.1.21.