Learning Outcomes
- Describe how a row operation affects the determinant of a matrix.
Section 5.1 Row Operations and Determinants (GT1)
Subsection 5.1.1 Class Activities
Activity 5.1.1.
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?
Activity 5.1.2.
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?
Observation 5.1.3.
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.
Observation 5.1.4.
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 \(1\)).
Remark 5.1.5.
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.
Activity 5.1.6.
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:
- 0
- 1
- 2
- 4
Activity 5.1.7.
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:
- 0
- 1
- 2
- 4
Activity 5.1.8.
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 \(\det([c\vec{v}\hspace{0.5em} \vec{w}])\text{:}\)
- \(\displaystyle \det([\vec{v}\hspace{0.5em} \vec{w}])\)
- \(\displaystyle c\det([\vec{v}\hspace{0.5em} \vec{w}])\)
- \(\displaystyle c^2\det([\vec{v}\hspace{0.5em} \vec{w}])\)
- Cannot be determined from this information.
Remark 5.1.9.
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.
Remark 5.1.10.
For example, \(\det([\vec{u}\hspace{0.5em} \vec{w}])\) represents the shaded area shown below.
Remark 5.1.11.
Similarly, \(\det([\vec{v}\hspace{0.5em} \vec{w}])\) represents the shaded area shown below.
Activity 5.1.12.
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 \(\det([\vec{u}+\vec{v}\hspace{0.5em} \vec{w}])\text{.}\)
- \(\displaystyle \det([\vec{u}\hspace{0.5em} \vec{w}])=\det([\vec{v}\hspace{0.5em} \vec{w}])\)
- \(\displaystyle \det([\vec{u}\hspace{0.5em} \vec{w}])+\det([\vec{v}\hspace{0.5em} \vec{w}])\)
- \(\displaystyle \det([\vec{u}\hspace{0.5em} \vec{w}])\det([\vec{v}\hspace{0.5em} \vec{w}])\)
- Cannot be determined from this information.
Definition 5.1.13.
The determinant is the unique function \(\det:M_{n,n}\to\IR\) satisfying these properties:
- \(\displaystyle \det(I)=1\)
- \(\det(A)=0\) whenever two columns of the matrix are identical.
- \(\det[\cdots\hspace{0.5em}c\vec{v}\hspace{0.5em}\cdots]= c\det[\cdots\hspace{0.5em}\vec{v}\hspace{0.5em}\cdots]\text{,}\) assuming no other columns change.
- \(\det[\cdots\hspace{0.5em}\vec{v}+\vec{w}\hspace{0.5em}\cdots]= \det[\cdots\hspace{0.5em}\vec{v}\hspace{0.5em}\cdots]+ \det[\cdots\hspace{0.5em}\vec{w}\hspace{0.5em}\cdots]\text{,}\) assuming no other columns change.
Note that these last two properties together can be phrased as “The determinant is linear in each column.”
Observation 5.1.14.
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*}
Remark 5.1.15.
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*}
Observation 5.1.16.
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*}
Fact 5.1.17.
To summarize, we’ve shown that the column versions of the three row-reducing operations a matrix may be used to simplify a determinant in the following way:
- Multiplying a column by a scalar multiplies the determinant by that scalar:\begin{equation*} c\det([\cdots\hspace{0.5em}\vec{v}\hspace{0.5em} \cdots])= \det([\cdots\hspace{0.5em}c\vec{v}\hspace{0.5em} \cdots]) \end{equation*}
- Swapping two columns changes the sign of the determinant:\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*}
- Adding a multiple of a column to another column does not change the determinant:\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*}
Activity 5.1.18.
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?
- \(\displaystyle 1\)
- \(\displaystyle 7\)
- \(\displaystyle 12\)
- Cannot be determined
Fact 5.1.19.
Since the transformation given by the standard matrix \(AB\) is obtained by applying the transformations given by \(A\) and \(B\text{,}\) it follows that
\begin{equation*}
\det(AB)=\det(A)\det(B)=\det(B)\det(A)=\det(BA)\text{.}
\end{equation*}
Remark 5.1.20.
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 \(A\text{:}\) \(\left[\begin{array}{cccc} 1 & 0 & c & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]A\)
Fact 5.1.21.
The determinants of row operation matrices may be computed by manipulating columns to reduce each matrix to the identity:
- Scaling a row: \(\det \left[\begin{array}{cccc} c & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] = c\det \left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] = c\)
- Swapping rows: \(\det \left[\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] = -1\det \left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] = -1\)
- Adding a row multiple to another row: \(\det \left[\begin{array}{cccc} 1 & 0 & c & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] = \det \left[\begin{array}{cccc} 1 & 0 & c-1c & 0\\ 0 & 1 & 0-0c & 0\\ 0 & 0 & 1-0c & 0 \\ 0 & 0 & 0-0c & 1 \end{array}\right] = \det(I)=1\)
Activity 5.1.22.
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*}
Activity 5.1.23.
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 \(\det(RC)\text{.}\)
Activity 5.1.24.
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 \(\det(RC)\text{.}\)
Activity 5.1.25.
Let \(A\) be any \(4 \times 4\) matrix with determinant \(2\text{.}\)
(a)
Let \(B\) be the matrix obtained from \(A\) by applying the row operation \(R_1 - 5 R_3 \to R_1\text{.}\) What is \(\mathrm{det}\,B\text{?}\)
- -4
- -2
- 2
- 10
(b)
Let \(M\) be the matrix obtained from \(A\) by applying the row operation \(R_3 \leftrightarrow R_1\text{.}\) What is \(\mathrm{det}\,M\text{?}\)
- -4
- -2
- 2
- 10
(c)
Let \(P\) be the matrix obtained from \(A\) by applying the row operation \(2 R_4 \to R_4\text{.}\) What is \(\mathrm{det}\,P\text{?}\)
- -4
- -2
- 2
- 10
Remark 5.1.26.
Recall that the column versions of the three row-reducing operations a matrix may be used to simplify a determinant:
- 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*}
- 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*}
- 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*}
Remark 5.1.27.
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: \(\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & c & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]\)
Fact 5.1.28.
Thus we can also use both row operations to simplify determinants:
- Multiplying rows by scalars:\begin{equation*} \det\left[\begin{array}{c}\vdots\\cR\\\vdots\end{array}\right]= c\det\left[\begin{array}{c}\vdots\\R\\\vdots\end{array}\right] \end{equation*}
- Swapping two rows:\begin{equation*} \det\left[\begin{array}{c}\vdots\\R\\\vdots\\S\\\vdots\end{array}\right]= -\det\left[\begin{array}{c}\vdots\\S\\\vdots\\R\\\vdots\end{array}\right] \end{equation*}
- Adding multiples of rows/columns to other rows:\begin{equation*} \det\left[\begin{array}{c}\vdots\\R\\\vdots\\S\\\vdots\end{array}\right]= \det\left[\begin{array}{c}\vdots\\R+cS\\\vdots\\S\\\vdots\end{array}\right] \end{equation*}
Activity 5.1.29.
Complete the following derivation for a formula calculating \(2\times 2\) determinants:
\begin{align*}
\det\left[\begin{array}{cc} a & b \\ c & d \end{array}\right]
&=
\unknown \det \left[\begin{array}{cc} 1 & b/a \\ c & d \end{array}\right]\\
&=
\unknown \det \left[\begin{array}{cc} 1 & b/a \\ c-c & d-bc/a \end{array}\right]\\
&=
\unknown \det \left[\begin{array}{cc} 1 & b/a \\ 0 & d-bc/a \end{array}\right]\\
&=
\unknown \det \left[\begin{array}{cc} 1 & b/a \\ 0 & 1 \end{array}\right]\\
&=
\unknown \det \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\\
&=
\unknown \det I\\
&=
\unknown
\end{align*}
Observation 5.1.30.
So we may compute the determinant of \(\left[\begin{array}{cc} 2 & 4 \\ 2 & 3 \end{array}\right]\) by using determinant properties to manipulate 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*}
Or we may use a formula:
\begin{equation*}
\det\left[\begin{array}{cc} 2 & 4 \\ 2 & 3 \end{array}\right]
=
(2)(3)-(4)(2)=-2
\end{equation*}
Subsection 5.1.2 Videos
Subsection 5.1.3 Slideshow
Slideshow of activities available at
https://teambasedinquirylearning.github.io/linear-algebra/2023/GT1.slides.html.Exercises 5.1.4 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2023/exercises/#/bank/GT1/.Subsection 5.1.5 Mathematical Writing Explorations
Exploration 5.1.31.
- Prove or disprove. The determinant is a linear operator on the vector space of \(n \times n\) matrices.
- Find a matrix that will double the area of a region in \(\mathbb{R}^2\text{.}\)
- Find a matrix that will triple the area of a region in \(\mathbb{R}^2\text{.}\)
- Find a matrix that will halve the area of a region in \(\mathbb{R}^2\text{.}\)
Subsection 5.1.6 Sample Problem and Solution
Sample problem Example B.1.22.
