# Lecture 4: Determinant

### Instructions

• This section covers the concepts listed below. You can click the button to go directly to that topic.
• After each video, there are notes for the material in the video.
• At the end of the page, there are exercises covering the material in the section.
• When you have finished the material below, you can go to the next section or return to the main page.

### Concepts

Determinants, Linear Independence, Spanning Set

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#### Video Notes

In this lecture, we introduce the determinant of a matrix and use it to determine when sets of vectors are linearly independent, when they are a spanning set, how this relates to properties of matrices.

Let's say we want to determine whether the vectors $$\begin{pmatrix}a\\c\end{pmatrix}$$ and $$\begin{pmatrix}b\\d\end{pmatrix}$$ are linearly independent or dependent. This means we need to determine how many solutions to:
\begin{align*}
x_1\begin{pmatrix}a\\c\end{pmatrix}
+ x_2\begin{pmatrix}b\\d\end{pmatrix}
= \begin{pmatrix}0\\0\end{pmatrix}.
\end{align*}
Another way to look at this is that we need to determine how many solutions there are to the system:
\begin{align*}
ax_1 + bx_2 &=0\\
cx_1 + dx_2 &=0.
\end{align*}
But this is the same as finding the nullspace of:
\begin{align*}
\begin{pmatrix}a & b \\c & d\end{pmatrix}.
\end{align*}
First, note that if both $$a$$ and $$c$$ are zero, the vectors are clearly dependent, so we assume at least one is not zero and we'll assume that $$a\neq 0$$. Then if we subtract $$\frac{c}{a}$$ times the first row from the second, we get:
\begin{align*}
\begin{pmatrix}a & b \\c & d\end{pmatrix}
\sim \begin{pmatrix}a & b \\ 0 & \frac{ad - bc}{a}\end{pmatrix}.
\end{align*}
If $$ad - bc = 0$$ then the matrix above has a free variable and so there is more than the zero vector in the null space (indeed, any vector of the form $$\begin{pmatrix}-\frac{b}{a}t\\t\end{pmatrix}$$ is in the null space). On the other hand, if $$ad - bc\neq 0$$ then the matrix can be reduced to the identity. This means that the only vector in the null space is $$\begin{pmatrix}0\\0\end{pmatrix}$$.

Therefore, we can relate several concepts in the next video.

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#### Video Notes

Theorem. The following are equivalent (meaning: if one is true, the rest are true):
1. $$ad - bc\neq 0$$
2. The vectors $$\begin{pmatrix}a\\c\end{pmatrix},\begin{pmatrix}b\\d\end{pmatrix}$$ are linearly independent;
3. There is only one solution to \begin{align*} ax_1 + bx_2 &=0\\cx_1 + dx_2 &=0\end{align*}
4. The only vector in the null space of the  matrix $$\begin{pmatrix}a & b\\c & d\end{pmatrix}$$ is the zero vector $$\begin{pmatrix}0\\0\end{pmatrix}$$.

For a $$2\times 2$$ matrix $$M = \begin{pmatrix}a&b\\c&d\end{pmatrix}$$, the quantity $$ab-bc$$ has a special name, the determinant:
\begin{align*}
\det\begin{pmatrix}a&b\\c&d\end{pmatrix} := ad - bc.
\end{align*}
If we want to solve the system:
\begin{align*}
\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}
=\begin{pmatrix}y_1\\y_2\end{pmatrix}
\end{align*}
and $$ad - bc \neq 0$$ then the row reduced matrix is:
\begin{align*}
\left(\begin{array}{cc|c}
a & b & y_1\\c & d & y_2
\end{array}\right)
\sim
\left(\begin{array}{cc|c}
a & b & y_1\\
0 & \frac{ad - bc}{a} & \frac{ay_2 - cy_1}{a}
\end{array}\right)
\sim \left(\begin{array}{cc|c}
a & b & y_1\\
0 & 1 & \frac{ay_2 - cy_1}{ad-bc}.
\end{array}\right)
\end{align*}
Then this gives the equivalent system:
\begin{align*}
ax_1 + bx_2 &= y_1\\
x_2 &= \frac{ay_2 - cy_1}{ad-bc},
\end{align*}
which has a unique solution. In other words, we've shown that if $$\det M\neq 0$$ then the equation $$M\boldsymbol{x} = \boldsymbol{y}$$ has a unique solution for any $$\boldsymbol{y}$$. On the other hand, if $$\det M = ad - bc = 0$$ and $$a\neq 0$$ then there is no solution to, for example:
\begin{align*}
\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}
=\begin{pmatrix}0\\1\end{pmatrix}
\end{align*}
since when row reduced, this becomes:
\begin{align*}
\left(\begin{array}{cc|c}
a & b & 0\\c & d & 1
\end{array}\right)
\sim
\left(\begin{array}{cc|c}
a & b & 0\\
0 & 0 & 1
\end{array}\right)
\end{align*}
which is the same as the system:
\begin{align*}
ax_1 + bx_2 = 0\\
0x_1 + 0x_2 = 1,
\end{align*}
but there is no solution to the second equation. In other words, if $$\det M = 0$$, then there is some vector that is not in the span of the columns of the matrix. So we have the following theorem:

Theorem. The vectors $$\begin{pmatrix}a\\c\end{pmatrix}$$ and $$\begin{pmatrix}b\\d\end{pmatrix}$$ form a spanning set for $$\mathbb{R}^2$$ if and only if $$ad - bc \neq 0$$.

Another way to say this is that we can modify the theorem above to:

Theorem. The following are equivalent (meaning: if one is true, the rest are true):
1. $$ad - bc\neq 0$$
2. The vectors $$\begin{pmatrix}a\\c\end{pmatrix},\begin{pmatrix}b\\d\end{pmatrix}$$ are linearly independent;
3. There is only one solution to \begin{align*} ax_1 + bx_2 &=0\\cx_1 + dx_2 &=0\end{align*}
4. The only vector in the null space of the  matrix $$\begin{pmatrix}a & b\\c & d\end{pmatrix}$$ is the zero vector $$\begin{pmatrix}0\\0\end{pmatrix}$$.
5. $$\begin{pmatrix}a\\c\end{pmatrix}$$ and $$\begin{pmatrix}b\\d\end{pmatrix}$$ form a spanning set for $$\mathbb{R}^2$$.

#### Example

For which values of $$\gamma$$ are the vectors $$\begin{pmatrix}2\gamma\\1\end{pmatrix}$$ and $$\begin{pmatrix}3\\-1\end{pmatrix}$$ linearly independent? Do the vectors form a spanning set for $$\mathbb{R}^2$$?

We compute the determinant of:
\begin{align*}
\det\begin{pmatrix}2\gamma & 3\\ 1 &-1\end{pmatrix}
= -2\gamma + 3.
\end{align*}
The vectors are linearly independent if and only if $$-2\gamma + 3\neq 0$$ and this happens if and only if $$\gamma \neq \frac{3}{2}$$. So the vectors are linearly independent as long as $$\gamma \neq \frac{3}{2}.$$ Similary,  the vectors form a spanning set as long as $$\gamma \neq \frac{3}{2}.$$

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L4V3

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#### Video Notes

All of these theorems are true for $$3\times 3$$ matrices as well. We just need to define the determinant:
\begin{align*}
\det
\begin{pmatrix}
a & b & c\\ d & e & f\\g & h & i
\end{pmatrix}
= a\det
\begin{pmatrix}
e & f\\ h & i
\end{pmatrix}
- b\det
\begin{pmatrix}
d & f\\ g & i
\end{pmatrix}
+c\det
\begin{pmatrix}
d & e\\ g & h
\end{pmatrix}.
\end{align*}
Recall that if $$\boldsymbol{v}_1,\boldsymbol{v}_2,\boldsymbol{v}_3 \in \mathbb{R}^3$$ then the matrix $$M = \begin{pmatrix}\boldsymbol{v}_1 & \boldsymbol{v}_2 & \boldsymbol{v}_3\end{pmatrix}$$ is the $$3\times 3$$ matrix whose columns are the vectors $$\boldsymbol{v}_1,\boldsymbol{v}_2,\boldsymbol{v}_3$$. We have the following theorem that we state with out proof:

Theorem. Let $$\boldsymbol{v}_1,\boldsymbol{v}_2, \boldsymbol{v}_3\in \mathbb{R}^3$$ and $$M = \begin{pmatrix}\boldsymbol{v}_1 & \boldsymbol{v}_2 & \boldsymbol{v}_3\end{pmatrix}$$. Then the following are equivalent:
1. $$\det M \neq 0$$
2. The vectors $$\boldsymbol{v}_1,\boldsymbol{v}_2,\boldsymbol{v}_3$$ are linearly independent.
3. There is exactly one solution to $$M\boldsymbol{x} = \boldsymbol{w}$$ for all $$\boldsymbol{w}\in\mathbb{R}^3$$.
4. The only vector in the null space of $$M$$ is the zero vector.
5. $$\boldsymbol{v}_1,\boldsymbol{v}_2,\boldsymbol{v}_3$$ form a spanning set for $$\mathbb{R}^3.$$
Example. For which values of $$\gamma$$ are the vectors $$\begin{pmatrix}1\\1\\1\end{pmatrix}, \begin{pmatrix}2\\1\\3\end{pmatrix}, \begin{pmatrix}4\gamma\\1\\2\end{pmatrix}$$ linearly independent?

Solution. We compute:
\begin{align*}
\det \begin{pmatrix}
1 & 2 & 4\gamma\\
1 & 1 & 1\\
1 & 3 & 2
\end{pmatrix}
&= \det\begin{pmatrix}1&1\\3&2\end{pmatrix}
-2\det\begin{pmatrix}1&1\\1&2\end{pmatrix}
+4\gamma\begin{pmatrix}1&1\\1&3\end{pmatrix}
\\&= -1 - 2 + 8 \gamma
\\& = -3 - 8\gamma.
\end{align*}
So the vectors are linearly independent whenever $$\gamma\neq -\frac{3}{8}$$.

Exercises

1. Determine whether the following are linearly independent or linearly dependent.
\begin{align*}
\begin{pmatrix}1\\1\end{pmatrix},\begin{pmatrix}-1\\0\end{pmatrix}
\end{align*}

We put these into a matrix and row–reduce:
\begin{align*}
\begin{pmatrix}
1&-1\\1&0
\end{pmatrix}
\xrightarrow{R2:=R2-R1}
\begin{pmatrix}
1&-1\\
0&1
\end{pmatrix}.
\end{align*}
There are no free variables, so this means there is only one solution to:
\begin{align*}
\begin{pmatrix}
1&-1\\
0&1
\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}
=\begin{pmatrix}0\\0\end{pmatrix}
\end{align*}
so they are linearly independent.

Also, the determinant is $$1$$ which shows they are linearly independent as well.

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L4E1

2. Determine whether the following are linearly independent or linearly dependent.
\begin{align*}
\begin{pmatrix}1\\1\end{pmatrix}, \begin{pmatrix}-1\\0\end{pmatrix},\begin{pmatrix}1\\-1\end{pmatrix}
\end{align*}

There are several related ways to think about this. First, if we put them in a matrix we get:
\begin{align*}
\begin{pmatrix}
1 & -1 & 1\\
1 & 0 & -1
\end{pmatrix}
\end{align*}
we know (with out row reducing) that there will be free variables when put into echelon form. This is because there are only two rows so there can be a maximum of 2 pivot columns. Since there are three columns, at least one must not be a pivot column. So, the vectors are linearly dependent.

We can also actually do the reduction:
\begin{align*}
\begin{pmatrix}
1 & -1 & 1\\
1 & 0 & -1
\end{pmatrix}
\xrightarrow{R2:=R2-R1}
\begin{pmatrix}
1&-1&1\\
0&1&-2
\end{pmatrix}.
\end{align*}
This is in echelon form and the third column isn't a pivot column, so they are linearly dependent.

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L4E2

3. Determine whether the following are linearly independent or linearly dependent.
\begin{align*}
\begin{pmatrix}1\\0\\1\end{pmatrix},\begin{pmatrix}1\\0\\-1\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix}
\end{align*}

Putting them in to a matrix we compute the determinant:
\begin{align*}
\det
\begin{pmatrix}
1&1&0\\
0&0&1\\
1&-1&0
\end{pmatrix}
&=1\det\begin{pmatrix}0&1\\-1&0\end{pmatrix}
-1\det\begin{pmatrix}0&1\\1&0\end{pmatrix}\\
&=1 - (-1)\\
&=2
\end{align*}
Since the determinant is not $$0$$, the vectors are linearly independent.

Another way would be to row reduce:
\begin{align*}
\begin{pmatrix}
1&1&0\\
0&0&1\\
1&-1&0
\end{pmatrix}
\xrightarrow{R2 \leftrightarrow R3}
\begin{pmatrix}
1&1&0\\
1&-1&0\\
0&0&1
\end{pmatrix}\\[4mm]
\xrightarrow{R2:=(-1/2)(R2-R3)}
\begin{pmatrix}
1&1&0\\
0&1&0\\
0&0&1
\end{pmatrix}.
\end{align*}
This is in echelon form and there are no free variables; so there is only one solution to
\begin{align*}
\begin{pmatrix}
1&1&0\\
1&-1&0\\
0&0&1
\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}
=\begin{pmatrix}0\\0\\0\end{pmatrix}.
\end{align*}
So the vectors are linearly independent.

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L4E3

4. Determine whether the following are linearly independent or linearly dependent.
\begin{align*}
\begin{pmatrix}1\\0\\1\end{pmatrix},\begin{pmatrix}1\\0\\-1\end{pmatrix}, \begin{pmatrix}2\\0\\0\end{pmatrix}
\end{align*}

Putting them into a matrix and finding the determinant:
\begin{align*}
\det
\begin{pmatrix}
1&1&2\\
0&0&0\\
1&-1&0
\end{pmatrix}
=1\det\begin{pmatrix}0&0\\-1&0\end{pmatrix}
-1\det\begin{pmatrix}0&0\\1&0\end{pmatrix}
+2\det\begin{pmatrix}0&0\\1&-1\end{pmatrix}
=0.
\end{align*}
So they are linearly dependent.

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L4E4

5. Determine whether the following are linearly independent or linearly dependent.
\begin{align*}
\begin{pmatrix}1\\0\\1\\1\end{pmatrix}, \begin{pmatrix}2\\0\\1\\2\end{pmatrix}
\begin{pmatrix}1\\1\\-2\\-2\end{pmatrix}, \begin{pmatrix}4\\1\\0\\2\end{pmatrix}
\end{align*}

We put them into a matrix and row reduce (swapping the second and fourth rows):
\begin{align*}
\begin{pmatrix}
1&2&1&4\\
1&2&-2&2\\
1&1&-2&0\\
0&0&1&1
\end{pmatrix}
\xrightarrow[R2:=R2-R1]{R3:=-(R3-R1)}
\begin{pmatrix}
1&2&1&4\\
0&0&-3&-2\\
0&1&3&4\\
0&0&1&1
\end{pmatrix}\\[4mm]
\xrightarrow{R2 \leftrightarrow R3}
\begin{pmatrix}
1&2&1&4\\
0&1&3&4\\
0&0&-3&-2\\
0&0&1&1
\end{pmatrix}\\[4mm]
\xrightarrow{R3 \leftrightarrow R4}
\begin{pmatrix}
1&2&1&4\\
0&1&3&4\\
0&0&1&1\\
0&0&-3&-2
\end{pmatrix}\\[4mm]
\xrightarrow{R4 = R4 + 3R3}
\begin{pmatrix}
1&2&1&4\\
0&1&3&4\\
0&0&1&1\\
0&0&0&1
\end{pmatrix}
\end{align*}
This matrix is in echelon form and there are no free variables. So there is only one solution to:
\begin{align*}
\begin{pmatrix}
1&2&1&4\\
1&2&-2&2\\
1&1&-2&0\\
0&0&1&1
\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}
=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}
\end{align*}
and so the vectors are linearly independent.

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L4E5

6. Determine whether the following are linearly independent or linearly dependent.
\begin{align*}
\begin{pmatrix}1\\0\\1\\1\end{pmatrix}, \begin{pmatrix}2\\0\\1\\2\end{pmatrix}
\begin{pmatrix}3\\1\\3\\3\end{pmatrix}, \begin{pmatrix}4\\0\\3\\4\end{pmatrix}
\end{align*}

We row reduce (note: below we move the third row to the bottom):
\begin{align*}
\begin{pmatrix}
1&2&3&4\\
1&1&3&3\\
1&2&3&4\\
0&0&1&0
\end{pmatrix}
\xrightarrow[R2 := R2 - R1]{R3 := R3 - R3}
\begin{pmatrix}
1&2&3&4\\
0&-1&0&-1\\
0&0&0&0\\
0&0&1&0
\end{pmatrix}\\[4mm]
\xrightarrow{R3\leftrightarrow R4}
\begin{pmatrix}
1&2&3&4\\
0&-1&0&-1\\
0&0&1&0\\
0&0&0&0
\end{pmatrix}
\end{align*}
This is in echelon form; the fourth variable is a free variable so the vectors are linearly dependent.

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L4E6

7. For which values of $$t$$ are the following vectors linearly independent?
\begin{align*}
\left\{e^{t}\begin{pmatrix}\cos 2t\\\cos 2t + \sin 2t\end{pmatrix},
e^{t}\begin{pmatrix}\sin 2t\\-\cos 2t + \sin 2t\end{pmatrix}\right\}
\end{align*}

We are going to put these into a matrix and find the determinant. Note that the determinant will be a function of $$t$$ and so we will find the values of $$t$$ that make it zero; those will be the values for which the vectors are linearly dependent and will be linearly independent for all other values of $$t$$.
\begin{align*}
\det
&\begin{pmatrix}
e^t\cos 2t & e^{t}\sin 2t\\
e^{t}(\cos 2t + \sin 2t) & e^{t}(-\cos 2t + \sin 2t)
\end{pmatrix}\\[4mm]
&= e^{2t}(-\cos^2 2t + \cos 2t\sin 2t)
-e^{2t}(\sin 2t \cos 2t - \sin^2 2t)
\\&= e^{2t}(-\cos^2 2t - \sin^2 2t + \cos 2t\sin 2t - \sin 2t \cos 2t)
\\&=-e^{2t}.
\end{align*}
Note that this is never $$0$$. So the vectors are always linearly independent.

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L4E7