Did you know that we can use the power of determinants to obtain theoretical formulas and geometrical interpretations, such as the area and volume of linear transformations?
Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher)
How?
By using a vital theorem called Cramer’s rule .
Understanding Cramer’s Rule
Cramer’s rule helps us study how the solution of \(A \vec{x}=\vec{b}\) is affected by changes in the entries of \(\vec{b}\).
If we let \(A\) be an invertible nxn matrix, then for any \(\vec{b}\) in \(\mathbb{R}^{n}\), the unique solution \(\vec{x}\) of \(\overrightarrow{A x}=\vec{b}\) has entries given by:
\begin{align*}
x_{i}=\frac{\operatorname{det} A_{i}(\vec{b})}{\operatorname{det} A}, \quad i=1,2, \ldots, n
\end{align*}
Okay, so this definition may not make much sense initially, but it’s pretty easy once you see it in action. So, let’s look at an example.
Example of Cramer’s Rule
Consider the following system. Use Cramer’s rule to solve.
\begin{align*}
\begin{gathered}
3 x_{1}-2 x_{2}=6 \\
-5 x_{1}+4 x_{2}=8
\end{gathered}
\end{align*}
First, let’s write the linear system as a matrix equation.
\begin{align*}
\underbrace{\left[\begin{array}{cc}
3 & -2 \\
-5 & 4
\end{array}\right]}_{A} \underbrace{\left[\begin{array}{c}
x_{1} \\
x_{2}
\end{array}\right]}_{x}=\underbrace{\left[\begin{array}{l}
6 \\
8
\end{array}\right]}_{b}
\end{align*}
Next, we will find the determinant of matrix A.
\begin{aligned}
\operatorname{det} A & \Rightarrow \operatorname{det}\left(\left[\begin{array}{cc}
3 & -2 \\
-5 & 4
\end{array}\right]\right) \\
& =(3)(4)-(-2)(-5) \\
& =2
\end{aligned}
Now, we will replace the columns of \(\mathrm{A}\) with \(\mathrm{b}\) and calculate each determinant separately.
\begin{aligned}
\operatorname{det} A_{1}(\vec{b}) & \Rightarrow \operatorname{det}\left(\left[\begin{array}{cc}6 & -2 \\ 8 & 4\end{array}\right]\right) \\
& =6(4)-(-2)(8) \\
& =40
\end{aligned}
\begin{aligned}
\operatorname{det} A_{2}(\vec{b}) & \Rightarrow \operatorname{det}\left(\left[\begin{array}{cc}3 & 6 \\ -5 & 8\end{array}\right]\right) \\
& =3(8)-(6)(-5) \\
& =54
\end{aligned}
Finally, we will divide the quantities to find the unique solution to the linear system.
\begin{aligned}
x_{1} & =\frac{\operatorname{det}\left(\left[\begin{array}{cc}
6 & -1 \\
8 & 4
\end{array}\right]\right)}{\operatorname{det}\left(\left[\begin{array}{cc}
3 & -1 \\
-5 & 4
\end{array}\right]\right)} \\
& =\frac{40}{2} \\
& =20
\end{aligned}
\begin{aligned}
x_{2} & =\frac{\operatorname{det}\left(\left[\begin{array}{cc}
3 & 6 \\
-5 & 8
\end{array}\right]\right)}{\operatorname{det}\left(\left[\begin{array}{cc}
3 & -1 \\
-5 & 4
\end{array}\right]\right)} \\
& =\frac{54}{2} \\
& =27
\end{aligned}
So, that wasn’t so bad?
Introduction to Inverse of Matrices Using Determinants
And Cramer’s rule leads to a fantastic general formula for the inverse of an nxn matrix.
Remember when we learned about how to find the inverse of a \(2 \times 2\) matrix, but sadly, we had to use the inverse algorithm for finding inverses of nxn matrices?
Well, with the help of determinants, we are able to generalize the inverse formula from \(2 \times 2\) matrices to \(3 \times 3\) matrices or larger.
If \(\mathrm{A}\) is an invertible nxn matrix, then \(A^{-1}=\frac{1}{\operatorname{det} A}(\operatorname{Adj\mathrm {A}})\), where \(\operatorname{Adj}(\mathrm{A})\) is called the Adjugate or Adjoint and is the transpose of the matrix of cofactors.
\begin{aligned}
A^{-1} & =\frac{1}{\operatorname{det} A}(A d j \mathrm{~A}) \\
& =\frac{1}{\operatorname{det} A}\left[\begin{array}{cccc}
C_{11} & C_{21} & \cdots & C_{n 1} \\
C_{12} & C_{22} & \cdots & C_{n 2} \\
\vdots & \vdots & \ddots & \vdots \\
C_{1 n} & C_{2 n} & \cdots & C_{n n}
\end{array}\right]
\end{aligned}
Granted, the adjugate looks scary at first glance, but it’s easy to find if you know the three simple steps.
Computing the Adjugate of a Matrix
Suppose \(A=\left[\begin{array}{ccc}2 & 1 & 3 \\ 1 & -1 & 1 \\ 1 & 4 & -2\end{array}\right]\), and we wish to find the adjugate.
First, we will calculate the cofactors for matrix A, using expansion by minors for every entry in matrix A and calculating the remaining determinant.
\begin{aligned}
& C_{11}=\left|\begin{array}{cc}
-1 & 1 \\
4 & -2
\end{array}\right| & =& -2 \\
& C_{12}=\left|\begin{array}{cc}
1 & 1 \\
1 & -2
\end{array}\right| & =& -3 \\
& C_{13}=\left|\begin{array}{cc}
1 & -1 \\
1 & 4
\end{array}\right| & =& 5 \\
& C_{21}=\left|\begin{array}{cc}
1 & 3 \\
4 & -2
\end{array}\right| & =& -14 \\
& C_{22}=\left|\begin{array}{cc}
2 & 3 \\
1 & -2
\end{array}\right| & =& -7 \\
& C_{23}=\left|\begin{array}{ll}
2 & 1 \\
1 & 4
\end{array}\right| & =& 7 \\
& C_{31}=\left|\begin{array}{cc}
1 & 3 \\
-1 & 1
\end{array}\right| & =& 4 \\
& C_{32}=\left|\begin{array}{ll}
2 & 3 \\
1 & 1
\end{array}\right| & =& -1 \\
& C_{33}=\left|\begin{array}{cc}
2 & 1 \\
1 & -1
\end{array}\right| & =& -3
\end{aligned}
Giving us \(\left[\begin{array}{ccc}-2 & -3 & 5 \\ -14 & -7 & 7 \\ 4 & -1 & -3\end{array}\right]\)
Next, we will apply the array of signs (i.e., seat assignments…seat “sign”ments) to our cofactors.
\begin{aligned}
\left[\begin{array}{ccc}
+(-2) & -(-3) & +(5) \\
-(-14) & +(-7) & -(7) \\
+(4) & -(-1) & +(-3)
\end{array}\right] & \Rightarrow\left[\begin{array}{ccc}
-2 & 3 & 5 \\
14 & -7 & -7 \\
4 & 1 & -3
\end{array}\right]
\end{aligned}
Lastly, we transpose the cofactor matrix.
\begin{aligned}
\operatorname{adj} \mathrm{A} & =\left[\begin{array}{ccc}
-2 & 14 & 4 \\
3 & -7 & 1 \\
5 & -7 & -3
\end{array}\right]
\end{aligned}
And that’s it! Find cofactors, apply signs, and then transpose!
Don’t worry! We’ll do several examples in the video so you can get the hang of it.
Applications of Determinants in Geometry
Now it’s time to see how determinants help us find area and volume for parallelograms (2D) and parallelepipeds (3D)
If \(A\) is a \(2 \times 2\) matrix, the area of the parallelogram determined by the columns of \(A\) is found by the absolute value of the determinant of \(\mathrm{A}\). And if \(\mathrm{A}\) is a \(3 \times 3\) matrix, the volume of the parallelepiped determined by the columns of \(A\) is calculated by the absolute value of the determinant of A.
In fact, in the video, we will also learn how translating a parallelogram does not change its area. Moreover, this idea can be used to describe geometric properties of linear transformations in the plane.
Let \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) is the linear transformation determined by the \(2 \times 2\) matrix \(\mathrm{A}\).
If \(\mathrm{S}\) is a parallelogram in \(\mathbb{R}^{2}\), then the area of \(T(S)=(|\operatorname{det} A|) \underbrace{(|\operatorname{det} S|}_{\text {area of } S}\).
And if \(\mathrm{T}\) is determined by a \(3 \times 3\) matrix \(A\), and if \(S\) is a parallelepiped in \(\mathbb{R}^{3}\), then volume of \(T(S)=(|\operatorname{det} A|)(\underbrace{(|\operatorname{det} S|}_{\text {volume of } S}\).
Example of Determinants in Linear Transformations
For example, let \(\mathrm{S}\) be the parallelogram determined by the vectors \(\overrightarrow{b_{1}}=\left[\begin{array}{c}3 \\ 1\end{array}\right]\) and \(\overrightarrow{b_{2}}=\left[\begin{array}{c}1 \\ 5\end{array}\right]\) and \(A=\left[\begin{array}{cc}2 & 0 \\ 1 & -1\end{array}\right]\).
Compute the area of the image of S under the mapping \(\vec{x} \mapsto A \vec{x}\).
First, we will create our \(\mathrm{S}\) matrix using our two given vectors
\begin{aligned}
S & =\left[\begin{array}{ll}
\vec{b} & \overrightarrow{b_{2}}
\end{array}\right] \\
& =\left[\begin{array}{ll}
3 & 1 \\
1 & 5
\end{array}\right]
\end{aligned}
Next, we will calculate the determinant of matrix \(A\) and \(S\) separately.
\begin{aligned}
\operatorname{det} A & =\operatorname{det}\left(\left[\begin{array}{cc}2 & 0 \\ 1 & -1\end{array}\right]\right)=-2 \\
\operatorname{det} S & =\operatorname{det}\left(\left[\begin{array}{cc}3 & 1 \\ 1 & 5\end{array}\right]\right)=14
\end{aligned}
Finally, we multiply the absolute value of these determinants to find the area of the image.
\begin{aligned}
\text{area of } T(S) & =(|\operatorname{det} A|)(|\operatorname{det} S|) \\
& =(|-2|)(|14|) \\
& =28
\end{aligned}
Therefore, the area of the parallelogram under the mapping has a value of 28 square units!
Next Steps
In this lesson, you will:
- Apply Cramer’s Rule to solve a given system
- Learn how to compute the Adjugate of a matrix to find the inverse of an nxn matrix
- Explore applications of determinants to calculate:
- The area of a parallelogram
- Volume of a parallelepiped
- The area or volume of an image under a specified mapping
Get ready for a great learning experience and jump right in!
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