Dated: 23-04-2025
Cramer’s Rule
Let \(A\) be an invertible \(n \times n\) matrix such that for any \(\vec b \in \mathbb R^n\), the unique solution of \(x\) of \(A \vec x = \vec b\) has entries given by
\[x_i = \frac {\det (A_i(b))}{\det (A)}, \quad i = 1, 2, \ldots, n\]
What is \(A_i(b)\)
It means a matrix \(A^\prime\) obtained by replacing the \(i^{th}\) column of \(A\) with \(\vec b\).
Example
\[
\begin{align*}
3x_1 - 2x_2 &= 6 \\
-5x_1 + 4x_2 &= 8
\end{align*}
\]
Solution
Re-writing in the matrix form \(A \vec x = \vec b\).
\[
\begin{bmatrix}
3 & -2 \\
-5 & 4
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
=
\begin{bmatrix}
6 \\
8
\end{bmatrix}
\]
\[
A = \begin{bmatrix}
3 & -2 \\
-5 & 4
\end{bmatrix},
x = \begin{bmatrix}
x_1 \\
x_2
\end{bmatrix},
b = \begin{bmatrix}
6 \\
8
\end{bmatrix}
\]
\[
\det A = \begin{bmatrix}
3 & -2 \\
-5 & 4
\end{bmatrix} = 12 - 10 = 2
\]
\[
A_1(b) = \begin{bmatrix}
6 & -2 \\
8 & 4
\end{bmatrix},
A_2(b) = \begin{bmatrix}
3 & 6 \\
-5 & 8
\end{bmatrix}
\]
\[
x_1 = \frac{\det A_1(b)}{\det A} = \frac{24 + 16}{2} = \frac{40}{2} = 20
\]
\[x_2 = \frac{\det A_2(b)}{\det A} = \frac{24 + 30}{2} = \frac{54}{2} = 27\]
Theorem
Let \(A\) be an invertible \(n \times n\) matrix such that
\[A^{-1} = \frac {adj \, A}{\det (A)}\]
What is \(adj \, A\)
It is called classical adjoint or adjugate.
\[
adj \, A =
\begin{bmatrix}
C_{11} & C_{21} & \cdots & C_{n1} \\
C_{12} & C_{22} & \cdots & C_{n2} \\
\vdots & \vdots & \ddots & \vdots \\
C_{1n} & C_{2n} & \cdots & C_{nn}
\end{bmatrix}
\]
Where \(C_{ij}\) is the cofactor of \(a_{ij}\) entry of \(A\).
Determinants as Area or Volume
Theorem
- For \(A_{2 \times 2}\), the area of a
parallelogram is given by the columns of \(A\) such that \(|\det (A)|\).
- For \(A_{3 \times 3}\), the area of a
parallelepiped is given by the columns of \(A\) such that \(|\det (A)|\).
Example
Calculate the area of the parallelogram determined by the points \((-2, -2), (0, 3), (4, -1)\) and \((6, 4)\).
Solution
Let
- \(A(-2, -2)\)
- \(B(0, 3)\)
- \(C(4, -1)\)
- \(D(6, 4)\)
Let's fix \(A\) and then find adjacent side lengths of the parallelogram.
\[
AB = \begin{bmatrix}
0 - (-2) \\
3 - (-2)
\end{bmatrix} = \begin{bmatrix}
2 \\
5
\end{bmatrix}, \quad
AC = \begin{bmatrix}
4 - (-2) \\
-1 - (-2)
\end{bmatrix} = \begin{bmatrix}
6 \\
1
\end{bmatrix}
\]
\[
= \left| \det \begin{bmatrix}
2 & 6 \\
5 & 1
\end{bmatrix} \right| = |2 - 30| = |-28| = 28
\]
Theorem
If \(T : \mathbb R^2 \to \mathbb R^2\) be a linear transformation determined by a \(2 \times 2\) matrix \(A\) and \(S\) is a parallelogram in \(\mathbb R^2\) then.
\[\text{area of }T(S) = |\det (A)| \cdot \text{area of } S\]
Same goes for 3 dimensions.
Example
Let \(a\) and \(b\) be positive numbers.
Find the area of the region \(E\) bounded by an ellipse whose equation is
\[
\frac{x_1^2}{a^2} + \frac{x_2^2}{b^2} = 1
\]
Solution
We are claiming that \(E\) is the image of a disk \(D\) under the linear transformation \(A: D \to E\) determined by the matrix
\[
A = \begin{bmatrix}
a & 0 \\
0 & b
\end{bmatrix}
\]
\[
A\vec u = \vec x \quad \text{ where } \vec u = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \in D, \quad \vec x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \in E
\]
\[
\implies \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
\]
\[
\implies \begin{bmatrix} au_1 \\ bu_2 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
\]
\[
\implies au_1 = x_1 \text{ and } bu_2 = x_2
\]
\[
\implies u_1 = \frac{x_1}{a} \text{ and } u_2 = \frac{x_2}{b}
\]
\[\vec u \in D \implies |\vec u| \le 1\]
\[
(u_1^2 - 0) + (u_2^2 - 0) \leq 1
\]
\[
\implies \left(\frac{x_1}{a}\right)^2 + \left(\frac{x_2}{b}\right)^2 \leq 1 \quad \because u_1 = \frac{x_1}{a}, u_2 = \frac{x_2}{b}
\]
\[\text{area of ellipse} = \text{area of } A(D)\]
\[= |\det(A)| \cdot \text{area of } D\]
\[ab \cdot \pi(1)^2 = \pi ab\]
References
Read more about notations and symbols.