Cramer's Rule, Volume, and Linear Transformations

Cramer’s Rule

• Theorem 7 (Cramer’s Rule): Let $A$ be an invertible $n \times n$ matrix. For any $\mathbf{b}$ in $\mathbb{R}^n$, the unique solution $\mathbf{x}$ of $A\mathbf{x} = \mathbf{b}$ has entries given by:

  • $x_i = \frac{\det(A_i(\mathbf{b}))}{\det(A)}$, for $i = 1, 2, \dots, n$
  • Where $A_i(\mathbf{b})$ is the matrix formed by replacing column $i$ of $A$ with the vector $\mathbf{b}$.

• Proof Logic for Cramer's Rule:

  • Denote the columns of $A$ as $\mathbf{a}_1, \dots, \mathbf{a}_n$ and the columns of the $n \times n$ identity matrix $I$ as $\mathbf{e}_1, \dots, \mathbf{e}_n$.
  • If $A\mathbf{x} = \mathbf{b}$, then by the definition of matrix multiplication:
    • $A \cdot I_i(\mathbf{x}) = A[\mathbf{e}_1 \dots \mathbf{x} \dots \mathbf{e}_n] = [A\mathbf{e}_1 \dots A\mathbf{x} \dots A\mathbf{e}_n] = [\mathbf{a}_1 \dots \mathbf{b} \dots \mathbf{a}_n] = A_i(\mathbf{b})$
  • By the multiplicative property of determinants:
    • $\det(A) \cdot \det(I_i(\mathbf{x})) = \det(A_i(\mathbf{b}))$
  • The determinant of $I_i(\mathbf{x})$ is simply $x_i$ (by cofactor expansion along row $i$).
  • Hence, $\det(A) \cdot x_i = \det(A_i(\mathbf{b}))$.
  • Since $A$ is invertible, $\det(A) \neq 0$, allowing us to solve for $x_i$.

• Example 1: Solving a System via Cramer’s Rule:

  • Consider the system:
    • $3x_1 - 2x_2 = 6$
    • $-5x_1 + 4x_2 = 8$
  • Matrix form $A\mathbf{x} = \mathbf{b}$:
    • $A = \begin{pmatrix} 3 & -2 \\ -5 & 4 \end{pmatrix}$, $\mathbf{b} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}$
  • Calculations:
    • $\det(A) = (3 \times 4) - (-2 \times -5) = 12 - 10 = 2$
    • $A_1(\mathbf{b}) = \begin{pmatrix} 6 & -2 \\ 8 & 4 \end{pmatrix} \implies \det(A_1(\mathbf{b})) = (6 \times 4) - (-2 \times 8) = 24 + 16 = 40$
    • $A_2(\mathbf{b}) = \begin{pmatrix} 3 & 6 \\ -5 & 8 \end{pmatrix} \implies \det(A_2(\mathbf{b})) = (3 \times 8) - (6 \times -5) = 24 + 30 = 54$
  • Solutions:
    • $x_1 = \frac{40}{2} = 20$
    • $x_2 = \frac{54}{2} = 27$

A Formula for the Inverse Matrix

• The Adjugate Matrix: The adjugate matrix (also called the classical adjoint) is the transpose of the matrix of cofactors formed from matrix $A$.

• Theorem 8: Let $A$ be an invertible $n \times n$ matrix. Then:

  • $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$

• Example 3: Finding the Inverse of a 3x3 Matrix:

  • Matrix $A = \begin{pmatrix} 2 & 1 & 3 \\ 1 & -1 & 1 \\ 1 & 4 & -2 \end{pmatrix}$
  • The nine cofactors ($C_{ij}$) are calculated as follows:
    • $C_{11} = -2$, $C_{12} = 3$, $C_{13} = 5$
    • $C_{21} = 14$, $C_{22} = -7$, $C_{23} = -7$
    • $C_{31} = 4$, $C_{32} = 1$, $C_{33} = -3$
  • The adjugate matrix is the transpose of the cofactor matrix:
    • $\text{adj}(A) = \begin{pmatrix} -2 & 14 & 4 \\ 3 & -7 & 1 \\ 5 & -7 & -3 \end{pmatrix}$
  • To find $\det(A)$, we can check the product $(\text{adj}(A))A$:
    • $(\text{adj}(A))A = \begin{pmatrix} 14 & 0 & 0 \\ 0 & 14 & 0 \\ 0 & 0 & 14 \end{pmatrix} = 14I$
  • Since $(\text{adj}(A))A = \det(A)I$, then $\det(A) = 14$.
  • The inverse is:
    • $A^{-1} = \frac{1}{14} \begin{pmatrix} -2 & 14 & 4 \\ 3 & -7 & 1 \\ 5 & -7 & -3 \end{pmatrix}$

Determinants as Area or Volume

• Theorem 9:

  • If $A$ is a $2 \times 2$ matrix, the area of the parallelogram determined by the columns of $A$ is $|\det(A)|$.
  • If $A$ is a $3 \times 3$ matrix, the volume of the parallelepiped determined by the columns of $A$ is $|\det(A)|$.\n

• Geometric Observations for the Proof:

  • The theorem is clearly true for any diagonal matrix. For example, in the $2 \times 2$ case with a diagonal matrix having entries $a$ and $d$, the area is $|ad|$.
  • Column operations: It is shown that column replacement does not change the area or volume.
  • Specifically, if $\mathbf{a}_1, \mathbf{a}_2$ are nonzero vectors, for any scalar $c$, the area of the parallelogram determined by $\mathbf{a}_1$ and $\mathbf{a}_2$ equals the area of the parallelogram determined by $\mathbf{a}_1$ and $\mathbf{a}_2 + c\mathbf{a}_1$.
  • Proof reasoning: This is because both parallelograms share the same base (from $0$ to $\mathbf{a}_1$) and the same altitude (since $\mathbf{a}_2 + c\mathbf{a}_1$ lies on a line parallel to the base).
  • Any matrix $A$ can be transformed into a diagonal matrix using column operations that do not change its determinant (up to a sign) or the volume of its associated shape.

• Definition of Volume for Parallelepipeds ($n=3$):

  • The volume is the area of the base in the plane $\text{Span}\{\mathbf{a}_1, \mathbf{a}_2\}$ multiplied by the altitude of $\mathbf{a}_3$ above that plane.
  • A column replacement operation where $\mathbf{a}_3$ is changed to $\mathbf{a}_3 + c\mathbf{a}_1$ does not change the altitude because the new vector still has the same perpendicular distance to the base plane.

• Example 4: Calculating Area for a Parallelogram not at the Origin:

  • Vertices of the parallelogram: $(1, 1), (3, 6), (7, 2), (9, 7)$.
  • To use the determinant method, translate the parallelogram to the origin by subtracting one vertex from all others.
  • Subtract $(1, 1)$ from each vertex:
    • $(1, 1) - (1, 1) = (0, 0)$
    • $(3, 6) - (1, 1) = (2, 5)$
    • $(7, 2) - (1, 1) = (6, 1)$
    • $(9, 7) - (1, 1) = (8, 6)$
  • The parallelogram is now determined by the vectors $\mathbf{v}_1 = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix} 6 \\ 1 \end{pmatrix}$.
  • The matrix is $A = \begin{pmatrix} 2 & 6 \\ 5 & 1 \end{pmatrix}$.
  • $\det(A) = (2 \times 1) - (6 \times 5) = 2 - 30 = -28$.
  • The area is $|\det(A)| = |-28| = 28$.

Linear Transformations and Geometric Change

• Theorem 10: Let $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be the linear transformation determined by a matrix $A$.

  • If $n=2$ and $S$ is a parallelogram in $\mathbb{R}^2$, then:
    • $\text{area}(T(S)) = |\det(A)| \cdot \text{area}(S)$
  • If $n=3$ and $S$ is a parallelepiped in $\mathbb{R}^3$, then:
    • $\text{volume}(T(S)) = |\det(A)| \cdot \text{volume}(S)$

• Proof Summary:

  • A parallelogram $S$ starting at the origin determined by $\mathbf{b}_1$ and $\mathbf{b}_2$ has the form $S = \{x_1\mathbf{b}_1 + x_2\mathbf{b}_2 : 0 \leq x_1, x_2 \leq 1\}$.
  • The image $T(S)$ consists of points $T(x_1\mathbf{b}_1 + x_2\mathbf{b}_2) = x_1T(\mathbf{b}_1) + x_2T(\mathbf{b}_2)$.
  • The image $T(S)$ is a parallelogram determined by the columns of the matrix $AB$, where $B = [\mathbf{b}_1 \dots \mathbf{b}_n]$.
  • The area of the image is $|\det(AB)|$, which equals $|\det(A) \cdot \det(B)|$.
  • Since $|\det(B)|$ is the area of $S$, then $\text{area}(T(S)) = |\det(A)| \cdot \text{area}(S)$.
  • This logic extends to arbitrary parallelograms $p + S$ because translation does not change area.

• Expansion to Other Regions:

  • Theorem 10 generalizes to any region with a finite area or volume, as these shapes can be approximated by unions of small squares or cubes.

• Example 5: Area of an Ellipse:

  • Standard equation for an ellipse $E$: $\frac{x_1^2}{a^2} + \frac{x_2^2}{b^2} = 1$, where $a, b > 0$.
  • The ellipse $E$ is the image of the unit disk $D$ ($u_1^2 + u_2^2 \leq 1$) under the transformation $T(\mathbf{u}) = A\mathbf{u}$ where:
    • $A = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$
  • Area calculation:
    • $\text{area}(E) = |\det(A)| \cdot \text{area}(D)$
    • $\det(A) = ab$
    • $\text{area}(D) = \pi(1)^2 = \pi$
    • $\text{area}(E) = ab\pi$