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3.3 Cramer's Rule, Volume, and Linear Transformations

Another way to solve A⃗x = ⃗b:

\[ A = \begin{bmatrix} 3 & -2 \\ -5 & 4 \end{bmatrix} \quad \vec{b} = \begin{bmatrix} 6 \\ 8 \end{bmatrix} \quad \vec{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \]

Identity: \[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

Replace first column with ⃗x: \[ I_1(\vec{x}) = \begin{bmatrix} x_1 & 0 \\ x_2 & 1 \end{bmatrix} \]

Multiply A by \( I_1(\vec{x}) \):

\[ \begin{bmatrix} 3 & -2 \\ -5 & 4 \end{bmatrix} \begin{bmatrix} x_1 & 0 \\ x_2 & 1 \end{bmatrix} = \begin{bmatrix} 6 & -2 \\ 8 & 4 \end{bmatrix} \]

because \( A\vec{x} = \vec{b} \)

Take determinant:

\[ \det \left( \begin{bmatrix} 3 & -2 \\ -5 & 4 \end{bmatrix} \begin{bmatrix} x_1 & 0 \\ x_2 & 1 \end{bmatrix} \right) = \det \left( \begin{bmatrix} 6 & -2 \\ 8 & 4 \end{bmatrix} \right) \]
\[ \det \left( \begin{bmatrix} 3 & -2 \\ -5 & 4 \end{bmatrix} \right) \det \left( \begin{bmatrix} x_1 & 0 \\ x_2 & 1 \end{bmatrix} \right) = \det \left( \begin{bmatrix} 6 & -2 \\ 8 & 4 \end{bmatrix} \right) \]
\[ (2)(x_1) = 40 \quad \rightarrow \quad x_1 = 20 \]

Repeat w/ replacing 2nd col of I to eventually find \( x_2 = 27 \)

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In general, the equation in green box is:

\[ \det(A) \cdot x_i = \det(A_i(\vec{b})) \]

A with \( i^{th} \) column replaced with \( \vec{b} \)

\[ x_i = \frac{\det(A_i(\vec{b}))}{\det(A)} \]

Cramer's Rule
(unique solution only)

Example

\[ A = \begin{bmatrix} 1 & 2 & 4 \\ 0 & 1 & 5 \\ -2 & -4 & -3 \end{bmatrix} \quad \vec{b} = \begin{bmatrix} -2 \\ 2 \\ 9 \end{bmatrix} \quad \vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \]

Find \( x_2 \):

\[ x_2 = \frac{\det \left( \begin{bmatrix} 1 & -2 & 4 \\ 0 & 2 & 5 \\ -2 & 9 & -3 \end{bmatrix} \right)}{\det \left( \begin{bmatrix} 1 & 2 & 4 \\ 0 & 1 & 5 \\ -2 & -4 & -3 \end{bmatrix} \right)} = \frac{-15}{5} = -3 \]
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Cramer's Rule for Matrix Inversion

Cramer's rule is also used to find \( A^{-1} \).

If \( B = A^{-1} \), then \( AB = I \).

\[ A \begin{bmatrix} \vec{b}_1 & \vec{b}_2 & \dots & \vec{b}_n \end{bmatrix} = \begin{bmatrix} \vec{e}_1 & \vec{e}_2 & \dots & \vec{e}_n \end{bmatrix} \]

Then solve \( A\vec{b}_i = \vec{e}_i \) by Cramer's rule.

Note: \( \vec{e}_1 = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} \) and \( \vec{e}_2 = \begin{bmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{bmatrix} \)

In the end, it turns out:

\[ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} C_{11} & C_{21} & \dots & C_{n1} \\ C_{12} & C_{22} & \dots & C_{n2} \\ \vdots & \vdots & & \vdots \\ C_{1n} & C_{2n} & \dots & C_{nn} \end{bmatrix} \]

The \( C_{ji} \) are cofactors of \( A \).

  • Signed determinants of submatrix formed by covering up \( j^{\text{th}} \) row and \( i^{\text{th}} \) column of \( A \).

The matrix of cofactors is called the adjugate (or classical adjoint) of \( A \), denoted as \( \text{adj}(A) \).

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Example

\[ A = \begin{bmatrix} 1 & 2 & 3 \\ -3 & -2 & -1 \\ 3 & 4 & 5 \end{bmatrix} \]

Calculating cofactors:

\[ \begin{vmatrix} -2 & -1 \\ 4 & 5 \end{vmatrix} = -6 \quad \text{and} \quad -\begin{vmatrix} -3 & -1 \\ 3 & 5 \end{vmatrix} = 12 \]
\[ \text{adj}(A) = \begin{bmatrix} -6 & 12 & -6 \\ 12 & -4 & 2 \\ 4 & -8 & 4 \end{bmatrix} \]

Since \( \det(A) = 0 \), \( A^{-1} \) does not exist.

Geometric Interpretation of Determinants

We saw last time: \( |\det[\vec{u} \quad \vec{v}]| \) is the area of a parallelogram in \( \mathbb{R}^2 \).

Coordinate graph showing a shaded parallelogram formed by vectors u and v in the x1-x2 plane.

Area of triangle with \( \vec{u}, \vec{v} \) sides is:

\[ \frac{|\det[\vec{u} \quad \vec{v}]|}{2} \]
Coordinate graph showing a shaded triangle formed by vectors u and v in the x1-x2 plane.
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Volume of a Parallelepiped in 3x3 Matrices

For a \(3 \times 3\) matrix, we get the volume of the parallelepiped:

\[ \left| \det \begin{bmatrix} \vec{a_1} & \vec{a_2} & \vec{a_3} \end{bmatrix} \right| \]
A 3D sketch of a parallelepiped formed by three vectors a1, a2, and a3.

Why?

Similar to the 2D case where a parallelogram has the same area as a rectangle:

Diagram showing a parallelogram being transformed into a rectangle of equal area.

Consider the diagonal matrix:

\[ \begin{bmatrix} 4 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 6 \end{bmatrix} \quad \det = 120 \]
A 3D coordinate system showing a rectangular prism with dimensions 4, 5, and 6.

Any row-equivalent matrix has the same absolute value of determinant.

A transformation from a skewed parallelepiped to a rectangular box via row operations.

row ops

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If \( | \det(B) | \) is area or volume where \( B = [ \vec{b_1} \quad \vec{b_2} \dots ] \), then \( AB \) can be interpreted as a linear transformation of the area/volume enclosed by the columns of \( B \).

A square in the x1-x2 plane being transformed into a parallelogram.

Example

Let \( \vec{b_1} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \), \( \vec{b_2} = \begin{bmatrix} -3 \\ 1 \end{bmatrix} \), and \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \).

Coordinate plot showing vectors b1 and b2 forming a parallelogram.
\[ \left| \det \begin{bmatrix} 1 & -3 \\ 2 & 1 \end{bmatrix} \right| = 7 \]
\[ AB = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & -1 \\ 11 & -5 \end{bmatrix} \]
Coordinate plot showing the transformed parallelogram after matrix multiplication.
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\[ ||AB|| = 14 = \det(A) \det(B) \]

Where \(\det(A)\) is the det of transformation matrix and \(\det(B)\) is the area in original form.

So, in general, if \(T(S) = AB\) is transformation of area/volume, then area/volume after transformation is

\[ ||A|| ||B|| = ||A|| \cdot \{\text{area/vol of } S\} \]

Singular Transformations

If \(\det(A) = 0\), \(\dim \text{Nul } A > 0\), so at least one axis is nullified by transformation.

A parallelogram transforming into a horizontal line segment or a vertical line segment, both with area zero.
area = 0