MIT 18.06SC Lecture 20: Cramer’s Rule, Inverse Matrix and Volume
Inverse Matrix Formula
The inverse of a matrix can be computed using cofactors and determinants, providing an explicit formula that reveals the geometric structure behind matrix inversion.
2×2 Matrices
\[ \begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix} \]
The general formula for the inverse is:
\[ A^{-1}=\frac{1}{\det A}C^\top \]
where \(C\) is the cofactor matrix.
Why Does This Formula Work?
We need to prove that \(AC^\top=(\det A)I\).
\[ \begin{bmatrix}a_{11}&\cdots&a_{1n}\\.&.&.\\a_{n1}&\cdots&a_{nn}\end{bmatrix} \begin{bmatrix}C_{11}&\cdots&C_{n1}\\.&.&.\\C_{1n}&\cdots&C_{nn}\end{bmatrix} \]
Proof:
Diagonal entries: From the cofactor formula, we have:
\[ a_{11}C_{11}+a_{12}C_{12}+\cdots+a_{1n}C_{1n}=\det A \]
So the diagonal entries of \(AC^\top\) are \(\det A\).
Off-diagonal entries: Row \(i\) of \(A\) times column \(j\) of \(C^\top\) (where \(i \neq j\)) is always 0.
Proof of off-diagonal being zero:
- Consider matrix \(A=\begin{bmatrix}A_1\\A_2\\A_3\end{bmatrix}\) and cofactor matrix \(C=\begin{bmatrix}C_1\\C_2\\C_3\end{bmatrix}\)
- If we compute \(A_1 \cdot C_2^\top\), construct a modified matrix:
\[ A_s=\begin{bmatrix}A_1\\A_1\\A_3\end{bmatrix} \]
- Then \(|A_s|\) equals \(A_1 \cdot C_2^\top\), because \(C_2\) is the cofactor computed by removing row 2, and now row 2 is replaced by \(A_1\)
- Since \(A_s\) has two identical rows, \(|A_s|=0\)
- Therefore, all off-diagonal entries are 0
This proves \(AC^\top=(\det A)I\), which gives us \(A^{-1}=\frac{1}{\det A}C^\top\).
Cramer’s Rule
Starting from:
\[ Ax=b\\ x=A^{-1}b=\frac{1}{\det A}C^\top b \]
Cramer’s Rule states:
\[ x_j=\frac{\det B_j}{\det A} \]
where \(B_j\) is matrix \(A\) with column \(j\) replaced by vector \(b\).
While this formula is mathematically elegant, it requires computing determinants for \(n+1\) matrices. In practice, elimination remains the more efficient method for solving \(Ax=b\).
Determinant as Volume
\(|\det A|\) equals the volume of a box (parallelepiped).
The absolute value of the determinant of an \(n \times n\) matrix equals the n-dimensional volume of the parallelotope spanned by its column vectors.
Sign of the Determinant
The sign is determined by the orientation (handedness) of the box:
- Positive for right-handed orientation
- Negative for left-handed orientation
Identity Matrix
\[ A=I \]
The identity matrix spans a unit cube with edge length 1, so the volume is 1. Correspondingly, \(\det I=1\).
Orthogonal Matrix
For an orthogonal matrix \(Q\):
\[ |\det Q|=1 \]
Proof:
\[ |QQ^\top|=|I|=1\\ |QQ^\top|=|Q||Q^\top|=|Q|^2=1 \]
Based on the determinant property \(|AB|=|A||B|\), we have \(|Q|^2=1\), so \(|Q|=\pm1\).
This means orthogonal transformations preserve volume.
Area of Parallelogram
For a 2D parallelogram formed by vectors \(\begin{bmatrix}a\\c\end{bmatrix}\) and \(\begin{bmatrix}b\\d\end{bmatrix}\):
\[ \text{Area} = |ad-bc| = \left|\begin{vmatrix}a&b\\c&d\end{vmatrix}\right| \]
Triangle Area
The area of a triangle is half the area of the parallelogram:
\[ \text{Triangle Area} = \left|\frac{1}{2}(ad-bc)\right| \]
When we only have the coordinates of the 3 vertices of the triangle \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\), we have 3 vectors in a 2D world. We can lift the dimension by using:
\[ \text{Triangle Area} = \frac{1}{2}\left|\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\right| \]
This projects the triangle into 3D space without changing its area (all points remain in the same plane with \(z=\) constant).
Scaling Property
Doubling an edge doubles the volume:
\[ \begin{vmatrix}ta&tb\\c&d\end{vmatrix}=t\begin{vmatrix}a&b\\c&d\end{vmatrix} \]
If we multiply row 1 by \(t\) and keep all other rows unchanged, then \(|A'|=t|A|\).
This reflects the geometric fact that scaling one edge of a parallelotope scales its volume proportionally.
Source: MIT 18.06SC Linear Algebra, Lecture 20