Cramer's Rule, Inverse Matrix, and Volume

We know a formula for and some properties of the determinant. Now we see how the determinant can be used.

Formula for A^-1

We know:

\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

Can we get a formula for the inverse of a 3 by 3 or n by n matrix? We expect that \(\frac{1}{det A}\) will be involved, as it is in the 2 by 2 matrix, and by lookin at the cofactor matrix \(\begin{bmatrix} d & -c \\ -b & a \end{bmatrix}\) we will guess that the cofactors will be involved.

In fact:

\[ A^{-1} = \frac{1}{det A} C^T \]

where C is the matrix of cofactors(Notice the transpose!). Cofactors of row one of a go into column 1 of \(A^{-1}\), and then we divide by the determinant.

The determinant of A involves products with n terms and the cofactor matrix involves products of n - 1 terms. A and \(\frac{1}{det A} C^T\) might cancel each other. This is much easier to see from our formula for the determinant than when using Gauss-Jordan elimination.

To more formally verify the formula, we'll check that \(AC^T = (det A) I\).

\[ AC^T = \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \\ \end{bmatrix} \begin{bmatrix} C_{11} & \cdots & C_{1n} \\ \vdots & \ddots & \vdots \\ C_{n1} & \cdots & C_{nn} \\ \end{bmatrix} \]

The entry in the first row and first column of the product matrix is:

\[ \sum_{j = 1}^{n} a_{1j}C_{j1} = det A \]

This is just the cofactor formula for the determinant. This happens for every entry on the diagonal of \(AC^T\).

To finish proving that \(AC^T = (det A)I\), we just need to check that the off-diagonal entries of \(AC^T\) are zero. In the 2 by 2 case, multiplying the entries in row 1 of A by the entries in column 2 of \(C^T\) gives \(a(-b) + b(a) = 0\). This is the determinant of \(A_{s} = \begin{bmatrix} a & b \\ a & b \end{bmatrix}\). In higher dimensions, the product of the first of A and the last column of \(C^T\) equals the determinant of a matrix whose first and last rows are identical. This happens with all the off diagonal matrices, which confirms that \(A^{-1} = \frac{1}{det A} C^T\).

This formula helps us answer questions about how the inverse changes when the matrix changes.

Cramer's Rule for x = A^-1 b

We know that if \(A\mathbf{x} = \mathbf{b}\) and A is nonsingular, then \(\mathbf{x} = A^{-1} \mathbf{b}\). Applying the formula \(A^{-1} = C^T / (det A)\) gives us:

\[ \mathbf{x} = \frac{1}{det A} C^T \mathbf{b} \]

Cramer's Rule gives us another way of looking at this question. To derive this rule we break \(\mathbf{x}\) down into its components. Because the i'th component of \(C^T\mathbf{b}\) is a sum of cofactors times some number, it is the determinant of some matrix \(B_{j}\).

\[ x_{j} = \frac{det B_{j}} {det A} \]

where \(B_{j}\) is the matrix created by starting with A and then replacing column j with \(\mathbf{b}\), so:

\[ B_{1} = \begin{bmatrix} \mathbf{b} & \text{last n - 1 columns of A} \end{bmatrix} \]

\[ B_{n} = \begin{bmatrix} \text{first n - 1 columns of A} & \mathbf{b} \end{bmatrix} \]

This agrees with our formula \(x_{1} = \frac{det B_{1}} {det A}\). When taking the determinant of \(B_{1}\) we get a sum whose first term is \(b_{1}\) times the cofactor \(C_{11}\) of A.

Computing inverses using Cramer's rule is usually less efficient than using elimination.

|A| = volume of box

Claim:

\(det A\) is the volume of the box (parallelepiped) whose edges are the column vectors of \(A\).

We could usually well use the row vectors, forming a different box with the same volume.

If \(A = I\), then the box is a unit cube and its volume is 1. Because this agrees with our claim, we can conclude that the volume obeys determinant property 1.

If \(A = Q\) is an orthogonal matrix then the box is a unit in a different orientation with volume = \(det Q = 1\). Because Q is an orthogonal matrix, \(Q^TQ = I\) and so \(det Q = \pm 1\).

Swapping two columns of A does not change the volume of the box or (remember that \(det A = det A^T\)) the absolute value of the determinant(property 2). If we show that the volume of the box also obeys property 3 we'll have proven \(|det A|\) equals the volume of the box.

If we double the length of one column of \(A\), we double the volume of the box formed by its columns. Volume satisfies property 3(a).

Property 3(b) says that the determinant is linear in the rows of the matrix:

\[ \begin{vmatrix} a + a' & b + b' \\ c & d \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} + \begin{vmatrix} a' & b' \\ c & d \end{vmatrix} \]

Although it's not needed for our proof, we can also see that determinants obey property 4. If two edges of a box are equal, the box flattens out and has no volume.

Important note: If you know the coordinates for the corners of a box, then computing the volume of the box is as easy as calculating a determinant. In particular, the area of a parallelogram with edges \(\begin{bmatrix} a \\ b \end{bmatrix}\) and \(\begin{bmatrix} c \\ d \end{bmatrix}\) is \(ad - bc\). The area of a triangle with edges \(\begin{bmatrix} a \\ b \end{bmatrix}\) and \(\begin{bmatrix} c \\ d \end{bmatrix}\) is half the area of that parallelogram, or \(\frac{1}{2} (ad - bc)\). The area of a triangle with vertices at \((x_{1}, y_{1}), (x_{2}, y_{2})\) and \((x_{3}, y_{3})\) is:

\[ \frac{1}{2} \begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \\ \end{vmatrix} \]