Inverting a $3 \times 3$ Matrix using the Cross Product

This method utilizes the cross product of the column vectors of a matrix to find its inverse.

1. Define the Column Vectors

Given a matrix $M$ with column vectors $a,\; b,$ and $c$:

$$M = \begin{pmatrix} | & | & | \\ a & b & c \\ | & | & | \end{pmatrix} = \begin{pmatrix} 1 & 3 & 1 \\ 0 & 4 & 1 \\ 2 & -1 & 0 \end{pmatrix}$$

Where:

$$a = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}, \quad b = \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix}, \quad c = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$$

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2. Calculate the Cross Products

We find the cross product of each pair of vectors to form the rows of the adjoint matrix:

$$\begin{aligned} b \times c &= \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} \\ c \times a &= \begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix} \\ a \times b &= \begin{pmatrix} -8 \\ 7 \\ 4 \end{pmatrix} \end{aligned}$$

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3. Finding the Determinant

The determinant of a $3 \times 3$ matrix can be found using the scalar triple product. This is equivalent to the dot product of one column vector with the cross product of the other two:

$\text{det}(M) = a \cdot (b \times c)$

Calculation:

$$\begin{aligned} \text{det}(M) &= \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} \\ &= (1 \times 1) + (0 \times -1) + (2 \times -1) \\ &= 1 + 0 - 2 \\ &= -1 \end{aligned}$$

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4. Construct the Inverse Matrix

The inverse matrix $M^{-1}$ is constructed by placing the transposed cross products over the determinant:

$$\begin{aligned} M^{-1} &= \frac{1}{\text{det}(M)} \begin{pmatrix} (b \times c)^T \\ (c \times a)^T \\ (a \times b)^T \end{pmatrix} \\ M^{-1} &= \frac{1}{-1} \begin{pmatrix} 1 & -1 & -1 \\ 2 & -2 & -1 \\ -8 & 7 & 4 \end{pmatrix} \end{aligned}$$