Matrix Inverse

85. Matrix Inverse

𝐴 \cdot 𝐴^{- 1} = 𝐴^{- 1} \cdot 𝐴 = 𝐼

Where $𝐼$ is the identity matrix (like the number $1$ in regular multiplication)

A matrix $𝐴$ is invertible (nonsingluar) if:

Row Reduction (Gauss-Jordan Elimination)

Augment $𝐴$ with the identity matrix $𝐼$ , and perform row operations until the left side becomes $𝐼$ . The right side will then be $𝐴^{- 1}$

[\begin{matrix} 𝐴 & 𝐼 \end{matrix}] ⟶ [\begin{matrix} 𝐼 & 𝐴^{- 1} \end{matrix}]

Example

𝐴 = [\begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix}]

Step 1. Set up augmented matrix

[𝐴 | 𝐼] = [\begin{matrix} 2 & 1 & 1 & 0 \\ 5 & 3 & 0 & 1 \end{matrix}]

Step 2. Make the pivot of the first column equal to 1

(𝑅_{1} \to \frac{1}{2} 𝑅_{1})

[\begin{matrix} 1 & 0.5 & 0.5 & 0 \\ 5 & 3 & 0 & 1 \end{matrix}]

Step 3: Eliminate the first column in row 2

𝑅_{2} \to 𝑅_{2} - 5 𝑅_{1}

[\begin{matrix} 1 & 0.5 & 0.5 & 0 \\ 0 & 0.5 & - 2.5 & 1 \end{matrix}]

Step 4: Make the pivot in row 2 a 1

𝑅_{2} \to 2 𝑅_{2}

[\begin{matrix} 1 & 0.5 & 0.5 & 0 \\ 0 & 1 & - 5 & 2 \end{matrix}]

Step 5: Eliminate the 0.5 above the pivot

𝑅_{1} \to 𝑅_{1} - 0.5 𝑅_{2}

[\begin{matrix} 1 & 0 & 3 & - 1 \\ 0 & 1 & - 5 & 2 \end{matrix}]

So the inverse is:

𝐴^{- 1} = [\begin{matrix} 3 & - 1 \\ - 5 & 2 \end{matrix}]