Finding the Inverse of a Matrix

2.7. Finding the Inverse of a Matrix#

Definition - Matrix Inverse Algorithm:

Create the augmented $n\times 2$ matrix denoted as $[~A~|~I~]$, where matrix $A$ has dimensions $n\times n$, and $[~A~|~I~]$ possesses double the number of columns, resulting in a $n\times 2$ configuration [Kuttler and Farah, 2020, Nicholson, 2018].
If feasible, apply row operations until a transformation is achieved, leading to an $n\times 2n$ matrix in the format of $[~I~|~B~]$.
Consequently, the matrix $B$ corresponds to the inverse of matrix $A$ ( $B = A^{-1} $), affirming the invertibility of $A$.
In instances where the transformation to a matrix in the structure of $[~I~|~B~]$ cannot be accomplished, it indicates that matrix $A$ lacks an inverse, denoting that A is non-invertible.

Example: Find the inverse of $A=\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 3 & -1\end{bmatrix}$ if it exists.

Solution:

Form the augmented $3\times 6$ matrix $[~A~|~I~]$ $$\begin{aligned} \left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & 0\\ 0 & 3 & -1 & 0 & 0 & 1 \end{array}\right]\end{aligned}$$
If possible do row operations until you obtain an $3\times 6$ matrix of the form $[~I~|~B~]$

\[\begin{align*} [~A~|~I~]&=\left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & 0\\ 0 & 3 & -1 & 0 & 0 & 1 \end{array}\right]\\ {-3R_{2}+R_{3}\rightarrow R_{3}} & \Rightarrow \left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & 0\\ 0 & 0 & -1 & 0 & -3 & 1 \end{array}\right] \\ {-R_{3}\rightarrow R_{3}} & \Rightarrow \left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 3 & -1 \end{array}\right] \\ {R_{1}-2R_{3}\rightarrow R_{1}} & \Rightarrow \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & -6 & 2\\ 0 & 1 & 0 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 3 & -1 \end{array}\right]=[~I~|~B~]. \end{align*}\]

Therefore,

\[\begin{align*} A^{-1}=B =\begin{bmatrix}1 & -6 & 2\\ 0 & 1 & 0\\ 0 & 3 & -1\end{bmatrix}. \end{align*}\]

Theorem: Inverses of Transposes and Products

Given an invertible matrix $A$, the relationship is as follows:

\[\begin{align*} (A^T)^{-1} = (A^{-1})^T. \end{align*}\]

When dealing with invertible matrices $A$ and $B, the corresponding expression is:

\[\begin{align*} (AB)^{-1} = B^{-1}A^{-1}. \end{align*}\]

In the context of invertible matrices $A$, $B$, and $C, the relationship is defined as:

\[\begin{align*} (ABC)^{-1} = C^{-1}B^{-1}A^{-1}. \end{align*}\]

Theorem: Inverses of Transposes and Products

Consider a square matrix $I$, which is the identity matrix. It is invertible, and its inverse is itself, as expressed by $I^{-1} = I$ [Kuttler and Farah, 2020, Nicholson, 2018].
If matrix $A$ is invertible, then the inverse of $A^{-1}$ is equal to the original matrix, denoted as $A^{-1})^{-1} = A$ [Kuttler and Farah, 2020, Nicholson, 2018].
Given an invertible matrix $A$, the power $A^k$ is also invertible, and its inverse is found as $(A^k)^{-1} = (A^{-1})^k$ [Kuttler and Farah, 2020, Nicholson, 2018].
In the case where matrix $A$ is invertible and $p$ is a nonzero real number, the scaled matrix $pA$ remains invertible, with its inverse expressed as $(pA)^{-1} = \dfrac{1}{p} A^{-1}$ [Kuttler and Farah, 2020, Nicholson, 2018].