2.7. Finding the Inverse of a Matrix

Definition - Matrix Inverse Algorithm:

1. 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].

2. If feasible, apply row operations until a transformation is achieved, leading to an \(n\times 2n\) matrix in the format of \([~I~|~B~]\).

3. Consequently, the matrix \(B\) corresponds to the inverse of matrix \(A\) ( \(B = A^{-1} \)), affirming the invertibility of \(A\).

4. 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:

1. Form the augmented \(3\times 6\) matrix \([~A~|~I~]\) $\(\begin{aligned} \left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & 0\\ 0 & 3 & -1 & 0 & 0 & 1 \end{array}\right]\end{aligned}\)$

2. 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,

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

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Theorem: Inverses of Transposes and Products

1. Given an invertible matrix \(A\), the relationship is as follows: \begin{align*} (A^T)^{-1} = (A^{-1})^T. \end{align*}

2. When dealing with invertible matrices \(A\) and $B, the corresponding expression is: \begin{align*} (AB)^{-1} = B^{-1}A^{-1}. \end{align*}

3. 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

1. 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].

2. 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].

3. 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].

4. 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].