### Systems of linear equations and matrices: Matrices

### The inverse of a matrix

The inverse of a matrix

If, for a given square matrix \(A\), a matrix \(B\) exists such that \(A\,B=B\,A=I\), then \(B\) is called the **inverse** of \(A\). The notation for this matrix is \( A^{-1}\). In this case we say that \(A\) is **invertible**, or **regular**.

If a matrix is not invertible, then we speak of a **singular matrix**.

- If \(A\, B = I\), then \(B\) the inverse of \(A\).
- If \(B\, A = I\), then \(B\) the inverse of \(A\).

Here are some calculation rules for the inverse of a matrix.

- Let \(A\) and \(B\) be \((n\times n)\)-matrices. If \(A\) and \(B\) are invertible, then their product, \(A\, B\), is also invertible, with inverse \[(A\, B)^{-1}=B^{-1} A^{-1}\]
- If \(A\) is an invertible matrix, then its transposed matrix, \(A^{\top}\), is also invertible, with inverse \[ \left(A^{\top}\right)^{-1}=\left(A^{-1}\right)^{\top}\]

For square matrices of dimension #2# we have an explicit expression for the inverse:

If \(A=\matrix{a & b \\ c & d}\), then \(A\) is invertible if and only if \(a\,d-b\,c\neq 0\). In this case \[A^{-1}=\frac{1}{a\,d-b\,c}\matrix{d & -b \\ -c & a}\]

The expression \(a\,d-b\,c\) is called the **determinant** of the matrix \(A=\matrix{a & b \\ c & d}\) and is denoted as \(\text{det}(A)\).

Finding the inverse of a matrix can be seen as the solving of a system of linear equations. This means that we have a method for finding it:

Gaussian elimination to invert a matrix In order to determine if the inverse of an \((n\times n)\)-matrix \(A\) exist and, if so, to calculate it, we write down the \((n\times 2n)\)-matrix \((A\,|\,I)\): \[(A\,|\,I)=\left(\begin{array}{cccc|cccc} a_{11} & a_{12} & \cdots & a_{1n} & 1 & 0 &\cdots & 0\\ a_{21} & a_{22} & \cdots & a_{2n} & 0 & 1 &\ddots & 0\\ \vdots & \vdots & & \vdots & \vdots & \ddots & \ddots & 0\\ a_{n1} & a_{n2} & \cdots & a_{nn} & 0 & \cdots &\cdots & 1\end{array}\right)\] We reduce this augmented matrix to reduced echelon form.

- If the reduced echelon form has the form \((I\,|\,B)\), meaning that the submatrix to the left of the vertical line is the identity matrix, then \(A\) has rank #n# and is invertible with inverse \(A^{-1}=B\).
- If the reduced echelon form is not of the form \((I\,|\,B)\), meaning that submatrix to the left of the vertical bar is not the identity matrix, then \(A\) has rank less than #n# and is not invertible.

We extend the matrix with an identity matrix. Row reduction then gives

\[\begin{aligned}\left(\begin{array}{ccc|ccc}

1&-1&0&1 & 0 & 0 \\

0&-16&0&0 &1 &0\\

1&-5&-16&0 &0 &1\\

\end{array}

\right)&\sim

\left(

\begin{array}{ccc|ccc}

1 &-1&0&1 & 0 & 0\\

0 &-16&0&0& 1 & 0 \\

0 &-4&-16&-1& 0 & 1 \\

\end{array}

\right)

&{\color{blue}{\begin{array}{ccc}

\mbox{}\\

\phantom{X}\\

R_3 -R_1

\end{array}}}\\

&\sim

\left(

\begin{array}{ccc|ccc}

1 &-1&0 &1 & 0 & 0 \\

0 &1 &0&0&-\frac{1}{16} &0 \\

0 &-4&-16&-1& 0 & 1 \\

\end{array}

\right)

&{\color{blue}{\begin{array}{ccc}

\mbox{}\\

-\frac{1}{16}R_2\\

\mbox{}

\end{array}}}\\

&\sim

\left(

\begin{array}{ccc|ccc}

1 &0 &0&1&-\frac{1}{16}&0 \\

0 &1 &0&0&-\frac{1}{16}&0 \\

0 &0 &-16&-1&-\frac{1}{4}&1\\

\end{array}

\right)

&{\color{blue}{\begin{array}{ccc}

{}R_1 +R_2\\

\mbox{}\\

R_3 +4 R_1

\end{array}}}\\

&\sim

\left(

\begin{array}{ccc|ccc}

1 &0 &0&1&-\frac{1}{16}&0\\

0 &1 &0&0&-\frac{1}{16}&0\\

0 &0 &1 &\frac{1}{16}&\frac{1}{64}&-\frac{1}{16} \\

\end{array}

\right)

&{\color{blue}{\begin{array}{ccc}

{}\\

{}\\

-\frac{1}{16}R_3

\end{array}}}\\

&\sim

\left(

\begin{array}{ccc|ccc}

1 &0 &0 &1&-\frac{1}{16}&0\\

0 &1 &0 &0&-\frac{1}{16}&0\\

0 &0 &1 &\frac{1}{16}&\frac{1}{64}&-\frac{1}{16} \\

\end{array}

\right)

&{\color{blue}{\begin{array}{ccc}

{}\phantom{X}\\

\phantom{X}\\

{}

\end{array}}}

\end{aligned}

\] The reduced echelon form implies that the given matrix is invertible. The inverse matrix is the submatrix to the right of the identity matrix of the reduced echelon form.

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