# Linear Algebra

## Inverse Matrices

 Let $$A\in M_{n\times n} (\mathbb{R})$$. If there exists $$B\in M_{n\times n} (\mathbb{R})$$ such that $$AB=I=BA$$, then $$A$$ is said to be invertible, and $$B$$ is called the inverse of $$A$$ (and $$A$$ is the inverse of $$B$$). The inverse of $$A$$ is denoted $$A^{-1}$$.

Example: The matrix $$\left[\begin{array}{rr} 2&-1\\-1&1\end{array}\right]$$ is the inverse of the matrix $$\left[\begin{array}{rr} 1&1\\1&2\end{array}\right]$$ because

$\left[\begin{array}{rr} 2&-1\\-1&1\end{array}\right]\left[\begin{array}{rr} 1&1\\1&2\end{array}\right]=\left[\begin{array}{rr} 1&0\\0&1\end{array}\right]=I$

and

$\left[\begin{array}{rr} 1&1\\1&2\end{array}\right]\left[\begin{array}{rr} 2&-1\\-1&1\end{array}\right]=\left[\begin{array}{rr} 1&0\\0&1\end{array}\right]=I$

 Properties of invertible matrices If $$A$$ and $$B$$ are invertible matrices and t is a non-zero real number, then $$(tA)^{-1}=\frac{1}{t}A^{-1}$$ $$(AB)^{-1}=B^{-1}A^{-1}$$ $$(A^T)^{-1}=(A^{-1})^T$$

## Finding the Inverse of a Matrix

 To find the inverse of a square matrix A, Row reduce the multi-augmented matrix $$\left[\begin{array}{c|c} A&I\end{array}\right]$$ so that the left block is in reduced row echelon form If the reduced row echelon form is $$\left[\begin{array}{c|c} I&B\end{array}\right]$$, then $$A^{-1}=B$$. If the reduced row echelon form of $$A$$ is not $$I$$, then $$A$$ is not invertible.

Key takeaways:

• Only square matrices can be invertible
• Not all square matrices are invertible
• Row reduction is a method to find the inverse
• Start with the augmented matrix $$\left[\begin{array}{c|c} A&I\end{array}\right]$$
• Use row reduction to turn the matrix into $$\left[\begin{array}{c|c} I&B\end{array}\right]$$
• Use multiplication of the original matrix and the calculated inverse to verify your solution.

Example: Determine whether $$A=\left[\begin{array}{ccc} 1&1&2\\1&2&2\\2&4&3\end{array}\right]$$ is invertible, and if it is, determine its inverse.

Solution Write the matrix $$\left[\begin{array}{c|c} A&I\end{array}\right]$$ and row reduce.

\begin{align} &\left[\begin{array}{rrr|rrr} 1&1&2&1&0&0\\1&2&2&0&1&0\\2&4&3&0&0&1\end{array}\right] \\ \sim&\left[\begin{array}{rrr|rrr} 1&1&2&1&0&0\\0&1&0&-1&1&0\\0&2&-1&-2&0&1\end{array}\right] \\ \sim&\left[\begin{array}{rrr|rrr} 1&0&2&2&-1&0\\0&1&0&-1&1&0\\0&0&-1&0&-2&1\end{array}\right] \\ \sim&\left[\begin{array}{rrr|rrr} 1&0&0&2&-5&2\\0&1&0&-1&1&0\\0&0&1&0&2&-1\end{array}\right]\end{align}

Hence, $$A$$ is invertible and $$A^{-1}=\left[\begin{array}{rrr} 2&-5&2\\-1&1&0\\0&2&-1\end{array}\right]$$ .

Check

$A=\left[\begin{array}{ccc} 1&1&2\\1&2&2\\2&4&3\end{array}\right]\left[\begin{array}{rrr} 2&-5&2\\-1&1&0\\0&2&-1\end{array}\right]=I$

$A=\left[\begin{array}{rrr} 2&-5&2\\-1&1&0\\0&2&-1\end{array}\right]\left[\begin{array}{ccc} 1&1&2\\1&2&2\\2&4&3\end{array}\right]=I$

 Using the determinant to find the inverse of a $$2\times 2$$ matrix Let $$A\in M_{2\times 2}(\mathbb{R})$$. Let $$A=\left[\begin{array}{cc} a&b\\c&d\end{array}\right]$$, if and only if $$ad-bc\neq 0$$ then $$A$$ is invertible and, $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{rr} d&-b\\-c&a\end{array}\right]$

The quantity $$ad-bc$$ is also what we call the determinant of the $$2\times 2$$ matrix which will be examined later.

Example: Determine whether $$A=\left[\begin{array}{rr} 1&2\\2&4\end{array}\right]$$ is invertible, and if it is, determine its inverse.

Solution: $$ad-bc = 1(4)-2(2)=0$$, therefore the $$A$$ is not invertible.

Example: Determine whether $$B=\left[\begin{array}{rr} 6&-4\\-4&7\end{array}\right]$$ is invertible, and if it is, determine its inverse.

Solution: $$ad-bc = 6(7)-(-4)(-4)=26$$

\begin{align}B^{-1}&=\frac{1}{26}\left[\begin{array}{rr} 7&4\\4&6\end{array}\right]\\ &=\left[\begin{array}{rr} \frac{7}{26}&\frac{4}{26}\\ \frac{4}{26}&\frac{6}{26}\end{array}\right] \\ &=\left[\begin{array}{rr} \frac{7}{26}&\frac{2}{13}\\ \frac{2}{13}&\frac{3}{13}\end{array}\right] \end{align}

Check $$BB^{-1}=I=B^{-1}B$$

Solving systems of linear equations by matrix inverse.

We saw earlier that a system of equations can be written in the form

$Ax=b$

Where matrix $$A$$ is the matrix of coefficients, vector $$x$$ represents each variable $$x_1, \ldots, x_n$$, and vector b the solution space.

We can use the inverse and solve for the variables

$x=A^{-1}b$

Example: Solve the given system of equations using the inverse of a matrix

\begin{align} 3x+8y &= 5\\4x+11y&=7\end{align}

Solution:

$A=\left[\begin{array}{cc} 3&8\\4&11\end{array}\right] \quad x=\left[\begin{array}{cc} x\\y \end{array}\right] \quad \left[\begin{array}{cc} 5\\7\end{array}\right]$

$$ad-bc=1$$, so the inverse is $$A^{-1}=\left[\begin{array}{cc} 11&-8\\-4&3\end{array}\right]$$

\begin{align} x&=A^{-1}b\\ &=\left[\begin{array}{cc} 11&-8\\-4&3\end{array}\right]\left[\begin{array}{cc} 5\\7\end{array}\right] \\&= \left[\begin{array}{cc} -1\\1\end{array}\right] \end{align}

The solution checks out when plugging x and y back into the original equations.