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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

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


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

Properties of invertible matrices

If \(A\) and \(B\) are invertible matrices and t is a non-zero real number, then

  1. \((tA)^{-1}=\frac{1}{t}A^{-1}\)
  2. \((AB)^{-1}=B^{-1}A^{-1}\)
  3. \((A^T)^{-1}=(A^{-1})^T\)

Finding the Inverse of a Matrix

To find the inverse of a square matrix A,

  1. 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
  2. If the reduced row echelon form is \(\left[\begin{array}{c|c} I&B\end{array}\right]\), then \(A^{-1}=B\).
  3. 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 \(A^{-1}=\left[\begin{array}{rrr} 2&-5&2\\-1&1&0\\0&2&-1\end{array}\right]\) .


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

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

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


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


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

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


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

\(ad-bc=1\), 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.

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Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.
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