| 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
\[\left[\begin{array}{rr} 1&1\\1&2\end{array}\right]\left[\begin{array}{rr} 2&-1\\-1&1\end{array}\right]=\left[\begin{array}{rr} 1&0\\0&1\end{array}\right]=I\]
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Properties of invertible matrices If \(A\) and \(B\) are invertible matrices and t is a non-zero real number, then
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To find the inverse of a square matrix A,
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Key takeaways:
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
\[A=\left[\begin{array}{ccc} 1&1&2\\1&2&2\\2&4&3\end{array}\right]\left[\begin{array}{rrr} 2&-5&2\\-1&1&0\\0&2&-1\end{array}\right]=I\]
\[A=\left[\begin{array}{rrr} 2&-5&2\\-1&1&0\\0&2&-1\end{array}\right]\left[\begin{array}{ccc} 1&1&2\\1&2&2\\2&4&3\end{array}\right]=I\]
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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.
