Keywords
[1] If the square matrix \(A\) has an inverse, then \(A^{-1}A=I\) and \(AA^{-1}=I\)
[2] The algorithm to test invertibility is elimination: \(A\) must have \(n\) nonzero pivots
[3] The algebra test for invertibility is the determinant of \(A\): \(det(A)\) must not be zero
[4] The equation that tests for invertibility is \(Ax=0\): \(x=0\) must be the only solution.
[5] If \(A\) and \(B\) (same size) are invertible, then \(AB\) is also invertible: \((AB)^{-1}=B^{-1}A^{-1}\).
[6] \(AA^{-1}=I\) is \(n\) equations for \(n\) columns of \(A^{-1}\). Gauss-Jordan eliminates \([AI]\) to \(IA^{-1}\)
Inverse matrix and Invertible
Suppose \(A\) is a square matrix. Intuitively, the inverse matrix \(A^{-1}\) (same size) is a matrix that undoes what \(A\) does. Hence, their product is the identity matrix which does nothing to a vector so \(A^{-1}A=x\).
The matrix \(A\) is invertible if there exists a matrix \(A^{-1}\) such that
\[A^{-1}A = I\ \text{and}\ AA^{-1}=I\]However, a matrix does not always have an inverse. Below are some notes about \(A^{-1}\).
[Note 1] The inverrse exists if and only if elimination produces \(n\) pivot
[Note 2] \(A\) cannot have two different inverses
[Note 3] If \(A\) is invertible, the only and only solution to \(Ax=b\) is \(x=A^{-1}b\)
[Note 4] Suppose there is a nonzero vector \(x\) such that \(Ax=0\). Then \(A\) is not invertible.
[Note 5] \(A\) is invertible if and only if its determinant is not zero
[Note 6] A diagonal matrix has an inverse provided no diagonal entries are zero
Inverse of a Product AB
The product \(AB\) has an inverse if and only if \(A\) and \(B\) of the same size are separately invertible. Note that the \(A^{-1}\) and \(B^{-1}\) come in the reverse order
If \(A\) and \(B\) are invertible then so is \(AB\). The inverse of \(AB\) is
\[(AB)^{-1} = B^{-1}A^{-1}\]The reasons it comes in reverse order is \((AB)(B^{-1}A^{-1})=I\) and \((B^{-1}A^{-1})(AB) = I\)
Calculate Inverse by Gauss-Jordan Elimination
Elimination goes directly to \(x\) without needing to calculate \(A^{-1}\). The Gauss-Jordan elimination is to solve \(AA^{-1}=I\), finding each column of \(A^{-1}\). \(A\) multiplies the first column of \(A^{-1}\) (call \(x_1\)) to give the first column of \(I\) (call \(e_1\)). So each of the columns \(x_1, x_2, x_3\) of \(A^{-1}\) is multiplied by \(A\) to produce a column of \(I\):
\[AA^{-1} = A[x_1\ x_2\ x_3] = [e_1\ e_2\ e_3] = I\]To invert \(A\), we have to solve three systems of equations: \(Ax_1 = e_1\), \(Ax_2=e_2\), and \(Ax_3=x_3\). Gauss-Jordan elimination finds \(A^{-1}\) by solving all n equations together.
The augmented matrix \([A\ I]\) is used where \(e_1, e_2, e_3\) are in the columns of \(I\). With the similar elimination strategy, we perform elimination on this augmented matrix. Our goal is to make the left-side \(A\) down to \(I\). What this means is that if \(A\) turns into \(A^{-1}\) by elimination, our elimination matrix is indeed \(A^{-1}\). Since the elimination is equally performed on \(A\) and \(I\), we get \(IA^{-1} = A^{-1}\) on the right side. Hence, finally we get our target \(A^{-1}\).
Let’s find the inverse of matrix \(K\).
\[K=\begin{bmatrix}2 & -1 & 0 \\\ -1 & 2 & -1 \\\ 0 & -1 & 2 \end{bmatrix}\][1] Construct Augmented matrix \([K\ I]\)
\[K=\begin{bmatrix}2 & -1 & 0 & 1 & 0 & 0 \\\ -1 & 2 & -1 & 0 & 1 & 0 \\\ 0 & -1 & 2 & 0 & 0 & 1\end{bmatrix}\][2] Perform elimination to get reduced echelon form \(R=I\).
\[K=\begin{bmatrix}1 & 0 & 0 & \frac{3}{4} & \frac{1}{2} & \frac{1}{4} \\\ 0 & 1 & 0 & \frac{1}{2} & 1 & \frac{1}{2} \\\ 0 & 0 & 1 & \frac{1}{4} & \frac{1}{2} & \frac{3}{4} \end{bmatrix}\]We got from \([K\ I]\) to \([I\ ?]\). Since \(K\) becomes \(I\), it turns out that we in fact multiplied the augmented matrix by the inverse matrix \(K^{-1}\).
Summary
\[\text{Multiply}\ [A\ I]\ \text{by}\ A^{-1}\ \text{to get}\ [I\ A^{-1}]\]Diagonally dominant matrices are invertible
Diagonally dominant matrices are invertible. Each \(a_{ii}\) on the diagonal is larger than the total sum along the rest of row \(i\). On every row,
\[\lvert a*{ii} \rvert > \sum*{j \neq i} \lvert a\_{ij} \vert\]For example,
\[A = \begin{bmatrix} 3 & 1 & 1 \\\ 1 & 3 & 1 \\\ 1 & 1 & 3 \end{bmatrix}\]\(A\) is a diagonally dominant matrix. Hence, \(A\) is invertible.
Exercises
Q7
If \(A\) has row 1 + row 2 = row 3, show that \(A\) is not invertible:
Set \(A = \begin{bmatrix} \vec{a_1} \\\ \vec{a_2} \\\ \vec{a_3} \end{bmatrix}\) and \(\vec{a_1} + \vec{a_2} = \vec{a_3}\)
(a) Explain why \(Ax = (0,0,1)\) cannot have a solution. Add eqn 1 + eqn 2
It must satisfy that \(\vec{a_1} \cdot x = 0\), \(\vec{a_2} \cdot x = 0\) and \(\vec{a_3} \cdot x = 1\)
However, since \(\vec{a_3} \cdot x = 1\) is \((\vec{a_1}+\vec{a_2}) \cdot x = 1\), it gives \(0=1\) which is contradictory.
(b) Which right sides \((b_1, b_2, b_3)\) might allow a solution to \(Ax = b\)?
\(b_1 + b_2 = b_3\). Otherwise, it gives contradition like (a).
(c) In elimination, what happens to equation 3?
Row 3 becomes a row of zeros so there’s no pivot. Hence, it’s not invertible.
References
[1] Introduction to Linear Algebra, 5th Edition