Learning Outcomes
- Determine if a matrix is invertible, and if so, compute its inverse.
Section 4.2 The Inverse of a Matrix (MX2)
Subsection 4.2.1 Warm Up
Activity 4.2.1.
Consider the matrices:
\begin{equation*}
A=\left[\begin{array}{ccc} 1 & 5 & -1 \\ 0 & 3 & 2 \end{array}\right],\ B=\left[\begin{array}{cccc} 7 & 2 & -1 & 1\\ 0 & 3 & 2 & -2\\ 1 & 1 & -1 & -3\end{array}\right].
\end{equation*}
Without using technology, what is the third column of the product \(AB\text{?}\)
Subsection 4.2.2 Class Activities
Activity 4.2.2.
Let \(A=\left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right]\text{.}\) Find a \(3 \times 3\) matrix \(B\) such that \(BA=A\text{,}\) that is,
\begin{equation*}
\left[\begin{array}{ccc} \unknown & \unknown & \unknown \\
\unknown & \unknown & \unknown
\\ \unknown & \unknown & \unknown \end{array}\right]
\left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right]
=
\left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right]
\end{equation*}
Check your guess using technology.
Definition 4.2.3.
The identity matrix \(I_n\) (or just \(I\) when \(n\) is obvious from context) is the \(n \times n\) matrix
\begin{equation*}
I_n = \left[\begin{array}{cccc} 1 & 0 & \cdots & 0 \\ 0 & 1 & \ddots & \vdots \\
\vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & 1 \end{array}\right].
\end{equation*}
It has a \(1\) on each diagonal element and a \(0\) in every other position.
Fact 4.2.4.
For any square matrix \(A\text{,}\) \(IA=AI=A\text{:}\)
\begin{equation*}
\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]
\left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right]
=
\left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right]
\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]
=
\left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right]
\end{equation*}
Activity 4.2.5.
Let \(T: \IR^n \rightarrow \IR^m\) be a linear map with standard matrix \(A\text{.}\) Sort the following items into three groups of statements: a group that means \(T\) is injective, a group that means \(T\) is surjective, and a group that means \(T\) is bijective.
- \(T(\vec x)=\vec b\) has a solution for all \(\vec b\in\IR^m\)
- \(T(\vec x)=\vec b\) has a unique solution for all \(\vec b\in\IR^m\)
- \(T(\vec x)=\vec 0\) has a unique solution.
- The columns of \(A\) span \(\IR^m\)
- The columns of \(A\) are linearly independent
- The columns of \(A\) are a basis of \(\IR^m\)
- Every column of \(\RREF(A)\) has a pivot
- Every row of \(\RREF(A)\) has a pivot
- \(m=n\) and \(\RREF(A)=I\)
Definition 4.2.6.
Let \(T: \IR^n \rightarrow \IR^n\) be a linear bijection with standard matrix \(A\text{.}\)
By item (B) from Activity 4.2.5 we may define an inverse map \(T^{-1} : \IR^n \rightarrow \IR^n\) that defines \(T^{-1}(\vec b)\) as the unique solution \(\vec x\) satisfying \(T(\vec x)=\vec b\text{,}\) that is, \(T(T^{-1}(\vec b))=\vec b\text{.}\)
Furthermore, let
\begin{equation*}
A^{-1}=[T^{-1}(\vec e_1)\hspace{1em}\cdots\hspace{1em}T^{-1}(\vec e_n)]
\end{equation*}
be the standard matrix for \(T^{-1}\text{.}\) We call \(A^{-1}\) the inverse matrix of \(A\text{,}\) and we also say that \(A\) is an invertible matrix.
Activity 4.2.7.
Let \(T: \IR^3 \rightarrow \IR^3\) be the linear bijection given by the standard matrix \(A=\left[\begin{array}{ccc} 2 & -1 & -6 \\ 2 & 1 & 3 \\ 1 & 1 & 4 \end{array}\right]\text{.}\)
(a)
To find \(\vec x = T^{-1}(\vec{e}_1)\text{,}\) we need to find the unique solution for \(T(\vec x)=\vec e_1\text{.}\) Which of these linear systems can be used to find this solution?
- \(\displaystyle \begin{array}{cccc} 2x_1 & -1x_2 & -6x_3 & =x_1 \\ 2x_1 & +1x_2 & +3x_3 & =0 \\ 1x_1 & +1x_2 & +4x_3 & =0 \end{array}\)
- \(\displaystyle \begin{array}{cccc} 2x_1 & -1x_2 & -6x_3 & =x_1 \\ 2x_1 & +1x_2 & +3x_3 & =x_2 \\ 1x_1 & +1x_2 & +4x_3 & =x_3 \end{array}\)
- \(\displaystyle \begin{array}{cccc} 2x_1 & -1x_2 & -6x_3 & =1 \\ 2x_1 & +1x_2 & +3x_3 & =0 \\ 1x_1 & +1x_2 & +4x_3 & =0 \end{array}\)
- \(\displaystyle \begin{array}{cccc} 2x_1 & -1x_2 & -6x_3 & =1 \\ 2x_1 & +1x_2 & +3x_3 & =1 \\ 1x_1 & +1x_2 & +4x_3 & =1 \end{array}\)
(b)
Use that system to find the solution \(\vec x=T^{-1}(\vec{e}_1)\) for \(T(\vec x)=\vec{e}_1\text{.}\)
(c)
Similarly, solve \(T(\vec x)=\vec{e}_2\) to find \(T^{-1}(\vec{e}_2)\text{,}\) and solve \(T(\vec x)=\vec{e}_3\) to find \(T^{-1}(\vec{e}_3)\text{.}\)
(d)
Use these to write
\begin{equation*}
A^{-1}= [T^{-1}(\vec e_1)\hspace{1em}
T^{-1}(\vec e_2)\hspace{1em}T^{-1}(\vec e_3)]\text{,}
\end{equation*}
the standard matrix for \(T^{-1}\text{.}\)
Activity 4.2.8.
Find the inverse \(A^{-1}\) of the matrix
\begin{equation*}
A=\left[\begin{array}{cccc} 0 & 0 & 0 & -1 \\ 1 & 0 & -1 & -4 \\ 1 & 1 & 0 & -4 \\ 1 & -1 & -1 & 2 \end{array}\right]
\end{equation*}
by computing how it transforms each of the standard basis vectors for \(\mathbb R^4\text{:}\) \(T^{-1}(\vec e_1)\text{,}\) \(T^{-1}(\vec e_2)\text{,}\) \(T^{-1}(\vec e_3)\text{,}\) and \(T^{-1}(\vec e_4)\text{.}\)
Activity 4.2.9.
Is the matrix \(\left[\begin{array}{ccc} 2 & 3 & 1 \\ -1 & -4 & 2 \\ 0 & -5 & 5 \end{array}\right]\) invertible?
- Yes, because its transformation is a bijection.
- Yes, because its transformation is not a bijection.
- No, because its transformation is a bijection.
- No, because its transformation is not a bijection.
Observation 4.2.10.
An \(n\times n\) matrix \(A\) is invertible if and only if \(\RREF(A) = I_n\text{.}\)
Activity 4.2.11.
Let \(T:\IR^2\to\IR^2\) be the bijective linear map defined by \(T\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)=\left[\begin{array}{c} 2x -3y \\ -3x + 5y\end{array}\right]\text{,}\) with the inverse map \(T^{-1}\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)=\left[\begin{array}{c} 5x+ 3y \\ 3x + 2y\end{array}\right]\text{.}\)
(a)
Compute \((T^{-1}\circ T)\left(\left[\begin{array}{c}-2\\1\end{array}\right]\right)\text{.}\)
(b)
If \(A\) is the standard matrix for \(T\) and \(A^{-1}\) is the standard matrix for \(T^{-1}\text{,}\) find the \(2\times 2\) matrix
\begin{equation*}
A^{-1}A=\left[\begin{array}{ccc}\unknown&\unknown\\\unknown&\unknown\end{array}\right].
\end{equation*}
Observation 4.2.12.
\(T^{-1}\circ T=T\circ T^{-1}\) is the identity map for any bijective linear transformation \(T\text{.}\) Therefore \(A^{-1}A=AA^{-1}\) equals the identity matrix \(I\) for any invertible matrix \(A\text{.}\)
Subsection 4.2.3 Individual Practice
Activity 4.2.13.
Now that we have defined the inverse of a matrix, we have the ability to solve matrix equations. In the following equations, \(A,B\) all denote square matrices of the same size and \(I\) denotes the identity matrix. For each equation, solve for \(X\text{.}\)
(a)
\(A^{-1}XA=B\)
(b)
\(AXA^{-1}=B\)
(c)
\(ABX=I\)
(d)
\(BAX=I\)
Subsection 4.2.4 Videos
Exercises 4.2.5 Exercises
Exercises available at
https://tbil.org/linear-algebra/preview/exercises/#/bank/MX2/.Subsection 4.2.6 Mathematical Writing Explorations
Exploration 4.2.14.
Assume \(A\) is an \(n \times n\) matrix. Prove the following are equivalent. Some of these results you have proven previously.- \(A\) row reduces to the identity matrix.
- For any choice of \(\vec{b} \in \mathbb{R}^n\text{,}\) the system of equations represented by the augmented matrix \([A|\vec{b}]\) has a unique solution.
- The columns of \(A\) are a linearly independent set.
- The columns of \(A\) form a basis for \(\mathbb{R}^n\text{.}\)
- The rank of \(A\) is \(n\text{.}\)
- The nullity of \(A\) is 0.
- \(A\) is invertible.
- The linear transformation \(T\) with standard matrix \(A\) is injective and surjective. Such a map is called an isomorphism.
Exploration 4.2.15.
- Assume \(T\) is a square matrix, and \(T^4\) is the zero matrix. Prove that \((I - T)^{-1} = I + T + T^2 + T^3.\) You will need to first prove a lemma that matrix multiplication distributes over matrix addition.
- Generalize your result to the case where \(T^n\) is the zero matrix.
Subsection 4.2.7 Sample Problem and Solution
Sample problem Example C.1.19.
