Learning Outcomes
- Determine if a matrix is invertible, and if so, compute its inverse.
Section 21.2 The Inverse of a Matrix (MX2)
Subsection 21.2.1 Warm Up
Activity 21.14.
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 21.2.2 Class Activities
Activity 21.15.
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 21.16.
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 21.17.
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 21.18.
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 21.19.
Let \(T: \IR^n \rightarrow \IR^n\) be a linear bijection with standard matrix \(A\text{.}\)
By item (B) from Activity 21.18 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 21.20.
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 21.21.
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 21.22.
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 21.23.
An \(n\times n\) matrix \(A\) is invertible if and only if \(\RREF(A) = I_n\text{.}\)
Activity 21.24.
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 21.25.
\(T^{-1}\circ T=T\circ T^{-1}\) 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 21.2.3 Individual Practice
Activity 21.26.
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 21.2.4 Videos
Exercises 21.2.5 Exercises
Exercises available at
https://tbil.org/preview/linear-algebra/exercises/#/bank/MX2/.Subsection 21.2.6 Mathematical Writing Explorations
Exploration 21.27.
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 21.28.
- 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 21.2.7 Sample Problem and Solution
Sample problem Example C.19.
