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 Class Activities
Activity 4.2.1.
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 4.2.2.
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 4.2.3.
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 4.2.4.
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 4.2.5.
Let \(T: \IR^n \rightarrow \IR^n\) be a linear bijection with standard matrix \(A\text{.}\)
By item (B) from Activity 4.2.4 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 4.2.6.
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 4.2.7.
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 4.2.8.
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 4.2.9.
An \(n\times n\) matrix \(A\) is invertible if and only if \(\RREF(A) = I_n\text{.}\)
Activity 4.2.10.
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 4.2.11.
\(T^{-1}\circ T=T\circ T^{-1}\) 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 4.2.2 Videos
Exercises 4.2.3 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2024e/exercises/#/bank/MX2/.Subsection 4.2.4 Mathematical Writing Explorations
Exploration 4.2.12.
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 4.2.13.
- 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 4.2.5 Sample Problem and Solution
Sample problem Example B.1.19.
