Learning Outcomes
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Determine if a matrix is invertible, and if so, compute its inverse and use it to solve an appropriate system of equations.
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.
-
\(T(\vec x)=\vec 0\) has a unique solution.
-
The columns of \(A\) are linearly independent
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Every column of \(\RREF(A)\) has a pivot
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Every row of \(\RREF(A)\) has a pivot
Definition 4.2.6.
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.
Let \(T\colon\IR^4\to\IR^4\) be the linear bijection given by the standard 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*}
(a)
Calculate \(T^{-1}(\vec e_2)\) by row reducing an appropriate augmented matrix.
(b)
Calculate \(T^{-1}(\vec e_4)\) by row reducing an appropriate augmented matrix.
(c)
Suppose we completed the previous two tasks by hand. Which of the following statements best describes the row operations we would use?
-
We had to use the same row operations in (a) as we did in (b).
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We could have used the same row operation in (a) as we did in (b).
-
The row operations used in (a) have nothing to do with with the row operations used in part (b).
(d)
So far, we have only considered augmented matrices with a single augmented column. Write down an augmented matrix with more than one augmented column whose RREF would help us find \(A^{-1}.\)
Observation 4.2.9.
Our exploration above yields a succinct way to calculate the inverse of a matrix. Indeed, if \(A\) is an invertible matrix, then we have
\begin{equation*}
\RREF\left[A|I\right]=\left[I|A^{-1}\right].
\end{equation*}
Activity 4.2.10.
Is the matrix \(\left[\begin{array}{ccc} 2 & 3 & 1 \\ -1 & -4 & 2 \\ 0 & -5 & 5 \end{array}\right]\) invertible?
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Yes, because its transformation is a bijection.
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Yes, because its transformation is not a bijection.
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No, because its transformation is a bijection.
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No, because its transformation is not a bijection.
Observation 4.2.11.
Observation 4.2.12.
Let \(T:\IR^n\to\IR^n\) be a linear bijection with standard matrix \(A\) and suppose \(\vec{b}\in\IR^n\text{.}\) By definition of the inverse map and inverse matrix, the vector \(\vec{x}=A^{-1}\vec{b}\) is the unique solution to the equation \(A\vec{x}=\vec{b}\text{.}\)
In other words, when the matrix \(A\) is invertible, we have a new method for solving the equation \(A\vec{x}=\vec{b}\text{:}\) we can first calculate \(A^{-1}\) and then calculate the product \(\vec{x}=A^{-1}\vec{b}\text{.}\)
Activity 4.2.13.
Consider the vector equation
\begin{equation*}
x_{1} \left[\begin{array}{c} 0 \\ 1 \\ 1\\1 \end{array}\right] + x_{2} \left[\begin{array}{c} 0 \\ 0 \\ 1\\-1 \end{array}\right] + x_{3} \left[\begin{array}{c} 0 \\ -1 \\ 0\\-1 \end{array}\right] +x_{4}\left[\begin{array}{c} -1 \\ -4 \\ -4\\2 \end{array}\right]= \left[\begin{array}{c} 1 \\ 1 \\ 1 \\ 1\end{array}\right]
\end{equation*}
with a unique solution.
(a)
Use technology to both verify that the coefficient matrix is invertible and calculate its inverse.
(b)
Explain and demonstrate how to use the inverse to find the unique solution to the given vector equation.
Activity 4.2.14.
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 4.2.15.
\(T^{-1}\circ T=T\circ T^{-1}\) 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 4.2.16.
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)
\(BAX=I\)
Activity 4.2.17.
Solving linear systems using matrix multiplication is most useful when we are working with one common coefficient matrix, and varying the right-hand side. That is, when we have \(A\vec{x}=\vec{b}\) for several different values of \(\vec{b}\text{.}\)
In the following, let \(A=\left[\begin{matrix}2 & -1 & -6\\ 2 & 1 & 3\\ 1 & 1 & 4\end{matrix}\right]\) and consider the following questions about various equations of the form \(A\vec{x}=\vec{b}\text{?}\)
(a)
Suppose that \(\vec{b}=\left[\begin{matrix} 1\\1\\1\end{matrix}\right]\text{.}\) If asked to solve the equation \(A\vec{x}=\vec{b}\text{,}\) which of the following approaches do you prefer?
-
Calculate \(\RREF[A|\vec{b}]\text{.}\)
(b)
Suppose that \(\vec{b}_1,\vec{b}_2,\vec{b}_3=\left[\begin{matrix} 1\\1\\1\end{matrix}\right],\left[\begin{matrix} 2\\1\\3\end{matrix}\right],\left[\begin{matrix} -1\\3\\5\end{matrix}\right]\text{.}\) If asked to solve each of the equations \(A\vec{x}=\vec{b}_1, A\vec{x}=\vec{b}_2, A\vec{x}=\vec{b}_3\text{,}\) which of the following approaches do you prefer?
-
Calculate \(A^{-1}\) and then compute \(\vec{x}=A^{-1}\vec{b}_1\text{,}\) \(\vec{x}=A^{-1}\vec{b}_2\text{,}\) and \(\vec{x}=A^{-1}\vec{b}_3\)
(c)
Suppose that \(\vec{b}_1,\dots, \vec{b}_{10}\) are 10 distinct vectors. If asked to solve each of the equations \(A\vec{x}=\vec{b}_1, \dots, A\vec{x}=\vec{b}_{10}\text{,}\) which of the following approaches do you prefer?
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Calculate \(A^{-1}\) and then compute \(\vec{x}=A^{-1}\vec{b}_1\text{,}\) ... \(\vec{x}=A^{-1}\vec{b}_{10}\text{.}\)
Subsection 4.2.4 Videos
Subsection 4.2.5 Exercises
Exercises available at
https://library.tbil.org/2025/linear-algebra/exercises/#/bank/MX2/
Subsection 4.2.6 Mathematical Writing Explorations
Exploration 4.2.18.
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.
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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.
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The columns of \(A\) are a linearly independent set.
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The columns of \(A\) form a basis for \(\mathbb{R}^n\text{.}\)
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The rank of \(A\) is \(n\text{.}\)
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The nullity of \(A\) is 0.
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\(A\) is invertible.
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The linear transformation \(T\) with standard matrix \(A\) is injective and surjective. Such a map is called an isomorphism.
Exploration 4.2.19.
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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.
Exploration 4.2.20.
Use row reduction to find the inverse of the following general matrix. Give conditions on which this inverse exists.
\begin{equation*}
\left[\begin{array}{ccc}1 & b & c \\ d & e & f \\ g & h & i \end{array}\right]
\end{equation*}
Exploration 4.2.21.
Assume that \(H\) is invertible, and that \(HG\) is the zero matrix. Prove that \(G\) must be the zero matrix. Would this still be true if \(H\) were not invertible?Exploration 4.2.22.
If \(H\) is invertible and \(r \in \mathbb{R}\text{,}\) what is the inverse of \(rH\text{?}\)Exploration 4.2.23.
If \(H\) and \(G\) are invertible, is \(H^{-1} + G^{-1} = (H+G)^{-1}\text{?}\)Exploration 4.2.24.
Prove that if \(A\text{,}\) \(P\text{,}\) and \(Q\) are invertible with \(PAQ = I\text{,}\) then \(A^{-1} = QP\text{.}\)Subsection 4.2.7 Sample Problem and Solution
Sample problem Example B.1.19.
