Linear Algebra for Team-Based Inquiry Learning

2023 Edition

Steven Clontz Drew Lewis
University of South Alabama

August 24, 2023

Section 4.2: The Inverse of a Matrix (MX2)

Activity 4.2.1 (~5 min)

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 4.2.1

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 4.2.2

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 4.2.2 (~15 min)

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.

  1. \(A\vec x=\vec b\) has a solution for all \(\vec b\in\IR^m\)

  2. \(A\vec x=\vec b\) has a unique solution for all \(\vec b\in\IR^m\)

  3. \(A\vec x=\vec 0\) has a unique solution.

  4. The columns of \(A\) span \(\IR^m\)

  5. The columns of \(A\) are linearly independent

  6. The columns of \(A\) are a basis of \(\IR^m\)

  7. Every column of \(\RREF(A)\) has a pivot

  8. Every row of \(\RREF(A)\) has a pivot

  9. \(m=n\) and \(\RREF(A)=I\)

Activity 4.2.3 (~5 min)

Let \(T: \IR^3 \rightarrow \IR^3\) be the linear transformation given by the standard matrix \(A=\left[\begin{array}{ccc} 2 & -1 & 0 \\ 2 & 1 & 4 \\ 1 & 1 & 3 \end{array}\right]\text{.}\)

Write an augmented matrix representing the system of equations given by \(T(\vec x)=\vec{0}\text{,}\) that is, \(A\vec x=\left[\begin{array}{c}0 \\ 0 \\ 0 \end{array}\right]\text{.}\) Then solve \(T(\vec x)=\vec{0}\) to find the kernel of \(T\text{.}\)

Definition 4.2.3

Let \(T: \IR^n \rightarrow \IR^n\) be a linear map with standard matrix \(A\text{.}\)

  • If \(T\) is a bijection and \(\vec b\) is any \(\IR^n\) vector, then \(T(\vec x)=A\vec x=\vec b\) has a unique solution.

  • So we may define an inverse map \(T^{-1} : \IR^n \rightarrow \IR^n\) by setting \(T^{-1}(\vec b)\) to be this unique solution.

  • Let \(A^{-1}\) be the standard matrix for \(T^{-1}\text{.}\) We call \(A^{-1}\) the inverse matrix of \(A\text{,}\) so we also say that \(A\) is invertible.

Activity 4.2.4 (~20 min)

Let \(T: \IR^3 \rightarrow \IR^3\) be the linear transformation given by the standard matrix \(A=\left[\begin{array}{ccc} 2 & -1 & -6 \\ 2 & 1 & 3 \\ 1 & 1 & 4 \end{array}\right]\text{.}\)

Part 1.

Write an augmented matrix representing the system of equations given by \(T(\vec x)=\vec{e}_1\text{,}\) that is, \(A\vec x=\left[\begin{array}{c}1 \\ 0 \\ 0 \end{array}\right]\text{.}\) Then solve \(T(\vec x)=\vec{e}_1\) to find \(T^{-1}(\vec{e}_1)\text{.}\)

Activity 4.2.4 (~20 min)

Let \(T: \IR^3 \rightarrow \IR^3\) be the linear transformation given by the standard matrix \(A=\left[\begin{array}{ccc} 2 & -1 & -6 \\ 2 & 1 & 3 \\ 1 & 1 & 4 \end{array}\right]\text{.}\)

Part 2.

Solve \(T(\vec x)=\vec{e}_2\) to find \(T^{-1}(\vec{e}_2)\text{.}\)

Activity 4.2.4 (~20 min)

Let \(T: \IR^3 \rightarrow \IR^3\) be the linear transformation given by the standard matrix \(A=\left[\begin{array}{ccc} 2 & -1 & -6 \\ 2 & 1 & 3 \\ 1 & 1 & 4 \end{array}\right]\text{.}\)

Part 3.

Solve \(T(\vec x)=\vec{e}_3\) to find \(T^{-1}(\vec{e}_3)\text{.}\)

Activity 4.2.4 (~20 min)

Let \(T: \IR^3 \rightarrow \IR^3\) be the linear transformation given by the standard matrix \(A=\left[\begin{array}{ccc} 2 & -1 & -6 \\ 2 & 1 & 3 \\ 1 & 1 & 4 \end{array}\right]\text{.}\)

Part 4.

Write \(A^{-1}\text{,}\) the standard matrix for \(T^{-1}\text{.}\)

Observation 4.2.4

We could have solved these three systems simultaneously by row reducing the matrix \([A\,|\,I]\) at once.

\begin{equation*} \left[\begin{array}{ccc|ccc} 2 & -1 & -6 & 1 & 0 & 0 \\ 2 & 1 & 3 & 0 & 1 & 0 \\ 1 & 1 & 4 & 0 & 0 & 1 \end{array}\right] \sim \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & -2 & 3 \\ 0 & 1 & 0 & -5 & 14 & -18 \\ 0 & 0 & 1 & 1 & -3 & 4 \end{array}\right] \end{equation*}

Activity 4.2.5 (~5 min)

Find the inverse \(A^{-1}\) of the matrix \(A=\left[\begin{array}{cc} 1 & 3 \\ 0 & -2 \end{array}\right]\) by row-reducing \([A\,|\,I]\text{.}\)

Activity 4.2.6 (~5 min)

Is the matrix \(\left[\begin{array}{ccc} 2 & 3 & 1 \\ -1 & -4 & 2 \\ 0 & -5 & 5 \end{array}\right]\) invertible? Give a reason for your answer.

Observation 4.2.5

An \(n\times n\) matrix \(A\) is invertible if and only if \(\RREF(A) = I_n\text{.}\)

Activity 4.2.7 (~10 min)

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 \(T^{-1}\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)=\left[\begin{array}{c} 5x+ 3y \\ 3x + 2y\end{array}\right]\text{.}\)

Part 1.

Compute \((T^{-1}\circ T)\left(\left[\begin{array}{c}-2\\1\end{array}\right]\right)\text{.}\)

Activity 4.2.7 (~10 min)

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 \(T^{-1}\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)=\left[\begin{array}{c} 5x+ 3y \\ 3x + 2y\end{array}\right]\text{.}\)

Part 2.

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 4.2.6

\(T^{-1}\circ T=T\circ T^{-1}\) 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 \(A\text{.}\)