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Linear Algebra for Team-Based Inquiry Learning: 2024 Early Edition

Steven Clontz, Drew Lewis

Section 4.3 Solving Systems with Matrix Inverses (MX3)

Subsection 4.3.1 Class Activities

Activity 4.3.1.

Consider the following linear system with a unique solution:
\begin{equation*} \begin{matrix} 3x_{1} & - & 2x_{2} & - & 2x_{3} & - & 4x_{4} & = & -7 \\ 2x_{1} & - & x_{2} & - & x_{3} & - & x_{4} & = & -1 \\ -x_{1} & & & + & x_{3} & & & = & -1 \\ & - & x_{2} & & & - & 2x_{4} & = & -5 \\ \end{matrix} \end{equation*}
(a)
Suppose we let
\begin{equation*} T\left(\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]\right)= \left[\begin{matrix} 3x_{1} & - & 2x_{2} & - & 2x_{3} & - & 4x_{4} \\ 2x_{1} & - & x_{2} & - & x_{3} & - & x_{4} \\ -x_{1} & & & + & x_{3} & & \\ & - & x_{2} & & & - & 2x_{4} \\ \end{matrix}\right]\text{.} \end{equation*}
Which of these choices would help us solve the given system?
  1. Compute \(T\left( \left[\begin{matrix} -7 \\ -1 \\ -1 \\ -5 \\ \end{matrix}\right]\right)\)
  2. Find \(\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]\) where \(T\left(\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]\right)= \left[\begin{matrix} -7 \\ -1 \\ -1 \\ -5 \\ \end{matrix}\right]\)
(b)
How can we express this in terms of matrix multiplication?
  1. \(\displaystyle \left[\begin{matrix} 3 & -2 & -2 & - 4 \\ 2 & -1 & -1 & -1 \\ -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & -2 \\ \end{matrix}\right] \left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right] = \left[\begin{matrix} -7 \\ -1 \\ -1 \\ -5 \\ \end{matrix}\right]\)
  2. \(\displaystyle \left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right] = \left[\begin{matrix} 3 & -2 & -2 & - 4 \\ 2 & -1 & -1 & -1 \\ -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & -2 \\ \end{matrix}\right] \left[\begin{matrix} -7 \\ -1 \\ -1 \\ -5 \\ \end{matrix}\right]\)
  3. \(\displaystyle \left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right] \left[\begin{matrix} 3 & -2 & -2 & - 4 \\ 2 & -1 & -1 & -1 \\ -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & -2 \\ \end{matrix}\right] = \left[\begin{matrix} -7 \\ -1 \\ -1 \\ -5 \\ \end{matrix}\right]\)
  4. \(\displaystyle \left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right] = \left[\begin{matrix} -7 \\ -1 \\ -1 \\ -5 \\ \end{matrix}\right] \left[\begin{matrix} 3 & -2 & -2 & - 4 \\ 2 & -1 & -1 & -1 \\ -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & -2 \\ \end{matrix}\right]\)
(c)
How could a matrix equation of the form \(A\vec x=\vec b\) be solved for \(\vec x\text{?}\)
  1. Multiply: \((\RREF A)(A\vec x)=(\RREF A)\vec b\)
  2. Add: \((\RREF A) + A\vec x=(\RREF A)+\vec b\)
  3. Multiply: \((A^{-1})(A\vec x)=(A^{-1})\vec b\)
  4. Add: \((A^{-1}) + A\vec x=(A^{-1})+\vec b\)
(d)
Find \(\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]\) using the method you chose in (c).

Remark 4.3.2.

The linear system described by the augmented matrix \([A \mid \vec b]\) has exactly the same solution set as the matrix equation \(A\vec x=\vec b\text{.}\)
When \(A\) is invertible, then we have both \([A \mid \vec b]\sim[I \mid \vec x]\) and \(\vec x=A^{-1}\vec b\text{,}\) which can be seen as
\begin{align*} && A\vec x&=\vec b\\ &\Rightarrow & A^{-1}A\vec x&=A^{-1}\vec b\\ &\Rightarrow &\vec x&=A^{-1}\vec b \end{align*}

Activity 4.3.3.

Consider the vector equation
\begin{equation*} x_{1} \left[\begin{array}{c} 1 \\ 2 \\ -2 \end{array}\right] + x_{2} \left[\begin{array}{c} -2 \\ -3 \\ 3 \end{array}\right] + x_{3} \left[\begin{array}{c} 1 \\ 4 \\ -3 \end{array}\right] = \left[\begin{array}{c} -3 \\ 5 \\ -1 \end{array}\right] \end{equation*}
with a unique solution.
(a)
Explain and demonstrate how this problem can be restated using matrix multiplication.
(b)
Use the properties of matrix multiplication to find the unique solution.

Subsection 4.3.2 Videos

Video coming soon to this YouTube playlist 1 .

Exercises 4.3.3 Exercises

Subsection 4.3.4 Mathematical Writing Explorations

Exploration 4.3.4.

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 4.3.5.

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 4.3.6.

If \(H\) is invertible and \(r \in \mathbb{R}\text{,}\) what is the inverse of \(rH\text{?}\)

Exploration 4.3.7.

If \(H\) and \(G\) are invertible, is \(H^{-1} + G^{-1} = (H+G)^{-1}\text{?}\)

Exploration 4.3.8.

Prove that if \(A\text{,}\) \(P\text{,}\) and \(Q\) are invertible with \(PAQ = I\text{,}\) then \(A^{-1} = QP\text{.}\)

Subsection 4.3.5 Sample Problem and Solution

Sample problem Example B.1.20.
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