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

Steven Clontz, Drew Lewis

Section 4.3 Solving Systems with Matrix Inverses (MX3)

Subsection 4.3.1 Warm Up

Activity 4.3.1.

Which of the following matrices is invertible? Find the inverse for the one that is invertible.
  1. \(\displaystyle \left[\begin{matrix}1 & -1 & 0\\ -1 & 1 & 0\\ 1 & 0 & 1\end{matrix}\right]\)
  2. \(\displaystyle \left[\begin{matrix}1 & -1 & 3\\ -1 & 1 & -1\\ 1 & 0 & -2\end{matrix}\right]\)

Subsection 4.3.2 Class Activities

Activity 4.3.2.

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 4.3.3.

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 4.3.4.

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 4.3.3 Cool Down

Activity 4.3.5.

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?
  1. Calculate \(\RREF[A|\vec{b}]\text{.}\)
  2. Calculate \(A^{-1}\) and then compute \(\vec{x}=A^{-1}\vec{b}\)
(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?
  1. Calculate \(\RREF[A|\vec{b}_1]\text{,}\) \(\RREF[A|\vec{b}_2]\text{,}\) and \(\RREF[A|\vec{b}_3]\)
  2. 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?
  1. Calculate \(\RREF[A|\vec{b}_1]\text{,}\) ... \(\RREF[A|\vec{b}_{10}]\text{.}\)
  2. Calculate \(A^{-1}\) and then compute \(\vec{x}=A^{-1}\vec{b}_1\text{,}\) ... \(\vec{x}=A^{-1}\vec{b}_{10}\text{.}\)

Subsection 4.3.4 Videos

Video coming soon to this YouTube playlist 1 .

Exercises 4.3.5 Exercises

Subsection 4.3.6 Mathematical Writing Explorations

Exploration 4.3.6.

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 4.3.7.

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 4.3.8.

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

Exploration 4.3.9.

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

Exploration 4.3.10.

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

Subsection 4.3.7 Sample Problem and Solution

Sample problem Example B.1.20.
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