August 15, 2025
Complete the following warm-up problems to re-familiarize yourself with concepts we’ll be leveraging today.
Let \(A = \left[\begin{array}{rr} 3 & 1 & -2\\ 1 & 0 & 5\end{array}\right]\), \(B = \left[\begin{array}{rrr} -1 & 1 & 0\\ 3 & 4 & -2\\ 2 & 0 & 1\end{array}\right]\), and \(C = \left[\begin{array}{rr} 2 & 1\\ -3 & 0\\ 0 & 1\end{array}\right]\).
Let \(A = \left[\begin{array}{rr} 3 & -6\\ -1 & 2\end{array}\right]\), \(B = \left[\begin{array}{rr} -1 & 1\\ 3 & 4\end{array}\right]\), and \(C = \left[\begin{array}{rr} -3 & -5\\ 2 & 1\end{array}\right]\). Verify that \(AB = AC\) but \(B \neq C\).
Our discussion of linear algebra, so far, has taken us through the following contexts
Our discussion of linear algebra, so far, has taken us through the following contexts
We saw that these types of equations are all analogous to one another, and that they can be solved in the same way.
Our discussion of linear algebra, so far, has taken us through the following contexts
We saw that these types of equations are all analogous to one another, and that they can be solved in the same way.
Solution Strategy:
Our discussion of linear algebra, so far, has taken us through the following contexts
We saw that these types of equations are all analogous to one another, and that they can be solved in the same way.
This is a solid strategy, but it is tedious and isn’t reusable. Solving an equation for \(\vec{b_1}\) and \(\vec{b_2}\) has required us to perform the full row-reduction twice(\(^*\)).
Solution Strategy:
Today, we’ll be focused on the vector equation \(A\vec{x} = \vec{b}\).
Because we’ve been able to argue equivalency between matrix equations, vector equations, and linear systems though, everything we discuss here is applicable to those other contexts as well!
Goal for Today: Develop a reusable strategy which will make the work we do in solving \(A\vec{x} = \vec{b_1}\) subsequently beneficial if we wish to also solve \(A\vec{x} = \vec{b_2},~A\vec{x} = \vec{b_3},~\cdots\)
We’ve been focused for some time now on solving the matrix equation \(A\vec{x} = \vec{b}\) (and its equivalent variations).
None of the approaches we’ve utilized so far are as simple as the approach you used to solve equations like \(2x = 4\) – just divide both sides by \(2\).
Matrices have some properties that are not familiar to us:
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Example (i): \(5x = 35\)
\[\begin{align} 5x &= 35 \end{align}\]
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Example (i): \(5x = 35\)
\[\begin{align} 5x &= 35 \end{align}\]
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Example (i): \(5x = 35\)
\[\begin{align} 5x &= 35\\ \implies \frac{1}{5}\left(5x\right) &= \implies \frac{1}{5}\left(35\right) \end{align}\]
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Example (i): \(5x = 35\)
\[\begin{align} 5x &= 35\\ \implies \frac{1}{5}\left(5x\right) &= \implies \frac{1}{5}\left(35\right)\\ \implies \left(\frac{1}{5}\cdot 5\right)x &= 7 \end{align}\]
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Example (i): \(5x = 35\)
\[\begin{align} 5x &= 35\\ \implies \frac{1}{5}\left(5x\right) &= \implies \frac{1}{5}\left(35\right)\\ \implies \left(\frac{1}{5}\cdot 5\right)x &= 7\\ \implies 1x &= 7 \end{align}\]
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Example (i): \(5x = 35\)
\[\begin{align} 5x &= 35\\ \implies \frac{1}{5}\left(5x\right) &= \implies \frac{1}{5}\left(35\right)\\ \implies \left(\frac{1}{5}\cdot 5\right)x &= 7\\ \implies 1x &= 7\\ \implies x &= 7 \end{align}\]
Note: The scalar \(\frac{1}{5}\) is the multiplicative inverse of the scalar \(5\) since \(\frac{1}{5}\cdot 5 = 5\cdot \frac{1}{5} = 1\), where \(1\) is the multiplicative identity.
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Example (i): \(5x = 35\)
\[\begin{align} 5x &= 35\\ \implies \frac{1}{5}\left(5x\right) &= \implies \frac{1}{5}\left(35\right)\\ \implies \left(\frac{1}{5}\cdot 5\right)x &= 7\\ \implies 1x &= 7\\ \implies x &= 7 \end{align}\]
Summary: We can solve the scalar equation \(ax = b\) by multiplying both sides of the equation by the multiplicative inverse of \(a\), provided that it exists!
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Equation (ii): \(2x = 0\)
\[\begin{align} 2x&= 0 \end{align}\]
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Equation (ii): \(2x = 0\)
\[\begin{align} 2x&= 0\\ \implies \frac{1}{2}\left(2x\right) &= \frac{1}{2}\left(0\right) \end{align}\]
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Equation (ii): \(2x = 0\)
\[\begin{align} 2x&= 0\\ \implies \frac{1}{2}\left(2x\right) &= \frac{1}{2}\left(0\right)\\ \implies \left(\frac{1}{2}\cdot 2\right)x &= 0 \end{align}\]
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Equation (ii): \(2x = 0\)
\[\begin{align} 2x&= 0\\ \implies \frac{1}{2}\left(2x\right) &= \frac{1}{2}\left(0\right)\\ \implies \left(\frac{1}{2}\cdot 2\right)x &= 0\\ \implies 1x &= 0 \end{align}\]
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Equation (ii): \(2x = 0\)
\[\begin{align} 2x&= 0\\ \implies \frac{1}{2}\left(2x\right) &= \frac{1}{2}\left(0\right)\\ \implies \left(\frac{1}{2}\cdot 2\right)x &= 0\\ \implies 1x &= 0\\ \implies x &= 0 \end{align}\]
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Equation (iii): \(0x = 2\)…
We run into a little bit of a problem here.
There is no multiplicative inverse of \(0\).
However, we do know that the product \(0x\) for any value of \(x\) will be \(0\).
This equation has no solutions
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Equation (iv): \(0x = 0\)…
Again, we run into a problem.
There is no multiplicative inverse of \(0\), so we cannot find a solution.
Inspection however, shows us that any value of \(x\) will satisfy this equation.
Consider the following four equations.
\[(i)~~5x = 35~~~~~~~~(ii)~~2x = 0~~~~~~~~(iii)~~0x = 2~~~~~~~~(iv)~~0x = 0\]
Summary: The scalar equation \(ax = b\) has a unique solution if the multiplicative inverse of \(a\) (that is, \(\frac{1}{a}\)) exists. Otherwise, the equation either has no solutions, or infinitely many solutions.
Recall that linear transformations (a special class of function) correspond to matrix multiplication.
You might remember the following two properties that some functions satisfy.
One-to-One: A function \(f\) is said to be one-to-one if \(f(x_1) = f(x_2)\) implies that \(x_1 = x_2\). That is, for each output there is exactly one input being mapped to it.
Onto: A function \(f\) is said to be onto if for every \(b\) in the codomain of \(f\), there is some \(x\) in the domain such that \(f\left(x\right) = b\).
These properties are important because if a function is both one-to-one and onto, then that function is invertible.
We can argue that a linear transformation \(T\left(\vec{x}\right) = A\vec{x}\) is
one-to-one if the matrix \(A\) has a pivot in every column.
onto if the matrix \(A\) has a pivot in every row .
Summary/Theorem: The above properties and discussion argue that the linear transformation \(T\left(\vec{x}\right) = A\vec{x}\) is invertible if and only if the matrix \(A\) has a pivot in every row and every column. This means that matrix multiplication via the matrix \(A\) is invertible in these scenarios.
Note: Only square matrices can have a pivot in every row and every column. Thus only square matrices are invertible, but not every square matrix is invertible.
Definition (Matrix Inverse): Given an \(n\times n\) matrix \(A\), a matrix \(B\) such that \(AB = BA = I_n\) is said to be the inverse of \(A\). In this case, we write \(B = A^{-1}\).
Definition (Singular and Non-Singular Matrices): A matrix whose inverse exists is said to be a non-singular matrix, whereas a matrix whose inverse does not exist is said to to be singular.
Note that the following is a shortcut for computing the inverse of a \(2\times 2\) matrix, but will not work for larger matrices.
Computing the Inverse of a \(2\times 2\) Matrix: Let \(A = \left[\begin{array}{rr} a & b\\ c & d\end{array}\right]\) be a \(2\times 2\) matrix and assume \(ad - bc \neq 0\), then \(A\) is invertible and \(A^{-1} = \frac{1}{ad - bc}\left[\begin{array}{rr} d & -b\\ -c & a\end{array}\right]\).
Note: If \(A = \left[\begin{array}{rr} a & b\\ c & d\end{array}\right]\) and \(ad - bc = 0\), then \(A\) is singular and \(A^{-1}\) does not exist.
Preview: The quantity \(ad - bc\) is called the determinant of the \(2\times 2\) matrix \(A = \left[\begin{array}{rr} a &b\\ c &d\end{array}\right]\). We write this as \(\det\left(A\right) = ad - bc\). We’ll discuss determinants in greater detail later in our course.
Example: Compute the inverses of the following \(2\times 2\) matrices if they exist. If they do not exist, discuss why.
Theorem (Properties of Invertible Matrices): The following properties about matrices and their inverses are true.
The property in the first bullet point above allows us a new straegy for solving matrix equations \(A\vec{x} = \vec{b}\).
Strategy (Invertible Matrices and Solving \(A\vec{x} = \vec{b}\)): If the matrix \(A\) is invertible, then the matrix equation \(A\vec{x} = \vec{b}\) has a unique solution for all \(\vec{b}\). That unique solution is \(\vec{x} = A^{-1}\vec{b}\).
Example: Consider the matrix \(A = \left[\begin{array}{rr} 5 & -4\\ 1 & 2\end{array}\right]\) and the vectors \(\vec{b_1} = \left[\begin{array}{r} -2\\ 7\end{array}\right]\), \(\vec{b_2} = \left[\begin{array}{r} -15\\ -9\end{array}\right]\), \(\vec{b_3} = \left[\begin{array}{r} -6\\ 10\end{array}\right]\). Find \(A^{-1}\) and use it to solve each of the matrix equations \(A\vec{x} = \vec{b_i}\) for \(i = 1, 2, 3\).
Strategy (Constructing the Inverse of a Matrix): The theorem above gives us a strategy for constructing the inverse of a matrix as long as the inverse exists.
Example: Construct the inverse of the matrix \(A = \left[\begin{array}{rrr} 1 & 3 & -2\\ 0 & 5 & 4\\ 2 & 1 & 8\end{array}\right]\) if it exists.
Example: Construct the inverse of the matrix \(A = \left[\begin{array}{rrr} 1 & 3 & -2\\ 0 & 5 & 4\\ 2 & 1 & 8\end{array}\right]\) if it exists.
\[\begin{align*} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 2 & 1 & 8 & 0 & 0 & 1\end{array}\right] \end{align*}\]
Example: Construct the inverse of the matrix \(A = \left[\begin{array}{rrr} 1 & 3 & -2\\ 0 & 5 & 4\\ 2 & 1 & 8\end{array}\right]\) if it exists.
\[\begin{align*} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 2 & 1 & 8 & 0 & 0 & 1\end{array}\right] &\substack{R_3\leftarrow R_3 + (-2R_1)\\ \longrightarrow} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 0 & -5 & 12 & -2 & 0 & 1\end{array}\right] \end{align*}\]
Example: Construct the inverse of the matrix \(A = \left[\begin{array}{rrr} 1 & 3 & -2\\ 0 & 5 & 4\\ 2 & 1 & 8\end{array}\right]\) if it exists.
\[\begin{align*} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 2 & 1 & 8 & 0 & 0 & 1\end{array}\right] &\substack{R_3\leftarrow R_3 + (-2R_1)\\ \longrightarrow} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 0 & -5 & 12 & -2 & 0 & 1\end{array}\right]\\ &\substack{R_3\leftarrow R_3 + R_2\\ \longrightarrow} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 0 & 0 & 16 & -2 & 1 & 1\end{array}\right] \end{align*}\]
Example: Construct the inverse of the matrix \(A = \left[\begin{array}{rrr} 1 & 3 & -2\\ 0 & 5 & 4\\ 2 & 1 & 8\end{array}\right]\) if it exists.
\[\begin{align*} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 2 & 1 & 8 & 0 & 0 & 1\end{array}\right] &\substack{R_3\leftarrow R_3 + (-2R_1)\\ \longrightarrow} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 0 & -5 & 12 & -2 & 0 & 1\end{array}\right]\\ &\substack{R_3\leftarrow R_3 + R_2\\ \longrightarrow} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 0 & 0 & 16 & -2 & 1 & 1\end{array}\right]\\ &\substack{R_2\leftarrow (1/5)R_2\\ R_3\leftarrow (1/16)R_3\\ \longrightarrow} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 1 & 4/5 & 0 & 1/5 & 0\\ 0 & 0 & 1 & -1/8 & 1/16 & 1/16\end{array}\right]\\ \end{align*}\]
Example: Construct the inverse of the matrix \(A = \left[\begin{array}{rrr} 1 & 3 & -2\\ 0 & 5 & 4\\ 2 & 1 & 8\end{array}\right]\) if it exists.
\[\begin{align*} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 2 & 1 & 8 & 0 & 0 & 1\end{array}\right] &\sim \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 1 & 4/5 & 0 & 1/5 & 0\\ 0 & 0 & 1 & -1/8 & 1/16 & 1/16\end{array}\right] \end{align*}\]
Example: Construct the inverse of the matrix \(A = \left[\begin{array}{rrr} 1 & 3 & -2\\ 0 & 5 & 4\\ 2 & 1 & 8\end{array}\right]\) if it exists.
\[\begin{align*} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 2 & 1 & 8 & 0 & 0 & 1\end{array}\right] &\sim \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 1 & 4/5 & 0 & 1/5 & 0\\ 0 & 0 & 1 & -1/8 & 1/16 & 1/16\end{array}\right]\\ &\substack{R_2\leftarrow R_2 + (-4/5R_3)\\ \longrightarrow} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 1 & 0 & 1/10 & 3/20 & -1/20\\ 0 & 0 & 1 & -1/8 & 1/16 & 1/16\end{array}\right]\\ \end{align*}\]
Example: Construct the inverse of the matrix \(A = \left[\begin{array}{rrr} 1 & 3 & -2\\ 0 & 5 & 4\\ 2 & 1 & 8\end{array}\right]\) if it exists.
\[\begin{align*} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 2 & 1 & 8 & 0 & 0 & 1\end{array}\right] &\sim \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 1 & 4/5 & 0 & 1/5 & 0\\ 0 & 0 & 1 & -1/8 & 1/16 & 1/16\end{array}\right]\\ &\substack{R_2\leftarrow R_2 + (-4/5R_3)\\ \longrightarrow} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 1 & 0 & 1/10 & 3/20 & -1/20\\ 0 & 0 & 1 & -1/8 & 1/16 & 1/16\end{array}\right]\\ &\substack{R_1\leftarrow R_1 + 2R_3\\ \longrightarrow} \left[\begin{array}{rrr|rrr} 1 & 3 & 0 & 3/4 & 1/8 & 1/8\\ 0 & 1 & 0 & 1/10 & 3/20 & -1/20\\ 0 & 0 & 1 & -1/8 & 1/16 & 1/16\end{array}\right]\\ \end{align*}\]
Example: Construct the inverse of the matrix \(A = \left[\begin{array}{rrr} 1 & 3 & -2\\ 0 & 5 & 4\\ 2 & 1 & 8\end{array}\right]\) if it exists.
\[\begin{align*} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 2 & 1 & 8 & 0 & 0 & 1\end{array}\right] &\sim \left[\begin{array}{rrr|rrr} 1 & 3 & 0 & 3/4 & 1/8 & 1/8\\ 0 & 1 & 0 & 1/10 & 3/20 & -1/20\\ 0 & 0 & 1 & -1/8 & 1/16 & 1/16\end{array}\right]\\ \end{align*}\]
Example: Construct the inverse of the matrix \(A = \left[\begin{array}{rrr} 1 & 3 & -2\\ 0 & 5 & 4\\ 2 & 1 & 8\end{array}\right]\) if it exists.
\[\begin{align*} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 & 0 & 0\\ 0 & 5 & 4 & 0 & 1 & 0\\ 2 & 1 & 8 & 0 & 0 & 1\end{array}\right] &\sim \left[\begin{array}{rrr|rrr} 1 & 3 & 0 & 3/4 & 1/8 & 1/8\\ 0 & 1 & 0 & 1/10 & 3/20 & -1/20\\ 0 & 0 & 1 & -1/8 & 1/16 & 1/16\end{array}\right]\\ &\substack{R_1\leftarrow R_1 + (-3R_2)\\ \longrightarrow} \left[\begin{array}{rrr|rrr} 1 & 0 & 0 & 9/20 & -13/40 & 11/40\\ 0 & 1 & 0 & 1/10 & 3/20 & -1/20\\ 0 & 0 & 1 & -1/8 & 1/16 & 1/16\end{array}\right]\\ \end{align*}\]
So \(\boxed{~\displaystyle{A^{-1} = \left[\begin{array}{rrr} 9/20 & -13/40 & 11/40\\ 1/10 & 3/20 & -1/20\\ -1/8 & 1/16 & 1/16\end{array}\right]}~}\). \(_\blacktriangledown\)
We can also use Python’s {sympy}
(or {numpy}
) modules to calculate matrix inverses.
As usual, don’t jump to Python use until you are perfectly comfortable with the manual method for constructing the inverse of a matrix.
{sympy}
{sympy}
We can also use Python’s {sympy}
(or {numpy}
) modules to calculate matrix inverses.
As usual, don’t jump to Python use until you are perfectly comfortable with the manual method for constructing the inverse of a matrix.
We can also use Python’s {sympy}
(or {numpy}
) modules to calculate matrix inverses.
As usual, don’t jump to Python use until you are perfectly comfortable with the manual method for constructing the inverse of a matrix.
{sympy}
Note. When working with {numpy}
, you’ll need to ensure that your array/matrix contains elements stored as floats rather than integers. You can do this by including at least one element written as a float (ie. 1.0
) or by using the dtype
argument.
Example 1: Find the inverse of the matrix \(\left[\begin{array}{rrr} 1 & 0 & 1\\ -2 & 2 & 4\\ 0 & 6 & -4\end{array}\right]\).
Example 2: Solve the system \(\left\{\begin{array}{rlr} 2x_1 + 3x_2 & = & 8\\ -x_1 + 5x_2 & = & 9\end{array}\right.\) by constructing the corresponding matrix equation and left-multiplying both sides by the inverse of the coefficient matrix.
Example 3: Consider the matrix \(A = \left[\begin{array}{rr} 5 & -4\\ 1 & 2\end{array}\right]\) and the vectors \(\vec{b_1} = \left[\begin{array}{r} -2\\ 7\end{array}\right]\), \(\vec{b_2} = \left[\begin{array}{r} -15\\ -9\end{array}\right]\), \(\vec{b_3} = \left[\begin{array}{r} -6\\ 10\end{array}\right]\).
You solved \(A\vec{x} = \vec{b_i}\) earlier by finding and using \(A^{-1}\). You could have also solved using our old methods by row reducing the augmented matrix \(\left[\begin{array}{c|ccc} A & \vec{b_1} & \vec{b_2} & \vec{b_3}\end{array}\right]\). Discuss any advantages/disadvantages that you see.
\[\Huge{\text{Complete Homework 7}}\] \[\Huge{\text{on MyOpenMath}}\]
\(\Huge{\text{Bases and}}\)
\(\Huge{\text{Coordinate Systems}}\)