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Observe that each component of the product vector corresponds to one of the equations in the system. Let \(\vec {b}=\begin{bmatrix}5\\1\\-4\end{bmatrix}\).
Then
This shows that the ordered pair \((2, -1)\) is a solution to the system. We conclude that \(\vec {x}=\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\-1\end{bmatrix}\) is a solution to the matrix equation in 1. A
quick verification confirms this
One way to obtain a solution is to convert this to a system of equations. It is not necessary to write the system down, but it
helps to think about it as you write out your solution vector.
We see that \(x_1\) and \(x_2\) are leading variables because they correspond to leading 1s in the reduced row-echelon form , while \(x_3\)
and \(x_4\) are free variables. We start by assigning parameters \(s\) and \(t\) to \(x_3\) and \(x_4\), respectively, then solve for \(x_1\) and \(x_2\).
The solution given in (1) is an example of a general solution because it accounts for all of the solutions to the system. Letting \(s\)
and \(t\) take on specific values produces particular solutions. For example, \(\begin{bmatrix}2\\-1\\1\\-1\end{bmatrix}\) is a particular solution that corresponds to \(s=1\),
\(t=-1\).
Singular and Nonsingular Matrices
Our examples so far involved non-square matrices. Square matrices, however, play a very important role in linear algebra.
This section will focus on square matrices.
We can immediately see that the solution vector is
\[\vec {x}=\begin{bmatrix}0\\-1\\1\end{bmatrix}\]
Observe that the left-hand side of the augmented matrix in Example 5 is the identity matrix \(I\). This means that
\(\mbox {rref}(A)=I\).
The elementary row operations that carried \(A\) to \(I\) were not dependent on the vector \(\vec {b}\). In fact, the same row reduction process
can be applied to the matrix equation \(A\vec {x}=\vec {b}\) for any vector \(\vec {b}\) to obtain a unique solution.
Given a matrix \(A\) such that \(\mbox {rref}(A)=I\), the system \(A\vec {x}=\vec {b}\) will never be inconsistent because we will never have a row like this: \(\left [\begin{array}{cccc|c} 0&0&\ldots& 0&1 \end{array}\right ]\). Neither will we
have infinitely many solutions because there will never be free variables. Matrices such as \(A\) deserve special
attention.
A square matrix \(A\) is said to be nonsingular provided that \(\mbox {rref}(A)=I\). Otherwise we say that \(A\) is singular.
Non-singular matrices have many useful properties.
The following statements are equivalent for an \(n\times n\) matrix \(A\).
1.
\(A\) is nonsingular
2.
\(A\vec {x}=\vec {b}\) has a unique solution for any \(\vec {b}\) in \(\RR ^n\)
3.
\(A\vec {x}=\vec {0}\) has only the trivial solution \(\vec {x}=\vec {0}\)
We will prove equivalence of the three statements by showing that
Proof of 1\(\Rightarrow \)2 Suppose \(\mbox {rref}(A)=I\). Given any vector \(\vec {b}\) in \(\RR ^n\), the augmented matrix \([A|\vec {b}]\) can be carried to its reduced row-echelon form \([I|\vec {b}^*]\).
Uniqueness of the reduced row-echelon form guarantees that \(\vec {b}^*\) is the unique solution of \(A\vec {x}=\vec {b}\).
Proof of 2\(\Rightarrow \)3 Suppose \(A\vec {x}=\vec {b}\) has a unique solution for all vectors \(\vec {b}\). Then \(A\vec {x}=\vec {0}\) has a unique solution. But \(\vec {x}=\vec {0}\) is always a solution to \(A\vec {x}=\vec {0}\).
Therefore \(\vec {x}=\vec {0}\) is the only solution.
Proof of 3\(\Rightarrow \)1 Suppose \(A\vec {x}=\vec {0}\) has only the trivial solution. This means that \(x_1=0, x_2=0,\dots ,x_n=0\) is the only solution of \(A\vec {x}=\vec {0}\). But then, we know that the
augmented matrix \([A|\vec {0}]\) can be reduced to \([I|\vec {0}]\). The same row operations will carry \(A\) to \(I\).
Not all square matrices are nonsingular. For example,
By Theorem 8, a matrix equation \(A\vec {x}=\vec {b}\) involving a singular matrix \(A\) cannot have a unique solution. The following example illustrates
the two scenarios that arise when solving equations that involve singular matrices.
2 When the vector \(\vec {b}\) is changed, the row operations that take \(A\) to its reduced row-echelon form produce a \(1\) in the last row of the
vector on the right, which shows that the system is inconsistent.
For each given matrix \(A\) and vector \(\vec {b}\), determine whether \(\vec {b}\) is a linear combination of the columns of \(A\). If possible, express \(\vec {b}\) as a
linear combination of the columns of \(A\).
Solving this equation amounts to finding \(\vec {x}=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\) such that \(A\vec {x}=\vec {b}\). The augmented matrix corresponding to this equation, together with its
reduced row-echelon form are
So, \(\vec {x}=\begin{bmatrix} 1\\-1\\2\end{bmatrix}\) is a solution to the matrix equation. We conclude that \(\vec {b}\) is a linear combination of the columns of \(A\), and
write
2 We begin by attempting to solve the matrix equation \(A\vec {x}=\vec {b}\). The augmented matrix corresponding to this equation, together with
its reduced row-echelon form are
Use an augmented matrix and elementary row operations to find coefficients \(x_1\) and \(x_2\) that make the expression true, or
demonstrate that such coefficients do not exist.
If coefficients \(x_1\) and \(x_2\) do not exist, enter NA in each answer box.
\[x_1=\answer {-1}, x_2=\answer {2}\]
Use an augmented matrix and elementary row operations to find coefficients \(x_1\) and \(x_2\) that make the expression true, or
demonstrate that such coefficients do not exist.
