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Section 1.5 Matrix Form and Vector Form

A linear system can be written in some equivalent ways, which are useful in different contexts: a system of linear equations, can be recorded in matrix equation \(A\mathbf{x}=\mathbf{b}\text{,}\) we also introduce vector equation shortly. This section demonstrates how these three forms arise from one another and why being able to move fluently among them is a core skill in linear algebra.

Subsection 1.5.1 From Linear System to Matrix Equation

We have learned from previous sections that any linear system can be represented as a matrix equation. Let’s apply this knowledge to our linear system:
\begin{align*} 0x_1 + 3x_2 - 6x_3 + 6x_4 + 4x_5 &= -5\\ 3x_1 - 7x_2 + 8x_3 - 5x_4 + 8x_5 &= 9\\ 3x_1 - 9x_2 + 12x_3 - 9x_4 + 6x_5 &= 15 \end{align*}
Following the method we learned earlier, we can organize the coefficients into a coefficient matrix \(A\) and write this system as the matrix equation \(A\mathbf{x} = \mathbf{b}\text{:}\)
\begin{equation*} \begin{bmatrix} 0 & 3 & -6 & 6 & 4 \\ 3 & -7 & 8 & -5 & 8 \\ 3 & -9 & 12 & -9 & 6 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix} = \begin{bmatrix}-5\\9\\15\end{bmatrix}. \end{equation*}

Subsection 1.5.2 From Vector Equation to Linear System

Definition 1.5.1. Vector in \(\mathbb{R}^n\).

A vector in \(\mathbb{R}^n\) is an ordered list of \(n\) real numbers, written as a column:
\begin{equation*} \mathbf{v} = \begin{bmatrix}v_1\\v_2\\\vdots\\v_n\end{bmatrix}, \quad v_1, v_2, \ldots, v_n \in \mathbb{R}. \end{equation*}
The numbers \(v_1, \ldots, v_n\) are called the entries (or components) of \(\mathbf{v}\text{.}\) Two vectors are equal if and only if all their corresponding entries are equal.
Given two vectors \(\mathbf{u}, \mathbf{v} \in \mathbb{R}^n\) and a scalar \(c \in \mathbb{R}\text{,}\) we define:
  • Addition: \(\mathbf{u} + \mathbf{v} = \begin{bmatrix}u_1+v_1\\\vdots\\u_n+v_n\end{bmatrix}\)
  • Scalar multiplication: \(c\mathbf{v} = \begin{bmatrix}cv_1\\\vdots\\cv_n\end{bmatrix}\)
An expression of the form \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k\) is called a linear combination of the vectors \(\mathbf{v}_1, \ldots, \mathbf{v}_k\) with coefficients (or weights) \(c_1, \ldots, c_k\text{.}\)
Problem: Solve the vector equation
\begin{equation*} x_1\begin{bmatrix}0\\3\\3\end{bmatrix} + x_2\begin{bmatrix}3\\-7\\-9\end{bmatrix} + x_3\begin{bmatrix}-6\\8\\12\end{bmatrix} + x_4\begin{bmatrix}6\\-5\\-9\end{bmatrix} + x_5\begin{bmatrix}4\\8\\6\end{bmatrix} = \begin{bmatrix}-5\\9\\15\end{bmatrix}. \end{equation*}
Solution Process: To solve this vector equation, we need to find scalars \(x_1, x_2, x_3, x_4, x_5\) such that the linear combination of the given vectors equals the target vector.
Let’s write out what this vector equation means. The left side is:
\begin{equation*} x_1\begin{bmatrix}0\\3\\3\end{bmatrix} + x_2\begin{bmatrix}3\\-7\\-9\end{bmatrix} + x_3\begin{bmatrix}-6\\8\\12\end{bmatrix} + x_4\begin{bmatrix}6\\-5\\-9\end{bmatrix} + x_5\begin{bmatrix}4\\8\\6\end{bmatrix} = \begin{bmatrix}0x_1 + 3x_2 - 6x_3 + 6x_4 + 4x_5\\3x_1 - 7x_2 + 8x_3 - 5x_4 + 8x_5\\3x_1 - 9x_2 + 12x_3 - 9x_4 + 6x_5\end{bmatrix}. \end{equation*}
For this to equal \(\begin{bmatrix}-5\\9\\15\end{bmatrix}\text{,}\) we need:
\begin{equation*} \begin{bmatrix}0x_1 + 3x_2 - 6x_3 + 6x_4 + 4x_5\\3x_1 - 7x_2 + 8x_3 - 5x_4 + 8x_5\\3x_1 - 9x_2 + 12x_3 - 9x_4 + 6x_5\end{bmatrix} = \begin{bmatrix}-5\\9\\15\end{bmatrix}. \end{equation*}
Two vectors are equal if and only if their corresponding components are equal. This gives us the linear system:
\begin{align*} 0x_1 + 3x_2 - 6x_3 + 6x_4 + 4x_5 &= -5\\ 3x_1 - 7x_2 + 8x_3 - 5x_4 + 8x_5 &= 9\\ 3x_1 - 9x_2 + 12x_3 - 9x_4 + 6x_5 &= 15 \end{align*}
This is exactly the same linear system we solved in the previous section! We found that the general solution is:
\begin{equation*} \begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix} = \begin{bmatrix}-24\\-7\\0\\0\\4\end{bmatrix} + x_3\begin{bmatrix}2\\2\\1\\0\\0\end{bmatrix} + x_4\begin{bmatrix}-3\\-2\\0\\1\\0\end{bmatrix} \end{equation*}
where \(x_3\) and \(x_4\) are free parameters.
This reveals a fundamental connection: the matrix equation \(A\mathbf{x} = \mathbf{b}\) is equivalent to the vector equation \(x_1\mathbf{v}_1 + x_2\mathbf{v}_2 + \cdots + x_5\mathbf{v}_5 = \mathbf{b}\).

Three Equivalent Forms.

We have now seen three equivalent ways to represent the same mathematical problem:
  1. Vector Equation: \(x_1\mathbf{v}_1 + x_2\mathbf{v}_2 + x_3\mathbf{v}_3 + x_4\mathbf{v}_4 + x_5\mathbf{v}_5 = \mathbf{b}\)
  2. Linear System: A system of linear equations in the variables \(x_1, x_2, x_3, x_4, x_5\)
  3. Matrix Equation: \(A\mathbf{x} = \mathbf{b}\) where \(A\) is the coefficient matrix
These three forms are mathematically equivalent, meaning they represent exactly the same problem and have the same solution set. Each form provides its own important perspective and computational advantages. However, what is even more important than any individual form is understanding the connections between them. The ability to translate fluently among these three representations is a fundamental skill in linear algebra, as it allows us to choose the most appropriate viewpoint for any given problem and to leverage the strengths of each approach.

Subsection 1.5.3 One Problem, Three Languages

The three forms we have seen—linear system, matrix equation, and vector equation—are not three different problems. They are three languages for the same problem, each chosen because it makes certain things easier to see or to say:
  • Linear system: closest to everyday reasoning. Each equation is a constraint written in plain arithmetic, immediately interpretable as a balance or a condition on the unknowns. It is the natural starting point for building intuition.
  • Matrix equation \(A\mathbf{x}=\mathbf{b}\): compact and computable. It packages all coefficients into a single object and allows us to state, manipulate, and solve problems with a few symbols. Row reduction, for instance, operates directly on \([A\mid\mathbf{b}]\text{.}\) This form is built for calculation.
  • Vector equation: algebraic and geometric at once. Asking whether \(\mathbf{b}\) is a linear combination of the columns of \(A\) reframes a computation as a question about span, geometry, and structure. This language will carry us deep into the theory of vector spaces.
The fact that one underlying idea can be expressed in these three distinct yet equivalent forms is precisely what gives linear algebra its vitality. Each form illuminates a different facet of the same truth, and the interplay among them generates the subject’s richness. Whenever we are stuck in one language, we can switch to another. This flexibility—not any single formula—is the real power of the subject.