In Exercises 7 And 8, Determine If The Vector B B B Is In The Span Of The Columns Of The Matrix A A A .7. $A=\begin{bmatrix}1 & 2 \ 3 & 4\end{bmatrix}, \quad B =\begin{bmatrix}5 \ 6\end{bmatrix}$8. $A=\begin{bmatrix}1 & 2 &

In Exercises 7 and 8, Determine if the Vector $b$ is in the Span of the Columns of the Matrix $A$

In linear algebra, the concept of span is crucial in understanding the relationship between vectors and matrices. The span of a set of vectors is the set of all linear combinations of those vectors. In this article, we will explore whether a given vector $b$ is in the span of the columns of a matrix $A$. We will use two exercises to illustrate this concept.

In this exercise, we are given a matrix $A$ and a vector $b$. We need to determine if $b$ is in the span of the columns of $A$.

Matrix $A$

The matrix $A$ is given by:

$$A=\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}$$

Vector $b$

The vector $b$ is given by:

$$b=\begin{bmatrix}5 \\ 6\end{bmatrix}$$

Determining if $b$ is in the Span of $A$

To determine if $b$ is in the span of $A$, we need to find out if there exists a linear combination of the columns of $A$ that equals $b$. In other words, we need to find scalars $x_1$ and $x_2$ such that:

$$x_1\begin{bmatrix}1 \\ 3\end{bmatrix} + x_2\begin{bmatrix}2 \\ 4\end{bmatrix} = \begin{bmatrix}5 \\ 6\end{bmatrix}$$

This can be written as a system of linear equations:

$$x_1 + 2x_2 = 5$$

$$3x_1 + 4x_2 = 6$$

We can solve this system using substitution or elimination. Let's use elimination.

Solving the System of Linear Equations

We can multiply the first equation by $-3$ and add it to the second equation to eliminate $x_1$:

$$-3(x_1 + 2x_2) = -3(5)$$

$$3x_1 + 4x_2 = 6$$

Adding the two equations, we get:

$$x_2 = -3$$

Now that we have found $x_2$, we can substitute it into one of the original equations to find $x_1$. Let's use the first equation:

$$x_1 + 2x_2 = 5$$

$$x_1 + 2(-3) = 5$$

$$x_1 = 11$$

Therefore, we have found scalars $x_1 = 11$ and $x_2 = -3$ such that:

$$11\begin{bmatrix}1 \\ 3\end{bmatrix} - 3\begin{bmatrix}2 \\ 4\end{bmatrix} = \begin{bmatrix}5 \\ 6\end{bmatrix}$$

This means that $b$ is in the span of $A$.

In this exercise, we are given a matrix $A$ and a vector $b$. We need to determine if $b$ is in the span of the columns of $A$.

Matrix $A$

The matrix $A$ is given by:

$$A=\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6\end{bmatrix}$$

Vector $b$

The vector $b$ is given by:

$$b=\begin{bmatrix}7 \\ 8\end{bmatrix}$$

Determining if $b$ is in the Span of $A$

To determine if $b$ is in the span of $A$, we need to find out if there exists a linear combination of the columns of $A$ that equals $b$. In other words, we need to find scalars $x_1$, $x_2$, and $x_3$ such that:

$$x_1\begin{bmatrix}1 \\ 4\end{bmatrix} + x_2\begin{bmatrix}2 \\ 5\end{bmatrix} + x_3\begin{bmatrix}3 \\ 6\end{bmatrix} = \begin{bmatrix}7 \\ 8\end{bmatrix}$$

This can be written as a system of linear equations:

$$x_1 + 2x_2 + 3x_3 = 7$$

$$4x_1 + 5x_2 + 6x_3 = 8$$

We can solve this system using substitution or elimination. Let's use elimination.

Solving the System of Linear Equations

We can multiply the first equation by $-4$ and add it to the second equation to eliminate $x_1$:

$$-4(x_1 + 2x_2 + 3x_3) = -4(7)$$

$$4x_1 + 5x_2 + 6x_3 = 8$$

Adding the two equations, we get:

$$x_2 + 3x_3 = -20$$

Now that we have found $x_2 + 3x_3$, we can substitute it into one of the original equations to find $x_1$. Let's use the first equation:

$$x_1 + 2x_2 + 3x_3 = 7$$

$$x_1 + 2(-20 - 3x_3) + 3x_3 = 7$$

$$x_1 - 40 - 6x_3 + 3x_3 = 7$$

$$x_1 - 43x_3 = 47$$

We can solve for $x_1$ in terms of $x_3$:

$$x_1 = 47 + 43x_3$$

Now that we have found $x_1$ in terms of $x_3$, we can substitute it into one of the original equations to find $x_2$. Let's use the second equation:

$$4x_1 + 5x_2 + 6x_3 = 8$$

$$4(47 + 43x_3) + 5x_2 + 6x_3 = 8$$

$$188 + 172x_3 + 5x_2 + 6x_3 = 8$$

$$5x_2 + 178x_3 = -180$$

We can solve for $x_2$ in terms of $x_3$:

$$x_2 = -180 - 178x_3$$

Therefore, we have found scalars $x_1 = 47 + 43x_3$, $x_2 = -180 - 178x_3$, and $x_3$ such that:

$$(47 + 43x_3)\begin{bmatrix}1 \\ 4\end{bmatrix} + (-180 - 178x_3)\begin{bmatrix}2 \\ 5\end{bmatrix} + x_3\begin{bmatrix}3 \\ 6\end{bmatrix} = \begin{bmatrix}7 \\ 8\end{bmatrix}$$

However, we notice that the system of linear equations has no solution. This means that $b$ is not in the span of $A$.

In this article, we have explored whether a given vector $b$ is in the span of the columns of a matrix $A$. We have used two exercises to illustrate this concept. In Exercise 7, we found that $b$ is in the span of $A$. In Exercise 8, we found that $b$ is not in the span of $A$. This demonstrates the importance of understanding the concept of span in linear algebra.

• [1] Strang, G. (1988). Linear Algebra and Its Applications. 3rd ed. Wellesley-Cambridge Press.
• [2] Lay, D. C. (2012). Linear Algebra and Its Applications. 4th ed. Addison-Wesley.
• [3] Anton, H. (2013). Elementary Linear Algebra. 10th ed. Wiley.

For further reading on linear algebra, we recommend the following resources:

• [1] Khan Academy: Linear Algebra
• [2] MIT OpenCourseWare: Linear Algebra
• [3] Wolfram MathWorld: Linear Algebra
  Q&A: Understanding the Span of a Matrix

In our previous article, we explored whether a given vector $b$ is in the span of the columns of a matrix $A$. We used two exercises to illustrate this concept. In this article, we will answer some frequently asked questions about the span of a matrix.

A: The span of a matrix is the set of all linear combinations of its columns. In other words, it is the set of all vectors that can be expressed as a linear combination of the columns of the matrix.

$$$$

A: To determine if a vector $b$ is in the span of a matrix $A$, you need to find out if there exists a linear combination of the columns of $A$ that equals $b$. In other words, you need to find scalars $x_1$, $x_2$, ..., $x_n$ such that:

$$x_1\begin{bmatrix}a_{11} \\ a_{21} \\ \vdots \\ a_{n1}\end{bmatrix} + x_2\begin{bmatrix}a_{12} \\ a_{22} \\ \vdots \\ a_{n2}\end{bmatrix} + \cdots + x_n\begin{bmatrix}a_{1n} \\ a_{2n} \\ \vdots \\ a_{nn}\end{bmatrix} = \begin{bmatrix}b_1 \\ b_2 \\ \vdots \\ b_n\end{bmatrix}$$

This can be written as a system of linear equations:

$$a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = b_1$$

$$a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = b_2$$

$$\vdots$$

$$a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}x_n = b_n$$

You can solve this system using substitution or elimination.

A: If the system of linear equations has no solution, then the vector $b$ is not in the span of the matrix $A$.

A: If the system of linear equations has multiple solutions, then the vector $b$ is in the span of the matrix $A$, but it is not unique.

A: Yes, a matrix can have a span of zero. This means that the only vector in the span of the matrix is the zero vector.

A: Yes, a matrix can have a span of all vectors. This means that every vector in the vector space is in the span of the matrix.

In this article, we have answered some frequently asked questions about the span of a matrix. We hope that this article has helped you to understand the concept of span and how to determine if a vector is in the span of a matrix.

• [1] Strang, G. (1988). Linear Algebra and Its Applications. 3rd ed. Wellesley-Cambridge Press.
• [2] Lay, D. C. (2012). Linear Algebra and Its Applications. 4th ed. Addison-Wesley.
• [3] Anton, H. (2013). Elementary Linear Algebra. 10th ed. Wiley.

For further reading on linear algebra, we recommend the following resources:

• [1] Khan Academy: Linear Algebra
• [2] MIT OpenCourseWare: Linear Algebra
• [3] Wolfram MathWorld: Linear Algebra