Problem 24
Question
Compute ax for each vector \(\mathbf{x}\) and scalar \(a\). Represent \(\mathbf{x}\) and ax in the plane, and explain graphically how you obtain \(a \mathrm{x}\) \(\mathbf{x}=\left[\begin{array}{r}3 \\ -1\end{array}\right]\) and \(a=-1\)
Step-by-Step Solution
Verified Answer
The vector \( a\mathbf{x} \) is \( \left[ \begin{array}{r} -3 \\ 1 \end{array} \right] \), which is a reflection of \( \mathbf{x} \) through the origin.
1Step 1: Understand Vector and Scalar Multiplication
We have a vector \( \mathbf{x} = \left[ \begin{array}{r} 3 \ -1 \end{array} \right] \) and a scalar \( a = -1 \). We need to compute the product \( a\mathbf{x} \) by multiplying each component of the vector \( \mathbf{x} \) by the scalar \( a \).
2Step 2: Multiply Each Component
To find \( a\mathbf{x} \), multiply each component of the vector \( \mathbf{x} \) by \( a \). Thus, the first component: \( 3 \times (-1) = -3 \), and the second component: \( -1 \times (-1) = 1 \). The resulting vector is \( \mathbf{ax} = \left[ \begin{array}{r} -3 \ 1 \end{array} \right] \).
3Step 3: Graphical Representation
Plot the vector \( \mathbf{x} \) in a coordinate plane starting from the origin \((0,0)\) and extending to \((3, -1)\). Next, plot the vector \( a\mathbf{x} = \left[ \begin{array}{r} -3 \ 1 \end{array} \right] \) by starting from the origin and extending to \((-3, 1)\). The transformation from \( \mathbf{x} \) to \( a\mathbf{x} \) is essentially a reflection through the origin, since the scalar \( a = -1 \) reverses the direction of each component of \( \mathbf{x} \).
4Step 4: Interpretation of the Graphical Transformation
This multiplication demonstrates that a vector multiplied by a scalar \( a = -1 \) results in the vector pointing in the opposite direction. On the graph, this effect is visible as each vertex of \( \mathbf{x} \) being reflected through the origin to become the corresponding vertex of \( a\mathbf{x} \).
Key Concepts
Scalar MultiplicationGraphical Representation of VectorsVector Transformation
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar. A vector has both a direction and magnitude, represented with components, such as in the vector \( \mathbf{x} = \left[ \begin{array}{r} 3 \ -1 \end{array} \right] \). A scalar is simply a single number, like \( a = -1 \). When you perform scalar multiplication, you apply the scalar to each component of the vector separately. This means we multiply each element in the vector by the scalar. For instance:
- First component: \( 3 \times (-1) = -3 \)
- Second component: \( -1 \times (-1) = 1 \)
Graphical Representation of Vectors
Understanding how vectors look in a plane helps us grasp vector operations, like scalar multiplication, visually. We can think of vectors as arrows pointing from the origin in the direction defined by their components. For the vector \( \mathbf{x} = \left[ \begin{array}{r} 3 \ -1 \end{array} \right] \), we draw an arrow starting from the origin \( (0,0) \) and ending at the point \( (3, -1) \).
- The tail of the vector is always at the origin.
- The head of the vector points to its components.
Vector Transformation
Scalar multiplication can transform vectors in several ways, depending on the scalar's value. For example, multiplying by \( a = -1 \) inverts the vector. Let's consider how this transformation occurs:
- Direction: The vector's direction is reversed with negative scalars. In our example, \( \mathbf{x} = \left[ \begin{array}{r} 3 \ -1 \end{array} \right] \), transformed to \( \mathbf{ax} = \left[ \begin{array}{r} -3 \ 1 \end{array} \right] \), flips direction.
- Magnitude: The length of the vector remains the same when multiplying by -1, which reflects the vector across the origin. However, other scalars scale vectors proportionally.
Other exercises in this chapter
Problem 23
In Problems 19-24, solve each system of linear equations. $$ \begin{array}{l} 2 x-y+3 z=3 \\ 2 x+y+4 z=4 \\ 2 x-3 y+2 z=2 \end{array} $$
View solution Problem 24
Find the angle between \(\mathbf{x}=[-1,2]^{\prime}\) and \(\mathbf{y}=[-2,4]\).
View solution Problem 24
Let $$A=\left[\begin{array}{rr} -1 & 0 \\ 1 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & 0 \\ -1 & -1 \end{array}\right], \quad C=\left[\begin{ar
View solution Problem 24
In Problems 19-24, solve each system of linear equations. $$ \begin{array}{r} 2 x+y-2 z=3 \\ 2 x-3 y-4 z=0 \\ x-5 y+3 z=-6 \end{array} $$
View solution