Problem 25
Question
In Problems , compute a \(\mathbf{x}\) for the given vector \(\mathbf{x}\) and scalar a. Represent \(\mathbf{x}\) and a \(\mathbf{x}\) in the plane, and explain graphically how you obtain \(a \mathbf{x}\). $$ \mathbf{x}=\left[\begin{array}{r} 0 \\ -2 \end{array}\right] \text { and } a=0.5 $$
Step-by-Step Solution
Verified Answer
\(a \mathbf{x} = \left[\begin{array}{r} 0 \\ -1 \end{array}\right]\); it is half the length of \(\mathbf{x}\).
1Step 1: Scalar Multiplication
To find \(a \mathbf{x}\), we will multiply each component of the vector \(\mathbf{x}\) by the scalar \(a\). The vector \(\mathbf{x}\) is \(\left[\begin{array}{r} 0 \ -2 \end{array}\right]\) and the scalar is \(a = 0.5\). Thus, \(a \mathbf{x} = 0.5 \times \left[\begin{array}{r} 0 \ -2 \end{array}\right] = \left[\begin{array}{r} 0 \times 0.5 \ -2 \times 0.5 \end{array}\right] = \left[\begin{array}{r} 0 \ -1 \end{array}\right]\).
2Step 2: Plotting the Original Vector
The vector \(\mathbf{x} = \left[\begin{array}{r} 0 \ -2 \end{array}\right]\) can be represented in the coordinate plane starting from the origin \((0,0)\) and ending at the point \((0, -2)\). This represents a vertical line going 2 units down from the origin on the y-axis.
3Step 3: Plotting the Scaled Vector
The vector \(a \mathbf{x} = \left[\begin{array}{r} 0 \ -1 \end{array}\right]\) is plotted similarly, starting from the origin \((0,0)\) and ending at the point \((0, -1)\). This is also a vertical line, but it ends only 1 unit down from the origin, showing that the vector has been scaled down by a factor of 0.5.
4Step 4: Graphical Explanation
By comparing \(\mathbf{x}\) and \(a \mathbf{x}\), we see that scaling a vector by a scalar affects its length. The original vector \(\mathbf{x}\) extended to \(y = -2\), while the scaled vector \(a \mathbf{x}\) reaches only \(y = -1\). This visually confirms that scaling \(\mathbf{x}\) by 0.5 halves its length.
Key Concepts
Scalar MultiplicationCoordinate PlaneGraphical Representation
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar value. In this process, each component of the vector is multiplied by the scalar. This operation modifies the magnitude of the vector, but not its direction.
Using our example, the vector \( \mathbf{x} = \left[ \begin{array}{r} 0 \ -2 \end{array} \right] \) is scaled by \( a = 0.5 \). We compute each component:
This example shows that scalar multiplication reduces the vector’s length if the scalar is between 0 and 1, effectively shrinking it.
Using our example, the vector \( \mathbf{x} = \left[ \begin{array}{r} 0 \ -2 \end{array} \right] \) is scaled by \( a = 0.5 \). We compute each component:
- The first component, 0, remains unchanged as \( 0 \times 0.5 = 0 \).
- The second component becomes \( -2 \times 0.5 = -1 \).
This example shows that scalar multiplication reduces the vector’s length if the scalar is between 0 and 1, effectively shrinking it.
Coordinate Plane
The coordinate plane is a two-dimensional space where vectors can be visualized graphically. It's defined by the x-axis (horizontal) and y-axis (vertical), intersecting at the origin \(0, 0\). Vectors are represented as arrows from this origin to a specific point.
For our vector \( \mathbf{x} = \left[ \begin{array}{r} 0 \ -2 \end{array} \right] \), it starts at the origin and points to the position (0, -2). This means \( \mathbf{x} \) moves downward parallel to the y-axis.
The scaled vector \( a \mathbf{x} = \left[ \begin{array}{r} 0 \ -1 \end{array} \right] \) also originates from (0,0) but ends at (0, -1). Both vectors lie on the y-axis, visually differing only in length.
For our vector \( \mathbf{x} = \left[ \begin{array}{r} 0 \ -2 \end{array} \right] \), it starts at the origin and points to the position (0, -2). This means \( \mathbf{x} \) moves downward parallel to the y-axis.
The scaled vector \( a \mathbf{x} = \left[ \begin{array}{r} 0 \ -1 \end{array} \right] \) also originates from (0,0) but ends at (0, -1). Both vectors lie on the y-axis, visually differing only in length.
Graphical Representation
Graphical representation of vectors helps us to see changes due to scalar multiplication. By comparing the lengths and directions on the graph, we gain insight into vector operations.
Our initial vector \( \mathbf{x} = \left[ \begin{array}{r} 0 \ -2 \end{array} \right] \) is plotted as a line extending 2 units downward. The scaled vector \( a \mathbf{x} = \left[ \begin{array}{r} 0 \ -1 \end{array} \right] \) shortens to only 1 unit, still pointing downward.
Our initial vector \( \mathbf{x} = \left[ \begin{array}{r} 0 \ -2 \end{array} \right] \) is plotted as a line extending 2 units downward. The scaled vector \( a \mathbf{x} = \left[ \begin{array}{r} 0 \ -1 \end{array} \right] \) shortens to only 1 unit, still pointing downward.
- This visual comparison shows the scalar \( a = 0.5 \) effectively halved the vector's size.
- Both vectors maintain their direction, demonstrating that scalar multiplication changes the magnitude but not the path.
Other exercises in this chapter
Problem 25
Find the augmented matrix and use it to solve the system of linear equations. $$ \begin{aligned} -x-2 y+3 z &=-9 \\ 2 x+y-z &=5 \\ 4 x-3 y+5 z &=-9 \end{aligned
View solution Problem 25
Find the angle between \(\mathbf{x}=[0,-1,3]^{\prime}\) and \(\mathbf{y}=[-3,1,1]^{\prime}\).
View solution Problem 26
$$ A=\left[\begin{array}{rr} -1 & 0 \\ 1 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & 3 \\ -1 & 1 \end{array}\right], \quad C=\left[\begin{array}
View solution Problem 26
Find the augmented matrix and use it to solve the system of linear equations. $$ \begin{array}{l} 3 x-2 y+z=4 \\ 4 x+y-2 z=-12 \\ 2 x-3 y+z=7 \end{array} $$
View solution