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Instructions:
Matrices and vectors are fundamental concepts in linear algebra, which is a branch of mathematics that deals with linear equations, linear transformations, and vector spaces.
A matrix is a rectangular array of numbers or other mathematical objects, such as complex numbers or functions. Matrices are usually denoted by capital letters and are often used to represent systems of linear equations, transformations of geometric objects, or data in a table format. A matrix with m rows and n columns is said to have dimensions m × n.
Matrices and vectors can be added, subtracted, multiplied, and transformed using various operations and algorithms, which make them useful for solving many practical problems in mathematics, physics, engineering, computer science, and other fields.
The line passing through (1, 1, −1) and perpendicular to the plane x + 2y − z = 3 intersects the plane 2x − y + z = 1 at what point?
To find the intersection point of the line passing through (1, 1, -1) and perpendicular to the plane x + 2y – z = 3 with the plane 2x – y + z = 1, we need to follow these steps:
- Find the direction vector of the line.
- Find the normal vector of the plane.
- Use the direction vector and a point on the line to write a parametric equation of the line.
- Find the intersection point by substituting the parametric equation of the line into the equation of the plane.
Step 1: Find the direction vector of the line The line passes through the point (1, 1, -1), so we can take this point as a reference point on the line. The direction vector of the line is perpendicular to the plane x + 2y – z = 3, so it is also perpendicular to the normal vector of the plane. The normal vector of the plane is (1, 2, -1). Therefore, the direction vector of the line is parallel to (1, 2, -1).
Step 2: Find the normal vector of the plane The equation of the plane is 2x – y + z = 1. The coefficients of x, y, and z give the normal vector of the plane, which is (2, -1, 1).
Step 3: Write a parametric equation of the line Let’s use the point (1, 1, -1) on the line as the reference point. We can write the parametric equation of the line as:
x = 1 + t y = 1 + 2t z = -1 – t
where t is a parameter that varies along the line.
Step 4: Find the intersection point To find the intersection point, we need to substitute the parametric equation of the line into the equation of the plane:
2x – y + z = 1
2(1 + t) – (1 + 2t) + (-1 – t) = 1
2 + 2t – 1 – 2t – 1 – t = 1
-2t = -2
t = 1
Substituting t = 1 into the parametric equation of the line, we get:
x = 1 + 1 = 2 y = 1 + 2(1) = 3 z = -1 – 1 = -2
Therefore, the intersection point of the line and the plane is (2, 3, -2).
Find an orthonormal basis for the solutions to the linear
equation
2×1 − x2 + x3 = 0.
Answer: The vectors $w_1=\frac{1}{\sqrt{3}}(1,1,-1)$ and $w_2=\frac{1}{\sqrt{2}}(0,1,1)$ form an orthonormal basis for the solution set.
Solution: Find one vector which is a solution to the equation, for example $(1,1,-1)$. Then, divide the vector by its length, obtaining the unit vector $w_1$. By inspection, find a vector $v_2$ which satisfies both the given equation and $w_1 \cdot v_2=0$. Then set $w_2=\frac{1}{| v_2} v_2$.
Let $A=\left[\begin{array}{llll}\mathbf{a}_1 & \mathbf{a}_2 & \mathbf{a}_3 & \mathbf{a}_4\end{array}\right]$ be the $3 \times 4$ matrix given in terms of its columns $\mathbf{a}_1=\left[\begin{array}{r}2 \ 0 \ -1\end{array}\right]$, $\mathbf{a}_2=\left[\begin{array}{l}1 \ 1 \ 1\end{array}\right], \mathbf{a}_3=\left[\begin{array}{r}3 \ -1 \ -3\end{array}\right]$, and $\mathbf{a}_4=\left[\begin{array}{l}3 \ 1 \ 0\end{array}\right]$. In each case below, either express $\mathbf{b}$ as a linear combination of $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$, and $\mathbf{a}_4$, or show that it is not such a linear combination. Explain what your answer means for the corresponding system $A \mathbf{x}=\mathbf{b}$ of linear equations.
a. $\mathbf{b}=\left[\begin{array}{l}1 \ 2 \ 3\end{array}\right]$
b. $\mathbf{b}=\left[\begin{array}{l}4 \ 2 \ 1\end{array}\right]$
Solution. b is a linear combination of $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$, and $\mathbf{a}_4$ if and only if the system $A \mathbf{x}=\mathbf{b}$ is consistent (that is, it has a solution). So in each case we carry the augmented matrix $[A \mid \mathbf{b}]$ of the system $A \mathbf{x}=\mathbf{b}$ to reduced form.
a. Here $\left[\begin{array}{rrrr|r}2 & 1 & 3 & 3 & 1 \ 0 & 1 & -1 & 1 & 2 \ -1 & 1 & -3 & 0 & 3\end{array}\right] \rightarrow\left[\begin{array}{rrrr|r}1 & 0 & 2 & 1 & 0 \ 0 & 1 & -1 & 1 & 0 \ 0 & 0 & 0 & 0 & 1\end{array}\right]$, so the system $A \mathbf{x}=\mathbf{b}$ has no solution in this case. Hence $\mathbf{b}$ is not a linear combination of $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$, and $\mathbf{a}_4$.
b. Now $\left[\begin{array}{rrrr|r}2 & 1 & 3 & 3 & 4 \ 0 & 1 & -1 & 1 & 2 \ -1 & 1 & -3 & 0 & 1\end{array}\right] \rightarrow\left[\begin{array}{rrrr|r}1 & 0 & 2 & 1 & 1 \ 0 & 1 & -1 & 1 & 2 \ 0 & 0 & 0 & 0 & 0\end{array}\right]$, so the system $A \mathbf{x}=\mathbf{b}$ is consistent.
Thus $\mathbf{b}$ is a linear combination of $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$, and $\mathbf{a}_4$ in this case. In fact the general solution is $x_1=1-2 s-t, x_2=2+s-t, x_3=s$, and $x_4=t$ where $s$ and $t$ are arbitrary parameters. Hence $x_1 \mathbf{a}_1+x_2 \mathbf{a}_2+x_3 \mathbf{a}_3+x_4 \mathbf{a}_4=\mathbf{b}=\left[\begin{array}{l}4 \ 2 \ 1\end{array}\right]$ for any choice of $s$ and $t$. If we take $s=0$ and $t=0$, this becomes $\mathbf{a}_1+2 \mathbf{a}_2=\mathbf{b}$, whereas taking $s=1=t$ gives $-2 \mathbf{a}_1+2 \mathbf{a}_2+\mathbf{a}_3+\mathbf{a}_4=\mathbf{b}$.
