Math / Algebra

Questionاجتماع لترشيج لنادي الشطرنج Forms - Google Drive Google Translate Translate Micrometer Instructional Techno... ChatGPT C.E 123 مكهبة - Goo.. Dashboard | Pre (b) 1 (c) 2 (d) 3. (11) Which of the following is a subspace of $P_{3}$ ? (a) $S=\left\{p \in P_{3}: p(1)=1\right\} \times$ $a x^{2}+b x+c$ (b) $S=\left\{p \in P_{3}: p(1)=0\right\}$ $a+b+c$ (c) $\beta=\left\{p \in P_{3}: p(0)=1\right\} \times$ (少) $S=\left\{p \in P_{3}: p(x)=x^{2}+a x+b, a, b \in \mathbb{R}\right\}$ (12) If the reduced row echelon form of $A$ is $\left[\begin{array}{lll}1 & 2 & 2 \\ 0 & 0 & 0\end{array}\right]$ and $a_{2}=(2,2)^{T}$ , (a) $\left[\begin{array}{lll}1 & 2 & 2 \\ 1 & 2 & 1\end{array}\right]$ (b) $\left[\begin{array}{lll}1 & 2 & 2 \\ 0 & 2 & 0\end{array}\right]$ $\begin{array}{lll} 2 & 4 & 4 \\ 2 & 4 & 1 \end{array}$ (c) $\left[\begin{array}{lll}1 & 2 & 2 \\ 1 & 2 & 2\end{array}\right]$ (d) $\left[\begin{array}{lll}4 & 2 & 2 \\ 4 & 2 & 2\end{array}\right]$ . Search

Studdy Solution

STEP 1

What is this asking? Out of the given options, which sets of polynomials are subspaces of the vector space of polynomials of degree 3 or less?
Also, given a reduced row echelon form of matrix $A$ and one of its columns, which of the provided matrices could be $A$ ? Watch out! Remember, a subspace must contain the zero vector and be closed under addition and scalar multiplication.
Don't forget the importance of linear independence when dealing with vector spaces!

STEP 2

1. Subspace conditions
2. Polynomial subspaces
3. Matrix A

STEP 3

To be a subspace, a set must satisfy three conditions: it must contain the zero vector, be closed under addition, and be closed under scalar multiplication.
Let's keep these in mind as we evaluate the options.

STEP 4

Let's analyze option (a): $S = \{p \in P_3 : p(1) = 1\}$ .
If we consider the zero polynomial $p(x) = 0$ , we see that $p(1) = 0 \neq 1$ .
So, this set doesn't contain the zero vector and is not a subspace.

STEP 5

Now, let's look at option (b): $S = \{p \in P_3 : p(1) = 0\}$ .
The zero polynomial satisfies $p(1) = 0$ , so it's in the set.
If we have two polynomials $p_1(x)$ and $p_2(x)$ in the set, then $p_1(1) = 0$ and $p_2(1) = 0$ .
Their sum, $(p_1 + p_2)(1) = p_1(1) + p_2(1) = 0 + 0 = 0$ , is also in the set.
Similarly, for any scalar $c$ , $(c \cdot p_1)(1) = c \cdot p_1(1) = c \cdot 0 = 0$ .
Thus, option (b) is a subspace.

STEP 6

For option (c): $S = \{p \in P_3 : p(0) = 1\}$ .
The zero polynomial gives $p(0) = 0 \neq 1$ .
So, this set doesn't contain the zero vector and is not a subspace.

STEP 7

Finally, option (d): $S = \{p \in P_3 : p(x) = x^2 + ax + b, a, b \in \mathbb{R}\}$ .
The zero polynomial is not in this set because the coefficient of $x^2$ is always 1, not 0.
Therefore, this is not a subspace.

STEP 8

We're given that the reduced row echelon form of $A$ is $\begin{bmatrix} 1 & 2 & 2 \\ 0 & 0 & 0 \end{bmatrix}$ and that the second column of $A$ , $a_2$ , is $\begin{bmatrix} 2 \\ 2 \end{bmatrix}$ .

STEP 9

Since the reduced row echelon form has a leading one in the first column, the first column of $A$ must be a scalar multiple of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ .
Also, the second column is $\begin{bmatrix} 2 \\ 2 \end{bmatrix}$ , which is not a scalar multiple of the first column of the reduced row echelon form.
The third column should be such that after row operations, it becomes $\begin{bmatrix} 2 \\ 0 \end{bmatrix}$ .

STEP 10

Let's denote the columns of $A$ as $a_1$ , $a_2$ , and $a_3$ .
We know $a_2 = \begin{bmatrix} 2 \\ 2 \end{bmatrix}$ .
The reduced row echelon form tells us that $a_3 = 2a_1 + 0a_2 = 2a_1$ .
Let's examine the options.

STEP 11

Option (a): $\begin{bmatrix} 1 & 2 & 2 \\ 1 & 2 & 1 \end{bmatrix}$ .
Here, $a_3$ is not $2a_1$ .
Incorrect.

STEP 12

Option (b): $\begin{bmatrix} 1 & 2 & 2 \\ 0 & 2 & 0 \end{bmatrix}$ .
Here, $a_3 = 2a_1$ .
This is a possibility.

STEP 13

Option (c): $\begin{bmatrix} 1 & 2 & 2 \\ 1 & 2 & 2 \end{bmatrix}$ .
Here, $a_3 = 2a_1$ .
This is also a possibility.

STEP 14

Option (d): $\begin{bmatrix} 4 & 2 & 2 \\ 4 & 2 & 2 \end{bmatrix}$ .
Here, $a_3$ is not $2a_1$ .
Incorrect.

STEP 15

If we row reduce option (b), we get the correct reduced row echelon form.
If we row reduce option (c), we don't.
So, the answer is (b).

STEP 16

The subspaces of $P_3$ are the set in option (b).
The matrix $A$ is the one shown in option (b) of the second question.

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Studdy Solution

STEP 1

STEP 2

1. Subspace conditions
2. Polynomial subspaces
3. Matrix A

STEP 3

To be a subspace, a set must satisfy three conditions: it must contain the zero vector, be closed under addition, and be closed under scalar multiplication.
Let's keep these in mind as we evaluate the options.

STEP 4

Let's analyze option (a): $S = \{p \in P_3 : p(1) = 1\}$ .
If we consider the zero polynomial $p(x) = 0$ , we see that $p(1) = 0 \neq 1$ .
So, this set doesn't contain the zero vector and is not a subspace.

STEP 5

STEP 6

For option (c): $S = \{p \in P_3 : p(0) = 1\}$ .
The zero polynomial gives $p(0) = 0 \neq 1$ .
So, this set doesn't contain the zero vector and is not a subspace.

STEP 7

Finally, option (d): $S = \{p \in P_3 : p(x) = x^2 + ax + b, a, b \in \mathbb{R}\}$ .
The zero polynomial is not in this set because the coefficient of $x^2$ is always 1, not 0.
Therefore, this is not a subspace.

STEP 8

We're given that the reduced row echelon form of $A$ is $\begin{bmatrix} 1 & 2 & 2 \\ 0 & 0 & 0 \end{bmatrix}$ and that the second column of $A$ , $a_2$ , is $\begin{bmatrix} 2 \\ 2 \end{bmatrix}$ .

STEP 9

STEP 10

Let's denote the columns of $A$ as $a_1$ , $a_2$ , and $a_3$ .
We know $a_2 = \begin{bmatrix} 2 \\ 2 \end{bmatrix}$ .
The reduced row echelon form tells us that $a_3 = 2a_1 + 0a_2 = 2a_1$ .
Let's examine the options.

STEP 11

Option (a): $\begin{bmatrix} 1 & 2 & 2 \\ 1 & 2 & 1 \end{bmatrix}$ .
Here, $a_3$ is not $2a_1$ .
Incorrect.

STEP 12

Option (b): $\begin{bmatrix} 1 & 2 & 2 \\ 0 & 2 & 0 \end{bmatrix}$ .
Here, $a_3 = 2a_1$ .
This is a possibility.

STEP 13

Option (c): $\begin{bmatrix} 1 & 2 & 2 \\ 1 & 2 & 2 \end{bmatrix}$ .
Here, $a_3 = 2a_1$ .
This is also a possibility.

STEP 14

Option (d): $\begin{bmatrix} 4 & 2 & 2 \\ 4 & 2 & 2 \end{bmatrix}$ .
Here, $a_3$ is not $2a_1$ .
Incorrect.

STEP 15

If we row reduce option (b), we get the correct reduced row echelon form.
If we row reduce option (c), we don't.
So, the answer is (b).

STEP 16

The subspaces of $P_3$ are the set in option (b).
The matrix $A$ is the one shown in option (b) of the second question.

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Studdy Solution

STEP 1

STEP 2

1. Subspace conditions2. Polynomial subspaces3. Matrix A

STEP 3

To be a subspace, a set must satisfy three conditions: it must contain the zero vector, be closed under addition, and be closed under scalar multiplication.Let's keep these in mind as we evaluate the options.

STEP 4

Let's analyze option (a): S={p∈P3:p(1)=1}S = \{p \in P_3 : p(1) = 1\}S={p∈P3​:p(1)=1}.If we consider the zero polynomial p(x)=0p(x) = 0p(x)=0, we see that p(1)=0≠1p(1) = 0 \neq 1p(1)=0=1.So, this set *doesn't* contain the zero vector and is **not** a subspace.

STEP 5

STEP 6

For option (c): S={p∈P3:p(0)=1}S = \{p \in P_3 : p(0) = 1\}S={p∈P3​:p(0)=1}.The zero polynomial gives p(0)=0≠1p(0) = 0 \neq 1p(0)=0=1.So, this set *doesn't* contain the zero vector and is **not** a subspace.

STEP 7

Finally, option (d): S={p∈P3:p(x)=x2+ax+b,a,b∈R}S = \{p \in P_3 : p(x) = x^2 + ax + b, a, b \in \mathbb{R}\}S={p∈P3​:p(x)=x2+ax+b,a,b∈R}.The zero polynomial is not in this set because the coefficient of x2x^2x2 is always 1, not 0.Therefore, this is **not** a subspace.

STEP 8

We're given that the reduced row echelon form of AAA is [122000]\begin{bmatrix} 1 & 2 & 2 \\ 0 & 0 & 0 \end{bmatrix}[10​20​20​] and that the second column of AAA, a2a_2a2​, is [22]\begin{bmatrix} 2 \\ 2 \end{bmatrix}[22​].

STEP 9

STEP 10

Let's denote the columns of AAA as a1a_1a1​, a2a_2a2​, and a3a_3a3​.We know a2=[22]a_2 = \begin{bmatrix} 2 \\ 2 \end{bmatrix}a2​=[22​].The reduced row echelon form tells us that a3=2a1+0a2=2a1a_3 = 2a_1 + 0a_2 = 2a_1a3​=2a1​+0a2​=2a1​.Let's examine the options.

STEP 11

Option (a): [122121]\begin{bmatrix} 1 & 2 & 2 \\ 1 & 2 & 1 \end{bmatrix}[11​22​21​].Here, a3a_3a3​ is not 2a12a_12a1​.Incorrect.

STEP 12

Option (b): [122020]\begin{bmatrix} 1 & 2 & 2 \\ 0 & 2 & 0 \end{bmatrix}[10​22​20​].Here, a3=2a1a_3 = 2a_1a3​=2a1​.This is a possibility.

STEP 13

Option (c): [122122]\begin{bmatrix} 1 & 2 & 2 \\ 1 & 2 & 2 \end{bmatrix}[11​22​22​].Here, a3=2a1a_3 = 2a_1a3​=2a1​.This is also a possibility.

STEP 14

Option (d): [422422]\begin{bmatrix} 4 & 2 & 2 \\ 4 & 2 & 2 \end{bmatrix}[44​22​22​].Here, a3a_3a3​ is not 2a12a_12a1​.Incorrect.

STEP 15

If we row reduce option (b), we get the correct reduced row echelon form.If we row reduce option (c), we don't.So, the answer is (b).

STEP 16

The subspaces of P3P_3P3​ are the set in option (b).The matrix AAA is the one shown in option (b) of the second question.

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1. Subspace conditions
2. Polynomial subspaces
3. Matrix A

To be a subspace, a set must satisfy three conditions: it must contain the zero vector, be closed under addition, and be closed under scalar multiplication.
Let's keep these in mind as we evaluate the options.

Let's analyze option (a): $S = \{p \in P_3 : p(1) = 1\}$ .
If we consider the zero polynomial $p(x) = 0$ , we see that $p(1) = 0 \neq 1$ .
So, this set doesn't contain the zero vector and is not a subspace.

For option (c): $S = \{p \in P_3 : p(0) = 1\}$ .
The zero polynomial gives $p(0) = 0 \neq 1$ .
So, this set doesn't contain the zero vector and is not a subspace.

Finally, option (d): $S = \{p \in P_3 : p(x) = x^2 + ax + b, a, b \in \mathbb{R}\}$ .
The zero polynomial is not in this set because the coefficient of $x^2$ is always 1, not 0.
Therefore, this is not a subspace.

We're given that the reduced row echelon form of $A$ is $\begin{bmatrix} 1 & 2 & 2 \\ 0 & 0 & 0 \end{bmatrix}$ and that the second column of $A$ , $a_2$ , is $\begin{bmatrix} 2 \\ 2 \end{bmatrix}$ .

Let's denote the columns of $A$ as $a_1$ , $a_2$ , and $a_3$ .
We know $a_2 = \begin{bmatrix} 2 \\ 2 \end{bmatrix}$ .
The reduced row echelon form tells us that $a_3 = 2a_1 + 0a_2 = 2a_1$ .
Let's examine the options.

Option (a): $\begin{bmatrix} 1 & 2 & 2 \\ 1 & 2 & 1 \end{bmatrix}$ .
Here, $a_3$ is not $2a_1$ .
Incorrect.

Option (b): $\begin{bmatrix} 1 & 2 & 2 \\ 0 & 2 & 0 \end{bmatrix}$ .
Here, $a_3 = 2a_1$ .
This is a possibility.

Option (c): $\begin{bmatrix} 1 & 2 & 2 \\ 1 & 2 & 2 \end{bmatrix}$ .
Here, $a_3 = 2a_1$ .
This is also a possibility.

Option (d): $\begin{bmatrix} 4 & 2 & 2 \\ 4 & 2 & 2 \end{bmatrix}$ .
Here, $a_3$ is not $2a_1$ .
Incorrect.

If we row reduce option (b), we get the correct reduced row echelon form.
If we row reduce option (c), we don't.
So, the answer is (b).

The subspaces of $P_3$ are the set in option (b).
The matrix $A$ is the one shown in option (b) of the second question.