Question

mark your answers in the bubble sheet. there are 18 questions. each correct answer is worth 1 point. every wrong or non - given answer is worth 0 points.
question 1
let ( s=left{\begin{bmatrix}1\\0\\2end{bmatrix},\begin{bmatrix}0\\1\\3end{bmatrix}
ight} ) be a linearly independent set in ( mathbb{r}^3 ). which of the following vectors can be added to ( s ) to form a basis for ( mathbb{r}^3 )?
(a) ( \begin{bmatrix}0\\0\\1end{bmatrix} )
(b) ( \begin{bmatrix}0\\0\\0end{bmatrix} )
(c) ( \begin{bmatrix}1\\1\\5end{bmatrix} )
(d) ( \begin{bmatrix}2\\2\\10end{bmatrix} )
(e) none of the above.

Explanation:

Step1: Recall Basis Requirement

A basis for \(\mathbb{R}^3\) must have 3 linearly independent vectors. Given \(S\) has 2 linearly independent vectors, we need a vector \(v\) such that \(S \cup \{v\}\) is linearly independent (i.e., \(v\) is not a linear combination of the vectors in \(S\)). Let the vectors in \(S\) be \(v_1=\begin{bmatrix}1\\0\\2\end{bmatrix}\) and \(v_2=\begin{bmatrix}0\\1\\3\end{bmatrix}\). We need to check for each option if \(a v_1 + b v_2 = w\) (where \(w\) is the option vector) has only the trivial solution \(a = b = 0\).

Step2: Check Option A

Let \(w=\begin{bmatrix}0\\0\\1\end{bmatrix}\). Set up the system:
\[ \begin{cases} a(1) + b(0) = 0 \\ a(0) + b(1) = 0 \\ a(2) + b(3) = 1 \end{cases} \]
From first two equations, \(a = 0\), \(b = 0\). Substituting into third equation: \(0 + 0 = 1\), which is false. So \(w\) is not a linear combination of \(v_1, v_2\). Wait, but let's check other options too.

Step3: Check Option B

\(w=\begin{bmatrix}0\\0\\0\end{bmatrix}\). The zero vector is a linear combination (with \(a = 0, b = 0\)), so adding it would make the set linearly dependent (since zero vector can't be in a linearly independent set). So B is out.

Step4: Check Option C

\(w=\begin{bmatrix}1\\1\\5\end{bmatrix}\). Set up \(a\begin{bmatrix}1\\0\\2\end{bmatrix}+b\begin{bmatrix}0\\1\\3\end{bmatrix}=\begin{bmatrix}1\\1\\5\end{bmatrix}\). This gives:
\[ \begin{cases} a = 1 \\ b = 1 \\ 2a + 3b = 5 \end{cases} \]
Substitute \(a = 1, b = 1\) into third equation: \(2(1)+3(1)=5\), which is \(5 = 5\). So \(w = v_1 + v_2\), so it's a linear combination. Thus, C is linearly dependent with \(S\).

Step5: Check Option D

\(w=\begin{bmatrix}2\\2\\10\end{bmatrix}\). Set up \(a\begin{bmatrix}1\\0\\2\end{bmatrix}+b\begin{bmatrix}0\\1\\3\end{bmatrix}=\begin{bmatrix}2\\2\\10\end{bmatrix}\). This gives:
\[ \begin{cases} a = 2 \\ b = 2 \\ 2a + 3b = 10 \end{cases} \]
Substitute \(a = 2, b = 2\): \(2(2)+3(2)=4 + 6 = 10\), which holds. So \(w = 2v_1 + 2v_2\), linear combination. So D is dependent.

Step6: Recheck Option A

Wait, for Option A: \(w=\begin{bmatrix}0\\0\\1\end{bmatrix}\). The system \(a\begin{bmatrix}1\\0\\2\end{bmatrix}+b\begin{bmatrix}0\\1\\3\end{bmatrix}=\begin{bmatrix}0\\0\\1\end{bmatrix}\) gives \(a = 0\), \(b = 0\), and \(2(0)+3(0)=0 eq1\). So no solution for \(a,b\) (except trivial, but trivial doesn't satisfy the third equation). Wait, no—linear independence: a set is linearly independent if the only solution to \(a v_1 + b v_2 + c w = 0\) is \(a = b = c = 0\). For \(w\) in A, consider \(a v_1 + b v_2 + c w = 0\). So:
\[ \begin{cases} a = 0 \\ b = 0 \\ 2a + 3b + c = 0 \implies c = 0 \end{cases} \]
Thus, \(w\) is linearly independent of \(v_1, v_2\). Wait, but earlier when checking if \(w\) is a linear combination of \(v_1, v_2\), we saw that \(a v_1 + b v_2 = w\) has no solution (since \(2a + 3b = 1\) with \(a = 0, b = 0\) fails), so \(w\) is not a linear combination. Wait, but let's re-express. Wait, maybe I made a mistake with Option A. Wait, let's check the rank. The matrix formed by \(v_1, v_2, w\) should have rank 3. For Option A: matrix \(M=\begin{bmatrix}1&0&0\\0&1&0\\2&3&1\end{bmatrix}\). The determinant of \(M\) is \(1(1\times1 - 0\times3) - 0 + 0 = 1 eq0\), so rank 3, hence linearly independent. Wait, but earlier when I checked \(a v_1 + b v_2 = w\), that's a different equation (looking for linear combination without the third vector's coefficient). Wait, no—linear independence of \(S \cup \{w\}\) means that the only solution to \(a v_1 + b v_2 + c w = 0\) is \(a = b = c = 0\). For Option A, \(c = 1\), so the equation is \(a v_1 + b v_2 + 1 \cdot w = 0\), which is \(a\begin{bmatrix}1\\0\\2\end{bmatrix}+b\begin{bmatrix}0\\1\\3\end{bmatrix}+\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\), leading to \(a = 0\), \(b = 0\), \(2a + 3b + 1 = 0 \implies 1 = 0\), which is impossible. Wait, no—wait, linear independence: the vectors \(v_1, v_2, w\) are linearly independent if the only solution to \(a v_1 + b v_2 + c w = 0\) is \(a = b = c = 0\). So for Option A, suppose \(a v_1 + b v_2 + c w = 0\). Then:

1. \(a = 0\) (from first component)
2. \(b = 0\) (from second component)
3. \(2a + 3b + c = 0 \implies c = 0\) (from third component)

Thus, only trivial solution, so \(v_1, v_2, w\) are linearly independent. Wait, but earlier when I thought \(a v_1 + b v_2 = w\), that's a different question (can \(w\) be written as a combination of \(v_1, v_2\)), but linear independence is about the combination to zero. So if \(w\) is not a linear combination of \(v_1, v_2\), then \(v_1, v_2, w\) are linearly independent. Wait, but for Option A, \(a v_1 + b v_2 = w\) has no solution (since \(2a + 3b = 1\) with \(a=0, b=0\) gives 0=1), so \(w\) is not a linear combination of \(v_1, v_2\), hence \(v_1, v_2, w\) are linearly independent. Wait, but let's check the determinant again. The matrix with columns \(v_1, v_2, w\) is \(\begin{bmatrix}1&0&0\\0&1&0\\2&3&1\end{bmatrix}\), determinant is \(1\times(1\times1 - 0\times3) - 0 + 0\times(0\times0 - 1\times3) = 1\), which is non-zero, so rank 3, hence linearly independent. So Option A should work? But wait, let's check Option C again. For Option C: matrix with columns \(v_1, v_2, w\) is \(\begin{bmatrix}1&0&1\\0&1&1\\2&3&5\end{bmatrix}\). Determinant: \(1(1\times5 - 1\times3) - 0 + 1(0\times3 - 1\times2) = 1(2) + 1(-2) = 0\), so rank 2, linearly dependent. Option D: matrix \(\begin{bmatrix}1&0&2\\0&1&2\\2&3&10\end{bmatrix}\). Determinant: \(1(1\times10 - 2\times3) - 0 + 2(0\times3 - 1\times2) = 1(4) + 2(-2) = 0\), rank 2. Option B: matrix with zero vector, determinant 0, rank at most 2. So Option A: matrix determinant 1, rank 3, so adding it to \(S\) gives a basis. Wait, but the initial thought was maybe wrong. Wait, let's confirm the definition: a basis for \(\mathbb{R}^n\) is a linearly independent set that spans \(\mathbb{R}^n\). Since \(\dim(\mathbb{R}^3)=3\), a set of 3 linearly independent vectors spans \(\mathbb{R}^3\). So if \(S \cup \{w\}\) has 3 linearly independent vectors, it's a basis. So for Option A, the three vectors are linearly independent, so it should form a basis. Wait, but let's check the linear combination again. Suppose \(a\begin{bmatrix}1\\0\\2\end{bmatrix}+b\begin{bmatrix}0\\1\\3\end{bmatrix}+c\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\). Then \(a=0\), \(b=0\), \(2a + 3b + c = 0 \implies c=0\). So linearly independent. Thus, Option A is correct. Wait, but maybe I made a mistake earlier. Let's re-express.

Wait, the vectors in \(S\) are \(\begin{bmatrix}1\\0\\2\end{bmatrix}\) and \(\begin{bmatrix}0\\1\\3\end{bmatrix}\). Let's check if Option A's vector \(\begin{bmatrix}0\\0\\1\end{bmatrix}\) is linearly independent. Suppose \(a\begin{bmatrix}1\\0\\2\end{bmatrix}+b\begin{bmatrix}0\\1\\3\end{bmatrix}+c\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\). Then:

- From the first component: \(a = 0\)
- From the second component: \(b = 0\)
- From the third component: \(2a + 3b + c = 0 \implies c = 0\)

Thus, only the trivial solution, so they are linearly independent. Therefore, adding Option A to \(S\) gives a basis for \(\mathbb{R}^3\).

Step7: Recheck Other Options

- Option B: Zero vector, adding it makes the set linearly dependent (since zero vector is a linear combination of any set), so invalid.
- Option C: As shown, determinant zero, linearly dependent (since \(1\begin{bmatrix}1\\0\\2\end{bmatrix}+1\begin{bmatrix}0\\1\\3\end{bmatrix}=\begin{bmatrix}1\\1\\5\end{bmatrix}\)), so invalid.
- Option D: \(2\begin{bmatrix}1\\0\\2\end{bmatrix}+2\begin{bmatrix}0\\1\\3\end{bmatrix}=\begin{bmatrix}2\\2\\10\end{bmatrix}\), so linear combination, invalid.

Thus, Option A is valid.

Answer:

A. \(

$$\begin{bmatrix}0\\0\\1\end{bmatrix}$$

\)