Let’s analyze each option in detail, keeping in mind that a set of vectors is linearly independent if the only solution to the linear combination equaling the zero vector is when all the coefficients are zero.

Given that $\{(a,b), (c,d)\}$ is linearly independent in $\mathbb{R}^2$, we know that there are no non-trivial solutions to the equation:

$x_1(a, b) + x_2(c, d) = (0, 0)$ which implies that $x_1 = x_2 = 0$.

Now, we consider each set in $\mathbb{R}^3$:

1. $\{(a,b,0), (c,d,0)\}$

In $\mathbb{R}^3$, this is equivalent to the following equation for linear dependence:

$x_1(a, b, 0) + x_2(c, d, 0) = (0, 0, 0)$ The third component for both vectors is 0, meaning the problem reduces to checking the linear independence of $(a,b)$ and $(c,d)$ in $\mathbb{R}^2$, which we know is already linearly independent. Thus, $\{(a,b,0), (c,d,0)\}$ is linearly independent in $\mathbb{R}^3$.

Conclusion: This set must be linearly independent.

2. $\{(a,b,0), (c,d,0), (0,0,1)\}$

Now we check if adding the vector $(0,0,1)$ preserves linear independence:

$x_1(a, b, 0) + x_2(c, d, 0) + x_3(0, 0, 1) = (0, 0, 0)$ The third component forces $x_3 = 0$. This reduces the problem to:

$x_1(a, b) + x_2(c, d) = (0, 0)$ Since $(a,b)$ and $(c,d)$ are linearly independent, we conclude that $x_1 = x_2 = 0$. Therefore, the entire set is linearly independent.

Conclusion: This set must be linearly independent.

3. $\{(a,b,0), (c,d,0), (1,0,0)\}$

Next, consider the vector $(1,0,0)$:

$x_1(a, b, 0) + x_2(c, d, 0) + x_3(1, 0, 0) = (0, 0, 0)$ The third component forces $x_1 = x_2 = 0$. Then we have $x_3(1, 0, 0) = (0, 0, 0)$, which implies $x_3 = 0$.

Thus, all coefficients are zero, so the set is linearly independent.

Conclusion: This set must be linearly independent.

4. $\{(a,b,0), (c,d,0), (0,1,0)\}$

Finally, consider the vector $(0,1,0)$:

$x_1(a, b, 0) + x_2(c, d, 0) + x_3(0, 1, 0) = (0, 0, 0)$ The third component forces $x_1 = x_2 = 0$, and the second component forces $x_3 = 0$. Hence, the set is linearly independent.

Conclusion: This set must be linearly independent.

Final Answer:

The following sets must be linearly independent in $\mathbb{R}^3$:

• $\{(a,b,0), (c,d,0)\}$
• $\{(a,b,0), (c,d,0), (0,0,1)\}$
• $\{(a,b,0), (c,d,0), (1,0,0)\}$
• $\{(a,b,0), (c,d,0), (0,1,0)\}$

Do you need more details or have any questions? Here are some follow-up questions you might consider:

1. How do we define linear independence in higher-dimensional spaces?
2. Why is the vector $(0, 0, 1)$ in $\mathbb{R}^3$ automatically independent of the other two vectors?
3. Can a set of more than two vectors in $\mathbb{R}^2$ ever be linearly independent?
4. How can we generalize the idea of linear independence to larger sets of vectors in $\mathbb{R}^n$?
5. What happens to linear independence if one of the vectors is a linear combination of the others?

Tip: Always check the third components in $\mathbb{R}^3$ when vectors are extended from $\mathbb{R}^2$. If they’re non-zero and the original vectors are independent, you are likely maintaining independence.