1.5: Suppose that $$ X_{1} $$ and $$ X_{2} $$ are solutions for the linear system $$ A X=0 $$. $$ k^{2} X_{1}-15 X_{2} $$ is also solution only if a) $$ k= \pm 4 $$ b) $$ k=16 $$ c) $$ k
eq \pm 4 $$ d) All of the above e) None of the above

Question

1.5: Suppose that $$ X_{1} $$ and $$ X_{2} $$ are solutions for the linear system $$ A X=0 $$. $$ k^{2} X_{1}-15 X_{2} $$ is also solution only if a) $$ k= \pm 4 $$ b) $$ k=16 $$ c) $$ k 
eq \pm 4 $$ d) All of the above e) None of the above

UpStudy AI · Accepted Answer

Since $$X_1$$ and $$X_2$$ are solutions to the homogeneous system
$$
AX = 0,
$$
we have 
$$
AX_1 = 0 \quad \text{and} \quad AX_2 = 0.
$$

Because the system is linear, any linear combination of solutions also satisfies the system. In other words, if $$c_1$$ and $$c_2$$ are scalars, then
$$
A (c_1 X_1 + c_2 X_2) = c_1 AX_1 + c_2 AX_2 = c_1 \cdot 0 + c_2 \cdot 0 = 0.
$$

In the given expression, the combination is
$$
k^2 X_1 - 15X_2.
$$
Here, the scalars are $$k^2$$ and $$-15$$. Since these are fixed numbers (with $$k$$ being any real number), the combination $$k^2X_1 - 15X_2$$ is automatically a solution regardless of the value of $$k$$. There is no restriction on $$k$$.

Thus, none of the answer choices a), b), c), or d) correctly restrict $$k$$ according to the properties of linear combinations of solutions. The correct answer is:

e) None of the above.
$$ k^2 X_1 - 15 X_2 $$ is always a solution, so the correct answer is e) None of the above.

1.5: Suppose that \( X_{1} \) and \( X_{2} \) are solutions for the linear system \( A X=0 \). \( k^{2} X_{1}-15 X_{2} \) is also solution only if a) \( k= \pm 4 \) b) \( k=16 \) c) \( k \neq \pm 4 \) d) All of the above e) None of the above

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Algebra Questions