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

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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 of the homogeneous system
$$
AX = 0,
$$
we have
$$
AX_1 = 0 \quad \text{and} \quad AX_2 = 0.
$$

A key property of homogeneous linear systems is linearity. This means that any linear combination of solutions is also a solution. Let
$$
Y = k^2 X_1 - 15 X_2.
$$
Then,
$$
AY = A(k^2 X_1 - 15 X_2) = k^2(AX_1) - 15(AX_2) = k^2 \cdot 0 - 15 \cdot 0 = 0.
$$
Thus, $$ Y $$ is a solution to $$ AX = 0 $$ for any value of $$ k $$.

Since no restriction on $$ k $$ is needed for $$ k^2 X_1 - 15 X_2 $$ to be a solution, the correct answer is:

e) None of the above.
$$ k $$ can be any real number, so none of the given options are correct.

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

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