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) \( k \neq \pm 4 \) d) All of the above e) None of the above
Ask by Bob Li. in Saudi Arabia
Mar 10,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( k \) can be any real number, so none of the given options are correct.
Solution
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.
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Simplify this solution Bonus Knowledge
To determine when \( k^{2} X_{1}-15 X_{2} \) is a solution for the linear system \( A X = 0 \), you can substitute the potential solutions into the equation and check if it holds true. Given that \( X_{1} \) and \( X_{2} \) are solutions, any linear combination of these solutions is also a solution. This will guide you in figuring out the values of \( k \). The correct answer is that the expression will indeed be a solution when \( k = \pm 4 \) ensuring \( k^2 = 16 \), but will not hold for other values outside of this due to the requirement that the linear combination remains a solution. Therefore, option (a) is the suitable choice.
