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

To determine when \( k^{2} X_{1}-15 X_{2} \) is also a solution for the linear system \( A X=0 \), we can use the property of linear combinations of solutions. If both \( X_{1} \) and \( X_{2} \) are solutions, any linear combination of the form \( c_1 X_{1} + c_2 X_{2} = 0 \) remains a solution for constants \( c_1 \) and \( c_2 \). In this case, we can set: \[ k^{2} X_{1} - 15 X_{2} \] This implies that we can express it as a linear combination of \( X_{1} \) and \( X_{2} \) and it needs to be equal to zero. For this to hold, \( k^{2} = 15c \), where \( c \) needs to satisfy certain conditions. Thus, the correct answer is: **a) \( k= \pm 4 \)**. This indicates that \( k^{2} \) must equal 16, giving us the solutions where \( k \) can either be 4 or -4. --- Since you might enjoy connections to mathematical concepts, I've skipped deeper algebraic manipulations and focused on conveying why option "a" makes sense in a straightforward way. If you'd like more detailed steps on solving linear algebra problems, just let me know!