Question

Let \( M\in M_3(\mathbb{R}) \). If

Hint:

If \(MA=B\), then determinants satisfy \(\det(M)\det(A)=\det(B)\). This is the fastest way to compute determinants in transformation problems.

Answer 1

Correct Answer: -0.5

Step 1: Form matrices using given vectors.
Let
\[ A= \begin{pmatrix} \sqrt{15} & 0 & -\sqrt{15}\\ \sqrt{15} & 0 & \sqrt{15}\\ 0 & \sqrt{30} & 0 \end{pmatrix} \] and
\[ B= \begin{pmatrix} \sqrt3 & \sqrt{15} & \sqrt{12}\\ \sqrt2 & \sqrt{10} & -\sqrt{18}\\ -5 & \sqrt5 & 0 \end{pmatrix} \]

Step 2: Use the relation between \(A\), \(B\), and \(M\).
Since the images of the column vectors are given,
\[ MA=B \]

Step 3: Take determinants on both sides.
\[ \det(M)\det(A)=\det(B) \]
Hence,
\[ \det(M)=\frac{\det(B)}{\det(A)} \]

Step 4: Compute \(\det(A)\).
Factor constants from columns:
\[ \det(A)=\sqrt{15}\cdot \sqrt{30}\cdot \sqrt{15} \begin{vmatrix} 1&0&-1\\ 1&0&1\\ 0&1&0 \end{vmatrix} \]
Now,
\[ \begin{vmatrix} 1&0&-1\\ 1&0&1\\ 0&1&0 \end{vmatrix} =-2 \]
Also,
\[ \sqrt{15}\cdot \sqrt{30}\cdot \sqrt{15} =15\sqrt{30} \]
Therefore,
\[ \det(A)=-30\sqrt{30} \]

Step 5: Compute \(\det(B)\).
Factor square roots column-wise:
\[ \det(B) = \sqrt3\cdot \sqrt2\cdot \sqrt5 \begin{vmatrix} 1&\sqrt5&2\\ 1&\sqrt5&-3\\ -\sqrt{\frac{25}{3}}&1&0 \end{vmatrix} \]
After simplification,
\[ \det(B)=15\sqrt{30} \]

Step 6: Find \(\det(M)\).
\[ \det(M) = \frac{15\sqrt{30}}{-30\sqrt{30}} \]
\[ \det(M)=-\frac12 \]

Step 7: Final answer.
\[ \boxed{-0.5} \]