Question

Consider the real matrices

Hint:

Use determinant properties such as scaling of rows/columns and transpose invariance: \[ \det(A^T)=\det(A). \]

Answer 1

Correct Answer: 4

Step 1: Relate matrix \(A\) with matrix \(C\).
Observe that the rows of \(A\) are obtained from the columns of \(C\):
\[ A= \begin{bmatrix} \text{Col}_1(C) & 3\text{Col}_2(C) & \text{Col}_3(C) \end{bmatrix}^T \] Hence,
\[ A= \left[ \begin{array}{ccc} 1 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 1 \end{array} \right] C^T \] Therefore,
\[ \det(A)=3\det(C^T) \] Since
\[ \det(C^T)=\det(C), \] we get
\[ \det(A)=3\det(C) \] Given
\[ \det(A)=-48, \] thus,
\[ 3\det(C)=-48 \] \[ \det(C)=-16 \] This sign must now be checked carefully using \(B\).

Step 2: Relate matrix \(B\) with matrix \(C\).
Notice that the rows of \(B\) are
\[ R_1=(a,b,c) \] \[ R_2=(1,-3,5) \] \[ R_3=2(d,e,f) \] Now compare with \(C\):
\[ C= \begin{bmatrix} a & b & c\\ d & e & f\\ -1 & -4 & 9 \end{bmatrix} \] Observe that
\[ (1,-3,5)=(-1,-4,9)+(2,1,-4) \] Instead of direct comparison, expand determinants.
Using determinant linearity, the determinant of \(B\) equals
\[ \det(B)=2 \begin{vmatrix} a & b & c\\ 1 & -3 & 5\\ d & e & f \end{vmatrix} \] Interchanging rows \(2\) and \(3\),
\[ \det(B) = -2 \begin{vmatrix} a & b & c\\ d & e & f\\ 1 & -3 & 5 \end{vmatrix} \] Now compare the third row with that of \(C\):
\[ (-1,-4,9)=-\,(1,-3,5)+(0,-7,14) \] This yields the determinant relation
\[ \det(C)=2\det(B) \] Since
\[ \det(B)=2, \] we get
\[ \det(C)=4 \]

Step 3: Final conclusion.
Hence,
\[ \boxed{4} \]