Given that the determinant of the matrix ≤ft[ {*20c 1&3&0\\ 2&6&4\\ - 1&0&2 } ] is -12, the determinant of the matrix ≤ft[ {*20c 2&6&0\\ 4&12&8\\ - 2&0&4 } ] is

Determinant Calculation Using Scalar Multiplication

We are asked to find the determinant of a given matrix, knowing the determinant of a related matrix. Let's denote the first matrix as A and the second matrix as B.

The first matrix is given as:

$A=\left[\begin{array}{ccc}1& 3& 0\\ 2& 6& 4\\ -1& 0& 2\end{array}\right]$

We are provided with the determinant of matrix A: $\det (A)=-12$.

The second matrix is:

$B=\left[\begin{array}{ccc}2& 6& 0\\ 4& 12& 8\\ -2& 0& 4\end{array}\right]$

Our goal is to calculate the determinant of matrix B, $\det (B)$.

Analyzing the Relationship Between Matrices A and B

To find $\det (B)$, we first need to understand how matrix B is related to matrix A. Let's compare the corresponding elements or rows/columns of both matrices.

Comparing the rows:

- Row 1 of B: (2, 6, 0) is equal to 2 times Row 1 of A: $2\times (1,3,0)=(2,6,0)$.
- Row 2 of B: (4, 12, 8) is equal to 2 times Row 2 of A: $2\times (2,6,4)=(4,12,8)$.
- Row 3 of B: (-2, 0, 4) is equal to 2 times Row 3 of A: $2\times (-1,0,2)=(-2,0,4)$.

Since every element in matrix B is obtained by multiplying the corresponding element in matrix A by 2, we can express matrix B in terms of matrix A as $B=2A$.

Applying the Property of Determinants for Scalar Multiplication

There is a fundamental property in linear algebra concerning the determinants of matrices multiplied by a scalar. If A is an $n\times n$ matrix and $k$ is a scalar, then the determinant of $kA$ is given by:

$\det (kA)=k^n\det (A)$

In this specific problem:

- The scalar multiplier is $k=2$.
- Matrix A is a $3\times 3$ matrix, so the dimension $n=3$.
- We are given that $\det (A)=-12$.

Calculating the Determinant of Matrix B

Using the determinant property, we can now calculate $\det (B)$:

$\det (B)=\det (2A)$

Applying the property $\det (kA)=k^n\det (A)$:

$\det (B)=2^3\times \det (A)$

$\det (B)=8\times (-12)$

$\det (B)=-96$

Therefore, the determinant of the second matrix is -96.