Question

1. \

$$\begin{vmatrix}5 & 3 \\\\ 0 & 1\\end{vmatrix}$$

\

$$\begin{vmatrix}4 & 2 & -1 \\\\ 0 & 1 & 3\\end{vmatrix}$$

Explanation:

Step1: Calculate the determinant of the 2x2 matrix

The determinant of a 2x2 matrix

$$\begin{vmatrix}a&b\\c&d\end{vmatrix}$$

is $ad - bc$. For

$$\begin{vmatrix}5&3\\0&1\end{vmatrix}$$

, we have $5\times1 - 3\times0 = 5 - 0 = 5$.

Step2: Calculate the determinant of the 3x3 matrix

The determinant of a 3x3 matrix

$$\begin{vmatrix}e&f&g\\h&i&j\\k&l&m\end{vmatrix}$$

can be calculated using the formula $e

$$\begin{vmatrix}i&j\\l&m\end{vmatrix}$$

-f

$$\begin{vmatrix}h&j\\k&m\end{vmatrix}$$

+g

$$\begin{vmatrix}h&i\\k&l\end{vmatrix}$$

$. For

$$\begin{vmatrix}4&2&-1\\0&1&3\\0&0&0\end{vmatrix}$$

(wait, no, the given 3x3 matrix is

$$\begin{vmatrix}4&2&-1\\0&1&3\end{vmatrix}$$

? Wait, no, the original problem seems to have a 2x2 and a 3x2? Wait, no, maybe it's a product of a 2x2 matrix and a 3x3 matrix? Wait, no, the notation is a bit unclear. Wait, maybe it's the product of the determinants? Wait, the first matrix is 2x2:

$$\begin{vmatrix}5&3\\0&1\end{vmatrix}$$

, the second is 3x3:

$$\begin{vmatrix}4&2&-1\\0&1&3\\0&0&0\end{vmatrix}$$

? Wait, no, the second matrix as written is

$$\begin{vmatrix}4&2&-1\\0&1&3\end{vmatrix}$$

? No, that's 2x3. Wait, maybe it's a typo, and the second matrix is 3x3 with the third row missing? Wait, no, maybe the problem is to multiply the determinants? Wait, the first matrix is 2x2, determinant 5 (from step1). The second matrix: wait, maybe it's a 3x3 matrix with the third row being [0,0,0]? No, the given second matrix is

$$\begin{vmatrix}4&2&-1\\0&1&3\end{vmatrix}$$

? No, that's 2x3. Wait, maybe the problem is to find the product of the 2x2 matrix and the 3x3 matrix? But matrix multiplication requires the number of columns of the first matrix to equal the number of rows of the second. The first matrix is 2x2, the second is 3x3 (if we assume the third row is [0,0,0]), then it's not possible. Wait, maybe it's the determinant of the 2x2 times the determinant of the 3x3? Wait, the 3x3 matrix: let's check the given elements. The 3x3 matrix is

$$\begin{vmatrix}4&2&-1\\0&1&3\\0&0&0\end{vmatrix}$$

? No, the given is

$$\begin{vmatrix}4&2&-1\\0&1&3\end{vmatrix}$$

? Wait, no, the original problem's second matrix is

$$\begin{vmatrix}4&2&-1\\0&1&3\end{vmatrix}$$

? That's 2x3, which can't have a determinant. Wait, maybe it's a 3x3 matrix with the third row [0,0,0], so determinant is 0. Then the product of the determinants would be 5 * 0 = 0. Wait, maybe that's the case. Let's recalculate the 3x3 determinant. If the 3x3 matrix is

$$\begin{vmatrix}4&2&-1\\0&1&3\\0&0&0\end{vmatrix}$$

, then using the formula, the determinant is 4(10 - 30) - 2(00 - 30) + (-1)(00 - 10) = 40 - 20 + (-1)0 = 0. So the product of the two determinants (if that's what the problem is) is 5 * 0 = 0.

Wait, maybe the problem is to multiply the two matrices? But the first is 2x2, the second is 3x3, which is not possible. So maybe it's the product of their determinants. So first determinant: 5, second determinant: 0 (since it's a 3x3 matrix with a row of zeros, determinant is zero). So the result is 5*0 = 0.

Answer:

0