Evaluating determinant of an implicit matrix

Question

I know that row operations does not change the determinant of a matrix but I also know that for example, A is a nxn matrix and if det(A) = 2 then, det(2A) = (2^n)*det(A).

So, how should I approach to this problem?

Answer 1

Hint:

Try the row operation R1 -> R1 - R2

Solution: You are left with a matrix with rows: $$ (v_2, 3v_1 + 3v_2, v_3, v_4) $$ Then do the operation R2 -> R2 - 3R1, it becomes $$ (v_2, 3v_1, v_3, v_4) $$whose determinant is $-3\times \det A$

(-1 because of the swap R1 <-> R2 and 3 because of the factor in front of $v_1$).

Answer 2

Determinants are multilinear operations, which implies:

$$\text{Det}\begin{bmatrix} \text{Row}_1 \\ \text{Row}_A \\ \text{Row}_3 \\ \text{Row}_4\end{bmatrix} + \text{Det}\begin{bmatrix} \text{Row}_1 \\ \text{Row}_B \\ \text{Row}_3 \\ \text{Row}_4\end{bmatrix} = \text{Det}\begin{bmatrix} \text{Row}_1 \\ \text{Row}_A + \text{Row}_B \\ \text{Row}_3 \\ \text{Row}_4\end{bmatrix} \tag 1$$

And that

$$\text{Det}\begin{bmatrix} \text{Row}_1 \\ c\cdot\text{Row}_A \\ \text{Row}_3 \\ \text{Row}_4\end{bmatrix} = c \cdot \text{Det}\begin{bmatrix} \text{Row}_1 \\ \text{Row}_A \\ \text{Row}_3 \\ \text{Row}_4\end{bmatrix} \tag 2$$

Determinants are also antisymmetric, which means that:

$$\text{Det}\begin{bmatrix} \text{Row}_A \\ \text{Row}_B \\ \text{Row}_3 \\ \text{Row}_4\end{bmatrix} = - \text{Det}\begin{bmatrix} \text{Row}_B \\ \text{Row}_A \\ \text{Row}_3 \\ \text{Row}_4\end{bmatrix} \tag 3$$

Use the fact that you know the value of $\text{Det}\begin{bmatrix} \text{Row}_1 \\ \text{Row}_2 \\ \text{Row}_3 \\ \text{Row}_4\end{bmatrix}$ along with (1) , (2) , and (3) to find the value of $\text{Det}\begin{bmatrix} 3\,\text{Row}_1 + 4\,\text{Row}_2\\ 3\,\text{Row}_1 + 3\,\text{Row}_2 \\ \text{Row}_3 \\ \text{Row}_4\end{bmatrix}$

You'll have to note that (2) implies that $$\text{Det}\begin{bmatrix} \text{Row}_A \\ \text{Row}_A \\ \text{Row}_3 \\ \text{Row}_4\end{bmatrix} = 0 \tag{2*}$$