Solution Check for Determinant

Question

Solution is 4.

Original matrix is simply [v1;v2;v3;v4]. It forms an identity matrix. Hence the only alteration of the determinant comes from row 1 operation where v1 is multiplied by 2. Then the determinant will also be multiplied by two so 2*2 =4.

We ignore the rest of row operations since they have no effect on the determinant.

Is this line of reasoning correct?

Answer 1

Note that: $$\begin{align}|A|=\begin{vmatrix}v_1\\v_2\\v_3\\v_4\end{vmatrix}&=2 \Rightarrow \\ \begin{vmatrix}2v_1\\4v_2\\v_3\\2v_4\end{vmatrix}&=32 \ \ \ \ \ \stackrel{\frac32R_1+R_2\to R_2\\ \frac14R_2+R_3\to R_3\\ \frac12R_1+R_4\to R_4}{\Rightarrow} \\ \begin{vmatrix}2v_1\\3v_1+4v_2\\v_2+v_3\\v_1+2v_4\end{vmatrix}&=32.\end{align}$$

Answer 2

We have

$$\begin{vmatrix}2v_1\\3v_1+4v_2\\v_2+v_3\\v_1+2v_4\end{vmatrix}=2\begin{vmatrix}v_1\\3v_1+4v_2\\v_2+v_3\\v_1+2v_4\end{vmatrix}=2\begin{vmatrix}v_1\\4v_2\\v_2+v_3\\2v_4\end{vmatrix}=16\begin{vmatrix}v_1\\v_2\\v_2+v_3\\v_4\end{vmatrix}=16\begin{vmatrix}v_1\\v_2\\v_3\\v_4\end{vmatrix}=32$$

To check we can try with

$$\begin{vmatrix}2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{vmatrix}=2$$

and

$$\begin{vmatrix}2&0&0&0\\3&4&0&0\\0&1&1&0\\1&0&0&2\end{vmatrix}=32$$