Matrix Determinant after row changes

Question

If a $4\times4$ matrix with rows $v1$, $v2$, $v3$ and $v4$ has a determinant of $-1$, then what would be the determinant of $\begin{pmatrix}2v1+3v4\\ v2\\ v3\\ 9v1+2v4\end{pmatrix}$?

Since the first row is being multiplied by $2$, I multiplied the initial determinant by $2$, giving $-2$. Then, since the fourth row is also being multiplied by $2$, I multiplied the determinant again by $2$, giving $-4$. Since rows being added to other rows does not change the determinant, I thought the final answer would be$-4$. But this is incorrect.

Any help?

Answer 1

It's true that adding a multiple one row of a matrix to a different one doesn't change the determinant, but that's not quite what you're doing here. Put \begin{align*} \alpha & = \pmatrix{v_1 \\ v_2 \\ v_3 \\ v_4} & \beta &= \pmatrix{2v_1 + 3v_4 \\ v_2 \\ v_3 \\ 9v_1 + 2v_4}. \end{align*} You can add $3v_4$ to the top row of $\alpha$ to form a new matrix \begin{align*} \alpha' &= \pmatrix{v_1 + 3v_4 \\ v_2 \\ v_3 \\ v_4} \end{align*} with $\det \alpha = \det \alpha'$. The result you cite, though, says that you preserve the determinant if you add a row of $\alpha'$ to $\alpha'$, not a row of the original matrix $\alpha$ to $\alpha'$. The vector $v_1$ is no longer a row of $\alpha'$, and it's not automatic that adding a multiple of it to the fourth row of $\alpha'$ preserves $\det \alpha'$. A correct computation would be: \begin{align*} \det \beta &= \frac{9}{2} \det \pmatrix{2v_1 + 3v_4 \\ v_2 \\ v_3 \\ 2v_1 + \frac{4}{9} v_4} = \frac{9}{2} \det \pmatrix{2v_1 + 3v_4 \\ v_2 \\ v_3 \\ -\frac{23}{9} v_4} = -\frac{23}{2} \det \pmatrix{2v_1 + 3v_4 \\ v_2 \\ v_3 \\ v_4} = -23 \det \alpha = 23. \end{align*}

More concisely, \begin{align*} \beta = \pmatrix{2 & 0 & 0 & 3 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 9 & 0 & 0 & 2}\, \alpha, \end{align*} and so $\det \beta = -23 \det \alpha = 23$.

Answer 2

As an alternative, since adding a scalar multiple of one row to another row does not change the value of the determinant we obtain

$$det\begin{pmatrix}2v_1+3v_4\\ v_2\\ v_3\\ 9v_1+2v_4\end{pmatrix} =det \begin{pmatrix}2v_1+3v_4\\ v_2\\ v_3\\ 7v_1-v_4\end{pmatrix}= \\=det \begin{pmatrix}23v_1\\ v_2\\ v_3\\ 7v_1-v_4\end{pmatrix} =det \begin{pmatrix}23v_1\\ v_2\\ v_3\\ -v_4\end{pmatrix} =-23\cdot det \begin{pmatrix}v_1\\ v_2\\ v_3\\ v_4\end{pmatrix} $$

Thus the determinant is $23$.

Answer 3

Since the determinant function is a multiplinear map, your determinant is the sum of these four determinants:$$\begin{vmatrix}2v_1\\v_2\\v_3\\9v_1\end{vmatrix},\ \begin{vmatrix}2v_1\\v_2\\v_3\\2v_4\end{vmatrix},\ \begin{vmatrix}3v_4\\v_2\\v_3\\9v_1\end{vmatrix}\text{, and }\begin{vmatrix}3v_4\\v_2\\v_3\\2v_4\end{vmatrix}.$$The first and the fourth ones are $0$, the second one is $-4$ and the third one is $27$. Therefore, the answer is $23$.