Consider the following matrices:
$P$ = $\begin{pmatrix}a&2d&1\\ b&2e&-2\\ c&2f&-1\end{pmatrix}$
$U$ = $\begin{pmatrix}a&b&c\\ 2&3&2\\ d&e&f\end{pmatrix}$
$V$ = $\begin{pmatrix}a&b&c\\ d&e&f\\ 1&5&3\end{pmatrix}$
If you are given that $det(P)$ = $10$ and $det(U)$ = $-3$, then find the value of $det(V)$.
I personally am finding this problem very confusing. I know the general rules of how row operations and row changes affect the determinant of a matrix but I am not sure which ones are being applied here.
For the first matrix, we are given arbitrary values for the first two columns and for the last two we are given arbitrary values for the rows instead. So I am quite confused on how to work with this.
Any help?
Best Answer
$$\mathbf P=\begin{pmatrix}a&2d&1\\b&2e&-2\\c&2f&-1\end{pmatrix},\,\mathbf U=\begin{pmatrix}a&b&c\\2&3&2\\d&e&f\end{pmatrix},\,\mathbf V=\begin{pmatrix}a&b&c\\d&e&f\\1&5&3\end{pmatrix}$$
For later use, we take the transpose of $\mathbf P$:
$$\mathbf P^\top=\begin{pmatrix}a&b&c\\2d&2e&2f\\1&-2&-1\end{pmatrix}$$
Recall that $\det\mathbf P=\det(\mathbf P^\top)$.
Denote by $\mathbf P_{i,j}$ the permutation matrix that, upon multiplication by a matrix $\mathbf A_{m\times n}$, swaps rows $i$ and $j$ in $\mathbf A$. So we can write
$$\mathbf P_{2,3}\mathbf U=\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}\mathbf U=\begin{pmatrix}a&b&c\\d&e&f\\2&3&2\end{pmatrix}$$
Recall that the determinant of any square matrix with any pair of its rows/columns swapped negates the value of the new matrix, so that $\det\mathbf U=-\det(\mathbf P_{i,j}\mathbf U)$ when $i\neq j$.
Now, observe that we can write the last row of $\mathbf V$ as a linear combination of the last rows of $\mathbf P^\top$ and $\mathbf P_{2,3}\mathbf U$:
$$\begin{pmatrix}1&5&3\end{pmatrix}=(-1)\begin{pmatrix}1&-2&-1\end{pmatrix}+\begin{pmatrix}2&3&2\end{pmatrix}$$
The determinant has the property that it is linear with respect to any given row, which is to say: focusing our attention on a single row, if we can write it as a linear combination of other row vectors, then we can expand the determinant as the sum of two component determinants. To illustrate in practice, we can write
$$\begin{vmatrix}a&b&c\\d&e&f\\1&5&3\end{vmatrix}=\begin{vmatrix}a&b&c\\d&e&f\\(-1)1&(-1)(-2)&(-1)(-1)\end{vmatrix}+\begin{vmatrix}a&b&c\\d&e&f\\2&3&2\end{vmatrix}$$
Aside: I highly recommend watching this lecture from MIT if you ever feel the need to brush up on the properties of the determinant. Strang does a great job of explaining them.
Next, we can pull out a factor of $-1$ from the first determinant, and simultaneously distribute a factor of $2$ along the second row of the first matrix,
$$\begin{vmatrix}a&b&c\\d&e&f\\1&5&3\end{vmatrix}=-\frac12\begin{vmatrix}a&b&c\\2d&2e&2f\\1&-2&-1\end{vmatrix}+\begin{vmatrix}a&b&c\\d&e&f\\2&3&2\end{vmatrix}$$
and we see that we have written $\det\mathbf V$ in terms of known determinants. We get
$$\det\mathbf V=-\frac12\det(\mathbf P^\top)+\det(\mathbf P_{2,3}\mathbf U)=-\frac12\det\mathbf P-\det\mathbf U=-5-(-3)=-2$$