-1

Let $D_1= \begin{vmatrix}a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix}$ and

$D_2= \begin{vmatrix}a_1+pb_1 & b_1+qc_1 & c_1+ra_1\\ a_2+pb_2 & b_2+qc_2 & c_2+ra_2\\ a_3+pb_3 & b_3+qc_3 & c_3+ra_3\\ \end{vmatrix}$, then how is $D_2= D_1(1+pqr)?$

user64742
  • 2,207
Ujjwal
  • 223

2 Answers2

3

Define $\mathbf a$ to be the column $(a_1,a_2,a_3)$ and similarly for the other letters. For simplicity I'll write a matrix as a row of columns. In this notation, your matrix is \begin{align*} (\mathbf a+p\mathbf b,\mathbf b+q\mathbf c,\mathbf c+r\mathbf a). \end{align*} As someone mentioned, the determinant is a trilinear form on the columns, meaning you can effectively "FOIL" them out like you would a product of polynomials. When you do this you get $2\times2\times2=8$ terms but note that some of them look like, for example, \begin{align*} \det(p\mathbf b,\mathbf b,\mathbf c). \end{align*} which equals zero since the first two columns are a multiple of each other. Therefore the only terms you have to worry about are \begin{align*} \det(\mathbf a,\mathbf b,\mathbf c)+\det(p\mathbf b,q\mathbf c,r\mathbf a). \end{align*} You can exchange the first and third, followed by second and third columns in the second term (each time picking up a minus sign) to get \begin{align*} \det(\mathbf a,\mathbf b,\mathbf c)+(-1)^2\det(r\mathbf a,p\mathbf b,q\mathbf c). \end{align*} Finally, taking out the constant factors, you get \begin{align*} \det(\mathbf a,\mathbf b,\mathbf c)+(-1)^2rpq\det(\mathbf a,\mathbf b,\mathbf c)=D_1(1+pqr) \end{align*} as desired.

Important note: In this very special case it did end up being the case that $\det(A+B)=\det A+\det B$. However this is only because of the peculiar relationship between the columns. In the vast majority of the cases, this will not hold so you should not skip any steps.

Funktorality
  • 3,221
  • 1
    A way to summarize your point of caution: A determinant is a multilinear function of its column vectors (and of its row vectors, of course), and is not in general a multilinear function of matrices. – Semiclassical Jun 17 '16 at 05:00
  • I should also say that while I think writing out an explicit matrix multiplication is simpler in the present case (hence my answer), the multilinearity of the determinant is definitely better for generalizing to higher dimensional cases. – Semiclassical Jun 17 '16 at 05:15
  • Your solution is prettier but maybe harder to do quickly in your head. – Funktorality Jun 17 '16 at 05:20
  • This answer copied pretty much what I did... – user64742 Jun 18 '16 at 20:36
  • @TheGreatDuck No, your first answer made use of the fallacy that $\det(A+B)=\det A+\det B$. – Funktorality Jun 18 '16 at 20:51
  • @StanCoreyCarter I had edited and fixed that before you posted... But it is probable we posted at the same time – user64742 Jun 18 '16 at 23:40
  • Are you talking about the post below? I have trouble making out what it says, but you don't seem to explain the key step of expanding the multilinear form. – Funktorality Jun 18 '16 at 23:45
3

In fact, the two matrices involved differ only by a matrix multiplication: $$\begin{pmatrix}a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{pmatrix} \begin{pmatrix} 1 & 0 & r\\ p & 1 & 0 \\ 0 & q & 1 \end{pmatrix}= \begin{pmatrix}a_1+pb_1 & b_1+qc_1 & c_1+ra_1\\ a_2+pb_2 & b_2+qc_2 & c_2+ra_2\\ a_3+pb_3 & b_3+qc_3 & c_3+ra_3\\ \end{pmatrix}.$$ But this matrix has determinant $1+pqr$, so $D_2$ must differ from $D_1$ by precisely this factor.

Semiclassical
  • 15,842