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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 $D_2= D_1(1+pqr)?$

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closed as off-topic by Did, Claude Leibovici, Watson, Dan Rust, Paramanand Singh Jun 17 '16 at 12:25

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  • $\begingroup$ Mind fixing the formatting slightly. I know I can solve it like your last one, but I need it to be better edited. Thank you. $\endgroup$ – The Great Duck Jun 17 '16 at 3:12
  • $\begingroup$ Okay thanks. Now its fully edited $\endgroup$ – Ujjwal Jun 17 '16 at 3:14
  • $\begingroup$ @thegreatduck the question is waiting to be looked upon.. $\endgroup$ – Ujjwal Jun 17 '16 at 3:31
  • $\begingroup$ That edit is wrong. D1 is the determinant. $\endgroup$ – The Great Duck Jun 17 '16 at 3:32
  • $\begingroup$ I was just slow in working it out, and writing it all. $\endgroup$ – The Great Duck Jun 17 '16 at 3:42
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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.

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    $\begingroup$ 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. $\endgroup$ – Semiclassical Jun 17 '16 at 5:00
  • $\begingroup$ 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. $\endgroup$ – Semiclassical Jun 17 '16 at 5:15
  • $\begingroup$ Your solution is prettier but maybe harder to do quickly in your head. $\endgroup$ – Funktorality Jun 17 '16 at 5:20
  • $\begingroup$ This answer copied pretty much what I did... $\endgroup$ – The Great Duck Jun 18 '16 at 20:36
  • $\begingroup$ @TheGreatDuck No, your first answer made use of the fallacy that $\det(A+B)=\det A+\det B$. $\endgroup$ – Funktorality Jun 18 '16 at 20:51
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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.

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Three main operations govern matrices in the elementary sense. Below are how each of these change the determinant:

Swapping two rows or columns => negates the determinant

Adding a multiple of one row or column to another => no change

Multiplying a column or row by a constant => multiply the derivative by the constant

By doing the following sequence of operations we change D1 to two mratrixes very close to D2:

Add rC1 to C3

Add qC3 to C2

Subtract qrC1 from C2

This gives a version of C2 with only a terms in the leftmost column.

Now using the algebraic rule a commentor mentioned, we see that this matrixes determinant (D1 by observation) will be added to the other matrixes determinant.

We will continue row operations where we left off...

Subtract C3/r from C1

Multiply C1 by r

Subtract C2/q from C1

Multiply C1 by qp

Now the second matrix has been built. By simple inspection we see it is equal to D1pqr.

By combining the two matrix determinants:

D2 = D1 + pqrD1

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    $\begingroup$ Determinant is not a linear operator, determinant is tri-linear in terms of the columns and rows $D(v_1 +u_1, v_2,v_3) = D ( v_1,v_2,v_3) + D( u_1,v_2,v_3)$ $\endgroup$ – clark Jun 17 '16 at 3:41
  • $\begingroup$ Hmm... Well then it is all flawed. The answer he gave could not be right under that premise. Are you sure? I thought the derivative was linear with regards to matrix additions? $\endgroup$ – The Great Duck Jun 17 '16 at 3:44
  • $\begingroup$ @clark i believe you are thinking of factoring out multiplication which would be "tri-linear" as you put it. $\endgroup$ – The Great Duck Jun 17 '16 at 3:46
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    $\begingroup$ For instance $$ 1=\left | \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array} \right | = \left | \begin{array}{cc} 0 & 0\\ 0 & 1 \end{array} \right |+ \left | \begin{array}{cc} 1 & 0\\ 0 & 0 \end{array} \right |=0 $$ $\endgroup$ – clark Jun 17 '16 at 3:46
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    $\begingroup$ Btw, I did not downvote. $\endgroup$ – clark Jun 17 '16 at 3:52

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