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Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose the row that they chose. This bothers me. As I should be able to do it as I see fit. I will post my work and someone point out the problem in my work. The matrix is as follows:

$$A = \left( \begin{matrix} 5&-7&2&2\\ 0&3&0&-4\\ -5&-8&0&3\\ 0&5&0&-6\\ \end{matrix} \right) $$

I decided to expand across row one and cross out columns as I found the minors. For the first minor obtaining: $$ \begin{pmatrix} 3 & 0 & -4 \\ -8 & 0 & 3 \\ 5 & 0 & -6 \\ \end{pmatrix} $$

M1 being row one column one we attain $-1^2 = 1$. This is to be multiplied by the determinate of the minor. Now finding the determinant I did:

3 times $$ \begin{pmatrix} 0 & 3 \\ 0 & -6 \\ \end{pmatrix} $$ giving $3(0-0)= 0$ then:

0 times $$ \begin{pmatrix} -8 & 3\\ 5 & -6\\ \end{pmatrix} $$

giving 0(48-15)=0

Then: 4 times $$ \begin{pmatrix} -8 & 0 \\ 5 & 0 \\ \end{pmatrix} $$ giving $4(0-0)=0$ adding the determinants we get $0+0+0=0$ So det M1 $= 0(1) = 0$

M2--> M(1,2)---> $-1^1+2= -1^3 = -1$

$$ \begin{pmatrix} 0 & 0 & -4 \\ -5 & 0 & 3 \\ 0 & 0 & -6 \\ \end{pmatrix} $$

o* $$ \begin{pmatrix} 0 & 3 \\ 0 & -6 \\ \end{pmatrix} $$ giving $0(0-0)=0$

obviously the next matrix will look the same as the top term in column two is a zero so the determinant for that will be $0$. Now finally

4 times $$ \begin{pmatrix} -8 & 0 \\ 5 & 0 \\ \end{pmatrix} $$ giving 4(0-0)= 0

So the Determinant of Minor 2 is (0+0+0)(-1)= 0 Now on to Minor number 3

M3 --> $-1^4 = 1$

$$ \begin{pmatrix} 0 & 3 & -4 \\ -5 & -8 & 3 \\ 0 & 5 & -6 \\ \end{pmatrix} $$

for the determinant:

0 times $$ \begin{pmatrix} -8 & 3 \\ 5 & -6 \\ \end{pmatrix} $$ which gives $0(48-15)=0$

-3 times $$ \begin{pmatrix} -5& 3 \\ 0 & -6 \\ \end{pmatrix} $$

which gives $-3(30-0)= -90$

it is redundant to go on from here because after the final computation for this minor I get -100 and as a result get det M3 = -190 and get determinant of zeros for the following determinant of M4. which gives: $0(5)+ 0(-7) + (-90)(2) + (0)(2)$ giving Det Ax $= -380.$ The book says its $20$ and when I did it in a calculator it got 20 but the problem is that both the book and calculator expand across the row with the most zeros but theoretically speaking NO MATTER WHICH row or column you choose to expand across you should get the same answer. So what is it? Is my computation wrong or is my assumption that you can expand across any row or column wrong? Isn't it only important if the determinant doesn't equal zero? or does the exact value matter in more advanced cases?

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2  
Expansion on any row or column is OK. But one has to be careful, e.g. the sign pattern used is always for the present (sub) determinant, the signs don't just come from the original matrix – coffeemath Jan 4 at 2:00
up vote 21 down vote accepted

Here is how you should write it down in practice. $$ A = \begin{pmatrix} 5 & -7 & 2 & 2 \\ 0 & 3 & 0 & -4 \\ -5 & -8 & 0 & 3 \\ 0 & 5 & 0 & -6 \\ \end{pmatrix} $$ so that $$ \det A = \begin{vmatrix} 5 & -7 & 2 & 2 \\ 0 & 3 & 0 & -4 \\ -5 & -8 & 0 & 3 \\ 0 & 5 & 0 & -6 \\ \end{vmatrix} = 2 \cdot \begin{vmatrix} 0 & 3 & -4 \\ -5 & -8 & 3 \\ 0 & 5 & -6 \\ \end{vmatrix} = 2 \cdot -(-5) \cdot \begin{vmatrix} 3 & -4 \\ 5 & -6 \\ \end{vmatrix} = 2 \cdot -(-5) \cdot (-18 -(-20)) = 20. $$ I expanded along the third column in the first case and then along the first column, because these are the ones with the most zeros, so it saves a lot of computations. I am assuming that's what's in your book.

If you want to expand along other columns, just keep track of the appropriate minors without making any computation mistakes (I guess that's the hard part ; the only trick is to make it slowly and be careful). So here we go : $$ \det A = \begin{vmatrix} 5 & -7 & 2 & 2 \\ 0 & 3 & 0 & -4 \\ -5 & -8 & 0 & 3 \\ 0 & 5 & 0 & -6 \\ \end{vmatrix} = 5 \begin{vmatrix} 3 & 0 & -4 \\ -8 & 0 & 3 \\ 5 & 0 & -6 \\ \end{vmatrix} -(-7) \begin{vmatrix} 0 & 0 & -4 \\ -5 & 0 & 3 \\ 0 & 0 & -6 \\ \end{vmatrix} + 2 \begin{vmatrix} 0 & 3 & -4 \\ -5 & -8 & 3 \\ 0 & 5 & -6 \\ \end{vmatrix} -2 \begin{vmatrix} 0 & 3 & 0 \\ -5 & -8 & 0 \\ 0 & 5 & 0 \\ \end{vmatrix}. $$ I don't know how you defined the determinant, but in any definition you chose it should be obvious that a determinant of a matrix with a column of zeros is zero. (If it is not clear to you, feel free to tell me your definition and I will happily answer in the comments.) So the only non-zero term in this sum is $$ 2 \begin{vmatrix} 0 & 3 & -4 \\ -5 & -8 & 3 \\ 0 & 5 & -6 \\ \end{vmatrix}. $$ Note that this is the same determinant as when I expanded along the third column. So you can finish this computation by looking at what I did before.

A useful trick to remember the signs in the Laplace expansion (that's the name of the trick of expanding along a row or a column) is the following matrix : $$ \begin{vmatrix} + & - & + & - \\ - & + & - & + \\ + & - & + & - \\ - & + & - & + \end{vmatrix} $$ It works for any determinant size, just make sure that the coordinate of the matrix in the top left is a $+$ sign.

Hope that helps,

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For M3 it is $-90+100$ rather than $-90-100$ which gives $10 \times 2=20$ as the final answer

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why is it -90 +100? could you show your work? That would be much appreciated – K. Gibson Jan 4 at 19:59
    
just for m3 thanks – K. Gibson Jan 4 at 20:00
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@K.Gibson $$-4\begin{vmatrix}-5 & -8 \\ 0 & 5\end{vmatrix} = -4\big((-5)(5)-(-8)(0)\big) = -4(-25)=100\ne -100$$ So you just made a sign error there. – Bye_World Jan 4 at 20:03
    
I see but I’m still not following according to the sign matrix it should be positive if im not incorrect: $$ \begin{matrix} + & - & + \\ - & + & - \\ + & - & + \\ \end{matrix} $$ So do you see where my discrepancy is since -4 is in the top right and it should positive? I got it wrong another problem too. Right number wrong sign so it must be me I’m just not understanding something. – K. Gibson Jan 4 at 21:36
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@K.Gibson + sign means the number should keep unchanged, so -4 is still -4, otherwise it will be -(-4) under - sign – Steve X Jan 4 at 21:50

Steve explained where you made a mistake in your calculations. And Patrick explained how you can save computations by judiciously choosing the rows/ columns you expand along. Just for fun, I'll explain a different way of evaluating the determinant. I'm just going to use the relationship between the elementary row/ column operations and the determinant.

Here are those relationships:

  1. Swapping two rows/ columns of a matrix will give a factor of $-1$ to the determinant. Let $a_k$ be the $k$th row (or column) of the matrix $A$. Then $$\det(A) = \det(a_1,\dots, a_i, \dots, a_j, \dots, a_n) = -\det(a_1,\dots, a_j, \dots, a_i, \dots, a_n)$$
  2. A common factor can be "pulled out" of a row/ column. $$\det(a_1,\dots,ka_i,\dots,a_n) = k\det(a_1,\dots,a_i,\dots,a_n)$$
  3. Adding a scalar muliple of one row/ column to another will not change the determinant at all. $$\det(a_1,\dots, a_i, \dots, a_j, \dots, a_n) = \det(a_1,\dots, a_i, \dots, a_j+ka_i, \dots, a_n)$$

Let's use these properties of the determinant to calculate the determinant of your matrix:

$$\begin{align}\begin{vmatrix} 5&-7&2&2\\ 0&3&0&-4\\ -5&-8&0&3\\ 0&5&0&-6\\ \end{vmatrix} &= \enspace\ \frac 12\begin{vmatrix} 5 & -7 & 2 & 2 \\ 0 & 6 & 0 & -8 \\ 0 & -15 & 2 & 5 \\ 0 & -1 & 0 & 2\end{vmatrix} &{\begin{pmatrix}R_2 \to 2R_2 \\ R_3 \to R_3+R_1 \\ R_4 \to R_4-R_2\end{pmatrix}} \\ &= -\frac 12\begin{vmatrix} 5 & -7 & 2 & 2 \\ 0 & -1 & 0 & 2 \\ 0 & -15 & 2 & 5 \\ 0 & 6 & 0 & -8\end{vmatrix} & \begin{pmatrix}R_2 \leftrightarrow R_4\end{pmatrix} \\ &= -\frac 12\begin{vmatrix} 5 & -7 & 2 & 2 \\ 0 & -1 & 0 & 2 \\ 0 & 0 & 2 & -25 \\ 0 & 0 & 0 & 4\end{vmatrix} & \begin{pmatrix}R_3\to R_3-15R_2 \\ R_4\to R_4+6R_2\end{pmatrix} \\ &\stackrel{(*)}= -\frac 12(5)(-1)(2)(4) \\ &= \enspace\ 20\end{align}$$

where $(*)$ is due to the fact that the determinant of a triangular matrix is the product of the diagonal elements.

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If you look at where the $0$'s are in your matrix, it is fairly easy to see that your matrix becomes block-triangular when transformed to (for instance) the ordered basis $[e_1,e_3,e_2,e_4]$, in other words the standard basis but with the second and third vectors interchanged. The change of basis swaps the second and third rows and the second and third columns, and gives you $$A'= \pmatrix{5&2&-7&2\\-5&0&-8&3\\0&0&3&-4\\0&0&5&-6}. $$ Since change of basis does not affect the determinant, and the determinant of a block-triangular matrix is the product of the determinants of the diagonal blocks, you get $$ \det(A)=\det(A') =\left|\matrix{5&2\\-5&0}\right|\times\left|\matrix{3&-4\\5&-6}\right| =10\times2=20. $$

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Write out the determinant. Alongside write the first 3 columns"Downward diagonal multiplication" gives ( O + O + 2OO + O ). "Upward diagonal multiplication" gives -(O + O + O + 18O) . det = 2O.

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The method is an extension of the "Diagonal Multiplication" used in evaluating a 2x2 determinant. – c.sidney Jan 4 at 16:50
    
First off, please use $0$ when you mean the number $0$. Do not use $O$ for this is the capital letter which has no inherent mathematical meaning. Secondly, your answer would benefit from a more thorough explanation (with matrices written out!) of what you are attempting to do. – NoseKnowsAll Jan 4 at 16:52
    
I downvoted because this technique only works for $2 \times 2$ and $3 \times 3$ matrices (because in this case all the possible permutations of two or three elements are involved using diagonals, in the $3 \times 3$ case you need to "extend" the matrix though to allow permutations with negative sign). In the $4 \times 4$ case this method does not work in general ; there are $24$ terms involved, not $8$. – Patrick Da Silva Jan 9 at 1:36

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