Help finding the determinant of a 4x4 matrix?
By James Williams
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?
$\endgroup$ 17 Answers
$\begingroup$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 or rows, 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 (along the first row):$$ \det A = \begin{vmatrix} 5 & -7 & 2 & 2 \\ 0 & 3 & 0 & -4 \\ -5 & -8 & 0 & 3 \\ 0 & 5 & 0 & -6 \\ \end{vmatrix} = 5 \begin{vmatrix} 3 & 0 & -4 \\ -8 & 0 & 3 \\ 5 & 0 & -6 \\ \end{vmatrix} -(-7) \begin{vmatrix} 0 & 0 & -4 \\ -5 & 0 & 3 \\ 0 & 0 & -6 \\ \end{vmatrix} + 2 \begin{vmatrix} 0 & 3 & -4 \\ -5 & -8 & 3 \\ 0 & 5 & -6 \\ \end{vmatrix} -2 \begin{vmatrix} 0 & 3 & 0 \\ -5 & -8 & 0 \\ 0 & 5 & 0 \\ \end{vmatrix}. $$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,
$\endgroup$ $\begingroup$For M3 it is $-90+100$ rather than $-90-100$ which gives $10 \times 2=20$ as the final answer
$\endgroup$ 5 $\begingroup$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:
- 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 $$\det(A) = \det(a_1,\dots, a_i, \dots, a_j, \dots, a_n) = -\det(a_1,\dots, a_j, \dots, a_i, \dots, a_n)$$
- A common factor can be "pulled out" of a row/ column. $$\det(a_1,\dots,ka_i,\dots,a_n) = k\det(a_1,\dots,a_i,\dots,a_n)$$
- 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.
$\endgroup$ $\begingroup$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 $$ \det(A)=\det(A') =\left|\matrix{5&2\\-5&0}\right|\times\left|\matrix{3&-4\\5&-6}\right| =10\times2=20. $$
$\endgroup$ $\begingroup$New Method assume this Matrics\begin{bmatrix} 1&2&2&-3\\4&2&5&0\\1&3&4&2\\-3&1&-1&1 \end{bmatrix}
rewrite the matrix after changing order of columns of original matrix from 1234 to 1342
\begin{bmatrix} 1&2&-3&2\\4&5&0&2\\1&4&2&3\\-3&-1&1&1 \end{bmatrix}
rewrite the matrix after changing order of columns of original matrix from 1234 to 1423
\begin{bmatrix}1&-3&2&2\\4&0&2&5\\1&2&3&4\\-3&1&1&-1 \end{bmatrix}
after that for each matrix divide it to 4 2x2 matrices and find the det of each normally to reach those in respective order
\begin{bmatrix} -6&15\\10&6\end{bmatrix}\begin{bmatrix} -3&-6\\11&-1\end{bmatrix}\begin{bmatrix} 12&6\\7&-7\end{bmatrix}
Now convert each of those 3 to values according to the rule of 2x2 determinant but with adding not subtracting for example the first one (-6)(6)+(15)(10)=114second one (-3)(-1)+(-6)(11)=-63Third one (12)(-7)+(7)(6)=-42
finally the det of the original matrix = 114-68-42=9
check my method it is my pure original work can not be found in today's books
$\endgroup$ $\begingroup$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.
$\endgroup$ 3 $\begingroup$You are all wrong. The determinant of this matrix is 380. You can check it here:
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