### Finding 2x2, 3x3 and even 5x5 Determinants (Ex1)

We'll show you how to find 2x2 and 3x3 determinants. Finding 3x3 determinants is just finding multiple 2x2 determinants, so we'll start with an example of a 2x2 determinant.

Find the determinant of $$A = \left[ {\begin{array}{*{20}{c}}1&7\\4&5\end{array}} \right]$$.

You just need to subtract the products of the diagonals. Be careful not to do the wrong order by mistake.

$\det (A) = (1)(5) - (4)(7)$

$\det (A) = - 23$

Easy! Now let's find the determinant of a 3x3 matrix. Find the determinant of $$A = \left[ {\begin{array}{*{20}{c}}1&7&3\\4&5&5\\7&8&5\end{array}} \right]$$.

What you need to do, is make a path going through any row or column. Then for each individual element in the path cover the row and column and find the determinant of the 2x2 matrix left. Be careful, you need to use the following cofactor matrix:

$\left[ {\begin{array}{*{20}{c}} + & - & + \\ - & + & - \\ + & - & + \end{array}} \right]$

That sounds wordy, but let's go across the first row, 1 7 and 3. See the matrix above, the first is added, second subtracted, and third is added again.

$\det (A) = 1\det \left( {\left[ {\begin{array}{*{20}{c}}5&5\\8&5\end{array}} \right]} \right) - 7\det \left( {\left[ {\begin{array}{*{20}{c}}4&5\\7&5\end{array}} \right]} \right) + 3\det \left( {\left[ {\begin{array}{*{20}{c}}4&5\\7&8\end{array}} \right]} \right)$

If you're confused how we got this, we start with covering the column and row in "1", so what's left is that first 2x2 determinant, then we cover up the column and row in "7" and that's subtracted, then we add the determinant left from covering the row and column of "3".

That means there are a lot of ways to find the 3x3 determinant of a matrix, you can go across any row or column. If you use the middle row to find the determinant, you get:

$\det (A) = - 4\det \left( {\left[ {\begin{array}{*{20}{c}}7&3\\8&5\end{array}} \right]} \right) + 5\det \left( {\left[ {\begin{array}{*{20}{c}}1&3\\7&5\end{array}} \right]} \right) - 5\det \left( {\left[ {\begin{array}{*{20}{c}}1&7\\7&8\end{array}} \right]} \right)$

That means that if you see a certain part of a matrix has a bunch of zeros, then you can use that to make computations easier. But for now, let's solve this using the first row we found earlier:

$\det (A) = 1\det \left( {\left[ {\begin{array}{*{20}{c}}5&5\\8&5\end{array}} \right]} \right) - 7\det \left( {\left[ {\begin{array}{*{20}{c}}4&5\\7&5\end{array}} \right]} \right) + 3\det \left( {\left[ {\begin{array}{*{20}{c}}4&5\\7&8\end{array}} \right]} \right)$

$\det (A) = 1( - 15) - 7( - 15) + 3( - 3)$

$\det (A) = 81$

And it's just that easy. Let's do another example that shows how little work you need to do if you take advantage of 0's in a matrix.

Find the determinant. $$A = \left[ {\begin{array}{*{20}{c}}1&7&3\\4&5&0\\7&8&0\end{array}} \right]$$

Remember the signs: $$\left[ {\begin{array}{*{20}{c}} + & - & + \\ - & + & - \\ + & - & + \end{array}} \right]$$

$\det (A) = 3\det \left( {\left[ {\begin{array}{*{20}{c}}4&5\\7&8\end{array}} \right]} \right) - 0\det \left( {\left[ {\begin{array}{*{20}{c}}1&7\\7&8\end{array}} \right]} \right) + 0\det \left( {\left[ {\begin{array}{*{20}{c}}1&7\\4&5\end{array}} \right]} \right)$

Since we multiply by the number we're traveling along, we now don't need to find those two 2x2 determinants. This reduces our work considerably! You could've indeed used any other row or column to get the determinant, but look how much easier to do this!

$\det (A) = 3( - 3)$

$\det (A) = - 9$

It's possible to find the determinant of a 4x4 matrix by reducing it into smaller 3x3 determinants, just like how we turned a 3x3 determinant into a bunch of 2x2 determinants.

A good tricky question on an exam is for your teacher to give a very high-ordered matrix, which you can only solve within time limits by purposely taking advantage of the 0's in a matrix. There was once a 7x7 matrix I was asked to find the determinant of, but let's do a 5x5 to keep things under control. To do this determinant under time constraints, you need to spot 0's and use them! Don't forget the plus and minus cofactor matrix!

Find the determinant, $$A = \left[ {\begin{array}{*{20}{c}}1&2&{18}&4&9\\4&6&2&3&0\\3&0&3&2&0\\4&0&8&0&0\\9&0&4&0&0\end{array}} \right]$$

Immediately make use of the last column of 0's. Now the determinant of the 5x5 matrix is made up of 4x4 matrices, or 1 4x4 matrix in this case.

$\det (A) = 9\det \left( {\left[ {\begin{array}{*{20}{c}}4&6&2&3\\3&0&3&2\\4&0&8&0\\9&0&4&0\end{array}} \right]} \right)$

Now use the second column of 0's. Remember, you put your hand over the 6's column and row to get the 3x3 matrix. Remember the positive and negative cofactor matrix, 6 here is negative.

$\det (A) = (9)( - 6)\det \left( {\left[ {\begin{array}{*{20}{c}}3&3&2\\4&8&0\\9&4&0\end{array}} \right]} \right)$

Now we make use of the last row of zeros. 2 here is positive according to the positive/negative cofactor matrix.

$\det (A) = (9)( - 6)(2)\det \left( {\left[ {\begin{array}{*{20}{c}}4&8\\9&4\end{array}} \right]} \right)$

$\det (A) = (9)( - 6)(2)[(4)(4) - (8)(9)]$

$\det (A) = 6048$

And there you have it!

Triple product examples (dot product of a cross product), including finding the volume of a parallelpiped:

Triple Product (Dot Product and Cross Product) Example 1

Triple Product (Dot Product and Cross Product) Example 2

Volume of Parallelpiped Triple Product Example 3

Volume of Parallelpiped Triple Product Example 4