Example 1 Compute the
determinant of each of the following matrices.
Solution
We don’t really
need to sketch in the diagonals for 2×2 matrices. The determinant is simply the product of the diagonal
running left to right minus the product of the diagonal running from right to left. So, here is the determinant for this matrix. The only thing we need to
worry about is paying attention to minus signs. It is easy to make a mistake
with minus signs in these computations if you aren’t paying attention.
det (A) = (3)(5) − (2)(−9) = 33.
Okay, with this
one we’ll copy the two columns over and sketch in the diagonals to make sure
we’ve got the idea of these down.
Now, just
remember to add products along the left to right diagonals and subtract
products along the right to left diagonals.
det (B)
= (3)(–1)(7) + (5)(8)(–11) + (4)(–2)(1) – (5)(–2)(7) – (3)(8)(1) – (4)(–1)(–11)
=
– 467
We’ll do this
one with a little less detail. We’ll copy the columns but not bother to
actually sketch in the diagonals this time.
= (2)(–8)(1) + (–6)(3)(–3)
+ (2)( 2)(1) – (–6)(2)(1) – (2)(3)(1) – (2)(–8)(–3)
= 0
Example 2 Compute the determinant of each of
the following matrices.
Solution
Here are these
determinants.
det (A) = (5)(–3)(4) = – 60
det (B)
= (6)( – 1) = – 6
det (C)
= (10)(0)(6)(5) = 0
Example
3 For the following matrix compute the cofactors C12
, C24 , and C32.
Solution
In order to
compute the cofactors we’ll first need the minor associated with each cofactor.
Remember that
in order to compute the minor we will remove the ith row and jth
column of A.
So, to compute M12
(which we’ll need for C12) we’ll need to compute the
determinate of the matrix we get by removing the 1st row and 2nd
column of A. Here is that work.
We’ve marked
out the row and column that we eliminated and we’ll leave it to you to verify
the determinant computation. Now we can get the cofactor.
C12 = (– 1)1 + 2 M12 = (– 1)3
(160) = – 160.
Let’s now move
onto the second cofactor. Here is the work for the minor.
The cofactor in
this case is,
C24 = (– 1)2 + 4 M24
= (– 1)6 (508) = 508
Here is the work for the final cofactor.
C32 = (– 1)3 + 2 M32
= (– 1)5 (150) = – 120.
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