We're asked to find the determinant of
the given matrix
using the co-factor expansion method.
To find the determinant
using the co-factor expansion method,
we can select any row or any column
and we multiple each element in the row or column
by the corresponding co-factor,
and then we find the sum of all the products.
So it is worthwhile to analyze the matrix
and select the row or column that contains
the most zeros,
because that will reduce the amount
of work to find the determinant.
So analyzing the matrix,
notice row 4 contains
three 0 entries,
and therefore it'll be less work
to use to co-factor expansion method
along row 4.
We begin with the first element in row 4,
which is in row 4, column 1.
So we begin with 0,
and then because the element is in
row 4, column 1,
we multiply by -1,
raised to power 4 plus 1.
Next, we eliminate row 4 and column 1
from the matrix,
and multiply by the determinant
of the remaining matrix.
Notice how we have a 3 by 3 matrix,
where the first row is 0, 0, -2,
the second row is 0, 5, -5,
the third row is 1, 4, 0.
So we have the 3 by 3 determinant of this matrix,
shown here.
So here we have A sub 4 comma 1 times
the corresponding co-factor.
And now we move along to the next element in row 4.
So we have plus,
next element is 0,
this element is in row 4, column 2,
and therefore we multiply by
1 raised to the power of 4 plus 2,
and then we multiply by the determinant
of the matrix formed by eliminating row 4
and column 2.
Notice how that gives us the 3 by 3 determinant
shown here,
where the first row is -1, 0, -2,
the second is 1, 5, -5,
and the third row is 0, 4, 0.
And we move on to the next element in row 4,
which is the element of -5 in row 4, column 3.
So we have plus
5
times
1 raised to the power of 4 plus 3.
Again 4 plus 3 is the exponent because
the -5 is in row 4, column 3.
And then we multiply
by the determinant of the matrix formed
by eliminating row 4 and column 3.
Which gives us the 3 by 3 determinant shown here,
where the first row is -1, 0, -2,
the second row is 1, 0, -5,
and the third row is 0, 1, 0.
Now we move on to the last element in row 4,
which is the element in row 4, column 4,
which again is another 0.
So we have plus 0 times
1 raised to the power of 4 plus 4.
Again the exponent is 4 plus 4 because
the element is in row 4, column 4.
Then we multiply by the determinant
of the matrix formed by eliminating
row 4,
column 4.
Which leaves us with the 3 by 3 determinant,
shown here.
Where the first row is -1, 0, 0,
the second row is 1, 0, 5,
and the third row is 0, 1, 4.
Now the reason we selected row 4 is because
three of these products contain a 0,
reducing the amount of work that we have to do.
This first product is 0,
the second product is 0,
and the forth product is 0.
Which gives us
0 plus 0
plus -5 times
1 raised to the power of 4 plus 3
times the 3 by 3 determinant
plus 0.
So of course we can drop the zeros,
and since -5 times
1
raised to the power of 7 is equal to positive 5,
we are left with positive 5
times the 3 by 3 determinant.
So now we need to evaluate this 3 by 3 determinant,
and again we'll use the method of co-factor expansion.
Because row 3 has 2 zeros,
we will perform co-factor expansion
along row 3.
Now we need to be careful here,
it's going to be 5 times
the value of this determinant,
so we have 5, and then an open parenthesis
and then the element in row 3 column 1 is a 0,
so that's why we have a 0 here,
times -1 raised to the power of 3 plus 1
because the zero is in row 3, column 1.
Next we eliminate
row 3, column 1,
which gives us a 2 by 2 determinant shown here.
Where the first row is 0, -2,
and second row is 0, -5.
Now we move along to the second element in row 3,
which is a 1.
Which gives us plus 1 times
1 raised to the power of 3 plus 2.
The exponent is 3 plus 2 because
the element of 1 is in row 3, column 2.
Next we eliminate the row and column
of this element,
or row 3 and column 2.
Which gives us a 2 by 2 determinant,
where row 1 is -1,
2,
the second row is 1, -5.
We move along to the last element in row 3,
which is the 0 in row 3, column 3.
So we have plus 0
times -1 raised to the power of 3 plus 3.
And we eliminate the row and column
of this element, or
row 3 and column 3,
which gives us the remaining 2 by 2 determinant
where the first row is -1, 0,
and the second row is 1, 0.
So two of these three products will be 0.
Notice how the first product has a factor of 0,
so this will be 0.
And the third product has a factor of 0,
so this will be 0.
This leaves us with 5
and then times -1 to the fifth power is -1
so 1 times -1
is -1.
And then we have the 2 by 2 determinant.
To evaluate the 2 by 2 determinant,
we have the product of the main diagonal
minus the product of the minor diagonal,
which gives us
5
minus -2,
inside parenthesis.
And of course 5 minus -2 is equivalent to
5 plus 2, or 7.
Which gives us 5 times -1 times 7,
or 5 times -7,
which is -35.
So the value of the 4 by 4 determinant
is -35.
I hope you found this helpful.