The cofactors of elements for every determinant

[Tonia1]

Find the cofactors of the elements in the second row of every determinant:
$$\begin{vmatrix}-2 & 0 & 1 \\ 1 & 2 & 0 \\ 4 & 2 & 1 \end{vmatrix}$$
I am going to guess that I need to look at each number in the second horizontal row to see what i and j are for finding the cofactors of the elements. I am a bit confused as to where to start this problem at. I am familiar with evaluating a 2X2 determinant but finding a cofactor of an element, evaluating a determinant, and understanding Cramer's Rule, is very confusing to me right now.

 

[I like Serena]

Find the cofactors of the elements in the second row of every determinant:
$$\begin{vmatrix}-2 & 0 & 1 \\ 1 & 2 & 0 \\ 4 & 2 & 1 \end{vmatrix}$$
I am going to guess that I need to look at each number in the second horizontal row to see what i and j are for finding the cofactors of the elements. I am a bit confused as to where to start this problem at. I am familiar with evaluating a 2X2 determinant but finding a cofactor of an element, evaluating a determinant, and understanding Cramer's Rule, is very confusing to me right now.

Hi Tonia, welcome to MHB! ;)

Let's start with the cofactor of the first element in the second row.
First we find the so called minor, which is the 2x2 determinant when we remove both the row and the column of the element.
That is:
$$\begin{vmatrix}\cancel{-2} & 0 & 1 \\ \cancel 1 & \cancel 2 & \cancel 0 \\ \cancel 4 & 2 & 1 \end{vmatrix}
= \begin{vmatrix}0 & 1 \\ 2 & 1 \end{vmatrix}
$$
The cofactor is the minor multiplied by -1 if it's in an 'alternate' position.
Top left is +1 and the element below 'alternates' to -1.
So the cofactor for the first element in the second row is:
$$-\begin{vmatrix}0 & 1 \\ 2 & 1 \end{vmatrix}$$

 

[Tonia1]

Hi Tonia, welcome to MHB! ;)

Let's start with the cofactor of the first element in the second row.
First we find the so called minor, which is the 2x2 determinant when we remove both the row and the column of the element.
That is:
$$\begin{vmatrix}\cancel{-2} & 0 & 1 \\ \cancel 1 & \cancel 2 & \cancel 0 \\ \cancel 4 & 2 & 1 \end{vmatrix}
= \begin{vmatrix}0 & 1 \\ 2 & 1 \end{vmatrix}
$$
The cofactor is the minor multiplied by -1 if it's in an 'alternate' position.
Top left is +1 and the element below 'alternates' to -1.
So the cofactor for the first element in the second row is:
$$-\begin{vmatrix}0 & 1 \\ 2 & 1 \end{vmatrix}$$

Why does your example show that the first cofactor must be negative even though the answer in the book says positive 2? I do not know why I did not get positive 2 for the first cofactor, or positive 4 for the 3rd cofactor.

I followed your explanation and got this for the 1st cofactor:
I crossed out -2, 1, 4 and 1, 2, 0 and then got this for the 2X2 determinant:
= -/0 1 top row 2 1 bottom row/ = -(0)(1) - (1)(2) = -0-2 = -2 (should be positive 2 like the answer in the book).

I got this for the 2nd cofactor:
I crossed out 0, 2, 2 and 1, 2, 0 and then got this for the 2X2 determinant: - (Vertical bar here) -2 1 (top horizontal row) 4 1 (bottom horizontal row) = - /(-2)(1)-4(1)/
= -/-2-4/ = -/-6/ = -6 (same as answer shown in book).

I got this for the 3rd cofactor:
I crossed out 1, 0, 1 and 1, 2, 0 and then got this for the 2X2 determinant: - /-2 0 (top row) 4 2 (bottom row)/ = -/(-2)(2) - (4)(0)/ = -/-4 - 0/ = -4 (should be positive 4 like the answer in the book).

 

[I like Serena]

Why does your example show that the first cofactor must be negative even though the answer in the book says positive 2? I do not know why I did not get positive 2 for the first cofactor, or positive 4 for the 3rd cofactor.

I followed your explanation and got this for the 1st cofactor:
I crossed out -2, 1, 4 and 1, 2, 0 and then got this for the 2X2 determinant:
= -/0 1 top row 2 1 bottom row/ = -(0)(1) - (1)(2) = -0-2 = -2 (should be positive 2 like the answer in the book).

That should be:
$$ -\begin{vmatrix}0 & 1 \\ 2 & 1\end{vmatrix} = -\Big((0)(1) - (1)(2)\Big) = -(0-2) = -(-2) = 2$$
Then it has the proper sign doesn't it?

I got this for the 2nd cofactor:
I crossed out 0, 2, 2 and 1, 2, 0 and then got this for the 2X2 determinant: - (Vertical bar here) -2 1 (top horizontal row) 4 1 (bottom horizontal row) = - /(-2)(1)-4(1)/
= -/-2-4/ = -/-6/ = -6 (same as answer shown in book).

I got this for the 3rd cofactor:
I crossed out 1, 0, 1 and 1, 2, 0 and then got this for the 2X2 determinant: - /-2 0 (top row) 4 2 (bottom row)/ = -/(-2)(2) - (4)(0)/ = -/-4 - 0/ = -4 (should be positive 4 like the answer in the book).

Same thing. We have to be careful to put parentheses where they need to be.
$$-/-4 - 0/=-(-4 - 0)=-(-4)=4$$
Then we find $4$ don't we?

 

[Tonia1]

That should be:
$$ -\begin{vmatrix}0 & 1 \\ 2 & 1\end{vmatrix} = -\Big((0)(1) - (1)(2)\Big) = -(0-2) = -(-2) = 2$$
Then it has the proper sign doesn't it?
Same thing. We have to be careful to put parentheses where they need to be.
$$-/-4 - 0/=-(-4 - 0)=-(-4)=4$$
Then we find $4$ don't we?

Okay, thanks. For some reason, I was thinking that the bars were like an absolute value sign, but it's not the same.

 

[Tonia1]

That should be:
$$ -\begin{vmatrix}0 & 1 \\ 2 & 1\end{vmatrix} = -\Big((0)(1) - (1)(2)\Big) = -(0-2) = -(-2) = 2$$
Then it has the proper sign doesn't it?
Same thing. We have to be careful to put parentheses where they need to be.
$$-/-4 - 0/=-(-4 - 0)=-(-4)=4$$
Then we find $4$ don't we?

I still don't understand how the 2nd cofactor is -6. This is what I did:
-/-2 1 (top row) 4 1 (bottom row)/ = -[(-2)(1) - (4)(1)] = -[-2 -4] = -(-6)
Why is the answer supposed to be -6 instead of positive 6? the two negatives should make it positive.

 

[I like Serena]

I still don't understand how the 2nd cofactor is -6. This is what I did:
-/-2 1 (top row) 4 1 (bottom row)/ = -[(-2)(1) - (4)(1)] = -[-2 -4] = -(-6)
Why is the answer supposed to be -6 instead of positive 6? the two negatives should make it positive.

The signs are not all negative - they alternate.
The signs of the cofactors are like this:
$$\begin{vmatrix}+&-&+ \\ -&+&- \\ +&-&+\end{vmatrix}$$
The second element of the second row has a $+$ instead of a $-$.