# The cofactors of elements for every determinant

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• Tonia1
In summary: 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).In summary, when solving a 2x2 determinant, you first find the minor, then the cofactor of the element in the second row. If the cofactor is negative, you must reverse the sign before solving for it.
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.

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Tonia said:
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:
$$-\begin{vmatrix}0 & 1 \\ 2 & 1 \end{vmatrix}$$

I like Serena said:
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).

Tonia said:
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?

Tonia said:
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?

I like Serena said:
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.

I like Serena said:
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.

Tonia said:
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 $-$.

## 1. What are cofactors and how do they relate to determinants?

Cofactors are numbers that are used to calculate the determinant of a matrix. They are found by using a specific formula involving the elements of the matrix, and they help determine the size and properties of the matrix.

## 2. How do I find the cofactors of a matrix?

To find the cofactors of a matrix, you must first calculate the minor of each element in the matrix. Then, you use the formula (-1)^{i+j}M_{ij} to find the cofactor of each element, where i and j represent the row and column of the element, respectively.

## 3. What is the significance of cofactors in solving systems of equations?

Cofactors are important in solving systems of equations because they allow us to calculate the determinant of a matrix, which is necessary for solving systems of equations. In addition, the value of the cofactors can provide information about the nature of the solutions to the system of equations.

## 4. Can the cofactors of a matrix be negative?

Yes, the cofactors of a matrix can be negative. This is because the sign of the cofactor is determined by the position of the element in the matrix, and not the value of the element itself.

## 5. How do cofactors differ from determinants?

Cofactors are individual values that are used to calculate the determinant of a matrix, whereas the determinant itself is a single number that represents the overall size and properties of the matrix. In other words, cofactors are components of the determinant.

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