Minors and Cofactors

Minors

The concept of minors is fundamental to evaluating determinants of higher orders.

Definition: The minor of an element $a_{ij}$ of a determinant is the determinant obtained by deleting the $i^{th}$ row and the $j^{th}$ column in which the element $a_{ij}$ lies. The minor of $a_{ij}$ is denoted by $M_{ij}$.

Important Note: If a determinant is of order $n$ ($n \ge 2$), then the minor of any element is a determinant of order $n-1$.

For example, consider $$\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}.$$ The element $a_{21}$ is $4$, which lies in the second row and first column. To find its minor, we delete row 2 and column 1. The remaining determinant is $$M_{21} = \begin{vmatrix} 2 & 3 \\ 8 & 9 \end{vmatrix} = (2 \times 9) - (3 \times 8) = 18 - 24 = -6.$$ So, the minor of $a_{21}$ is $-6$.

This idea is very important because minors are the building blocks used to define cofactors and to expand determinants of order 3 or higher.

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Cofactors

Cofactors are closely related to minors, but with an additional sign attached depending on the position of the element.

Definition: The cofactor of an element $a_{ij}$, denoted by $A_{ij}$, is defined as $$A_{ij} = (-1)^{i+j} M_{ij},$$ where $M_{ij}$ is the minor of $a_{ij}$.

This means:

• If $i+j$ is even, then $(-1)^{i+j}=+1$, so $A_{ij}=M_{ij}$.
• If $i+j$ is odd, then $(-1)^{i+j}=-1$, so $A_{ij}=-M_{ij}$.

The signs of cofactors in a $3 \times 3$ determinant follow the checkerboard pattern $$\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}.$$

Using the previous example, we found that $$M_{21} = -6.$$ Now, $$A_{21} = (-1)^{2+1}M_{21} = (-1)^3(-6) = -1 \times (-6) = 6.$$ So the cofactor of $a_{21}$ is $6$.

Thus, a cofactor is simply the minor with the appropriate sign attached.

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Properties involving Elements and Cofactors

Let $$\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$$ be a determinant of order 3.

Property 1 (Expansion of a Determinant)

The sum of the products of the elements of any row (or any column) with their corresponding cofactors is equal to the value of the determinant.

For example:

• Expanding along the first row, $$\Delta = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}.$$
• Expanding along the second column, $$\Delta = a_{12}A_{12} + a_{22}A_{22} + a_{32}A_{32}.$$

Property 2 (Zero Property)

If the elements of one row (or one column) are multiplied by the cofactors of another row (or another column), then the sum is zero.

For example, if we take the elements of the first row and the cofactors of the second row, then $$a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0.$$ Similarly, $$a_{21}A_{11} + a_{22}A_{12} + a_{23}A_{13} = 0.$$

This property is extremely useful in determinant theory and later in matrices, especially while proving identities and finding inverses of matrices.

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Example 1: Minor and Cofactor in a 2x2 Matrix

Find the minor and cofactor of the element $6$ in the determinant $\Delta = \begin{vmatrix} 1 & -2 \\ 4 & 6 \end{vmatrix}$.

Solution: The element $6$ is in the second row and second column, so it is $a_{22}$.

To find the minor $M_{22}$, delete the second row and second column. The only remaining entry is $1$, so $$M_{22} = |1| = 1.$$ Now use the cofactor formula: $$A_{22} = (-1)^{2+2}M_{22} = (-1)^4 \cdot 1 = 1 \cdot 1 = 1.$$

Answer: $$M_{22} = 1, \qquad A_{22} = 1.$$

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Example 2: Finding all Minors and Cofactors of a 2x2 Matrix

Find all the minors and cofactors of the determinant $\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$.

Solution: The elements are $$a_{11}=2, \quad a_{12}=-4, \quad a_{21}=0, \quad a_{22}=3.$$

Now find the minors: $$M_{11} = |3| = 3,$$ $$M_{12} = |0| = 0,$$ $$M_{21} = |-4| = -4,$$ $$M_{22} = |2| = 2.$$

Next, find the cofactors using $A_{ij}=(-1)^{i+j}M_{ij}$: $$A_{11} = (-1)^{1+1}M_{11} = (+1)(3) = 3,$$ $$A_{12} = (-1)^{1+2}M_{12} = (-1)(0) = 0,$$ $$A_{21} = (-1)^{2+1}M_{21} = (-1)(-4) = 4,$$ $$A_{22} = (-1)^{2+2}M_{22} = (+1)(2) = 2.$$

Answer: Minors: $$M_{11}=3, \quad M_{12}=0, \quad M_{21}=-4, \quad M_{22}=2.$$ Cofactors: $$A_{11}=3, \quad A_{12}=0, \quad A_{21}=4, \quad A_{22}=2.$$

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Example 3: Minor and Cofactor in a 3x3 Matrix

Find the minor and cofactor of the element $a_{32}$ in the determinant $\begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}$.

Solution: The element $a_{32}$ means the entry in the third row and second column, which is $5$.

To find the minor $M_{32}$, delete row 3 and column 2. The remaining $2 \times 2$ determinant is $$M_{32} = \begin{vmatrix} 2 & 5 \\ 6 & 4 \end{vmatrix} = (2 \times 4) - (5 \times 6) = 8 - 30 = -22.$$ Now compute the cofactor: $$A_{32} = (-1)^{3+2}M_{32} = (-1)^5(-22) = -1 \times (-22) = 22.$$

Answer: $$M_{32} = -22, \qquad A_{32} = 22.$$

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Example 4: Evaluating Determinant using Cofactors of a Row

Evaluate the determinant $\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$ using the cofactors of the elements of the second row.

Solution: To expand along the second row, use $$\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}.$$ Here, $$a_{21}=2, \quad a_{22}=0, \quad a_{23}=1.$$

Now calculate the cofactors: $$A_{21} = (-1)^{2+1} \begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = - \big((3)(3) - (8)(2)\big) = -(9-16)=7,$$ $$A_{22} = (-1)^{2+2} \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = +(15-8)=7,$$ $$A_{23} = (-1)^{2+3} \begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = -\big((5)(2)-(3)(1)\big)=-(10-3)=-7.$$ Now substitute: $$\Delta = (2)(7) + (0)(7) + (1)(-7) = 14 + 0 - 7 = 7.$$

Answer: $$\Delta = 7.$$

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Example 5: Evaluating Determinant using Cofactors of a Column

Evaluate $\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}$ using cofactors of elements of the third column.

Solution: Expanding along the third column, $$\Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}.$$ Here, $$a_{13}=yz, \quad a_{23}=zx, \quad a_{33}=xy.$$

Now compute the cofactors: $$A_{13} = (-1)^{1+3} \begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix} = +(z-y)=z-y,$$ $$A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix} = -(z-x)=x-z,$$ $$A_{33} = (-1)^{3+3} \begin{vmatrix} 1 & x \\ 1 & y \end{vmatrix} = +(y-x)=y-x.$$ So, $$\Delta = yz(z-y) + zx(x-z) + xy(y-x).$$ Expanding, $$\Delta = yz^2 - y^2z + x^2z - xz^2 + xy^2 - x^2y.$$ Now factor systematically: $$\Delta = x^2(z-y) + y^2(x-z) + z^2(y-x).$$ This standard cyclic expression factorizes as $$\Delta = (x-y)(y-z)(z-x).$$

We can also verify the sign by checking a simple case. Let $x=1, y=2, z=3$. Then the determinant becomes 2, and $$(x-y)(y-z)(z-x) = (1-2)(2-3)(3-1) = (-1)(-1)(2)=2,$$ so the factorization is correct.

Answer: $$\Delta = (x-y)(y-z)(z-x).$$

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Example 6: Verifying the Zero Property

For the determinant $\Delta = \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}$, verify that $a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} = 0$.

Solution: From the first row, $$a_{11}=2, \quad a_{12}=-3, \quad a_{13}=5.$$ We now calculate the cofactors of the third row.

First, $$A_{31} = (-1)^{3+1} \begin{vmatrix} -3 & 5 \\ 0 & 4 \end{vmatrix} = +( (-3)(4) - (5)(0) ) = -12.$$ Second, $$A_{32} = (-1)^{3+2} \begin{vmatrix} 2 & 5 \\ 6 & 4 \end{vmatrix} = -\big((2)(4) - (5)(6)\big) = -(8-30)=22.$$ Third, $$A_{33} = (-1)^{3+3} \begin{vmatrix} 2 & -3 \\ 6 & 0 \end{vmatrix} = +( (2)(0) - (-3)(6) ) = 18.$$ Now evaluate the expression: $$a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} = (2)(-12) + (-3)(22) + (5)(18).$$ Compute each term: $$= -24 - 66 + 90 = -90 + 90 = 0.$$ Hence the result is zero, which verifies the property that elements of one row multiplied by cofactors of another row sum to zero.

Answer: The expression is equal to $0$.

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Example 7: Cofactors of an Identity Matrix

Find all cofactors of the identity matrix $I_3 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$.

Solution: For the identity matrix, $$I_3 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}.$$

Now find each cofactor: $$A_{11} = (+1) \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1,$$ $$A_{12} = (-1) \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = 0,$$ $$A_{13} = (+1) \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} = 0,$$ $$A_{21} = (-1) \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = 0,$$ $$A_{22} = (+1) \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1,$$ $$A_{23} = (-1) \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} = 0,$$ $$A_{31} = (+1) \begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix} = 0,$$ $$A_{32} = (-1) \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} = 0,$$ $$A_{33} = (+1) \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1.$$ Thus the cofactor matrix is itself the identity matrix.

Answer: $$A_{11}=1, \quad A_{22}=1, \quad A_{33}=1,$$ and all other cofactors are $0$.

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Example 8: Minor of a Diagonal Element

In the diagonal matrix $D = \begin{vmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{vmatrix}$, find the minor $M_{22}$.

Solution: The element $a_{22}$ is $b$. To find $M_{22}$, delete the second row and second column. The remaining determinant is $$M_{22} = \begin{vmatrix} a & 0 \\ 0 & c \end{vmatrix}.$$ Now evaluate: $$M_{22} = ac - 0 = ac.$$

Answer: $$M_{22} = ac.$$

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Example 9: Minors and Cofactors involving Trigonometry

Find $M_{12}$ and $A_{12}$ for $\Delta = \begin{vmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{vmatrix}$.

Solution: The element $a_{12}$ is $\sin\theta$, which lies in the first row and second column.

To find the minor $M_{12}$, delete row 1 and column 2. The remaining entry is $-\sin\theta$. Therefore, $$M_{12} = |-\sin\theta| = -\sin\theta.$$ Now find the cofactor: $$A_{12} = (-1)^{1+2}M_{12} = (-1)^3(-\sin\theta) = -1 \times (-\sin\theta) = \sin\theta.$$

Answer: $$M_{12} = -\sin\theta, \qquad A_{12} = \sin\theta.$$

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Example 10: Theoretical Application

If $\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$ and $A_{ij}$ is the cofactor of $a_{ij}$, write the expression for the determinant $\Delta$ expanding along the third column.

Solution: The elements of the third column are $$a_{13}, \quad a_{23}, \quad a_{33}.$$ Their corresponding cofactors are $$A_{13}, \quad A_{23}, \quad A_{33}.$$ By the cofactor expansion property, the determinant expanded along the third column is $$\Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}.$$

Answer: $$\Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}.$$