Minors and cofactors

Every entry of a determinant hides a smaller determinant inside it. Delete the row and column that entry sits in, and what remains is its minor; attach a checkerboard sign to that minor and you get its cofactor. These two ideas, sitting in the CBSE Class 12 Determinants chapter right after the $3\times3$ expansion formula and right before the adjoint of a matrix, are not extra vocabulary bolted onto material you already know — they are that material, renamed and made precise. The same terms $a_1\begin{vmatrix}b_2&c_2\\b_3&c_3\end{vmatrix}$ you have already been computing are minors, and the alternating $+,-,+$ signs you have already been applying are cofactor signs. Naming them separately matters because cofactors — not raw minors — are exactly the ingredients that get assembled into the adjoint of a matrix, the very next topic, where the sign is not optional.

The core insight

The cofactor of an entry $a_{ij}$ is its signed minor, $C_{ij} = (-1)^{i+j}M_{ij}$ , where $M_{ij}$ is the determinant left after deleting row $i$ and column $j$ , and a determinant equals the sum of any row's (or column's) entries times their cofactors.

The core idea

Minor $M_{ij}$ . For a square matrix, delete the row and the column that the entry $a_{ij}$ lies in; the determinant of whatever square block survives is the minor $M_{ij}$ . For a $3\times3$ matrix $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$ , deleting row 1 and column 2 to find the minor of $a_{12}$ leaves

M_{12} = \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix}.

where: $a_{ij}$ is the entry in row $i$ , column $j$ ; $M_{ij}$ is the $2\times2$ determinant of the remaining entries once row $i$ and column $j$ are struck out. A $3\times3$ matrix has nine minors, one for each entry, and each is an ordinary $2\times2$ determinant you already know how to evaluate.

Cofactor $C_{ij}$ . The cofactor dresses the minor with a sign that depends only on the position $(i,j)$ , never on the entry's value:

C_{ij} = (-1)^{i+j}M_{ij}.

where: the sign is $+$ when $i+j$ is even and $-$ when $i+j$ is odd, giving the checkerboard pattern $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$ starting with $+$ at position $(1,1)$ . So $C_{ij}$ either equals $M_{ij}$ outright or equals $-M_{ij}$ ; the minor is never itself signed, only the cofactor is.

Link to the determinant. Once every cofactor of a row (or column) is known, the determinant is just a weighted sum:

|A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}

where: the sum runs across one full row (row 1 shown; any row or any column works identically), each entry multiplied by its own cofactor and the three products added. This is exactly the familiar row-1 expansion $a_1\begin{vmatrix}b_2&c_2\\b_3&c_3\end{vmatrix} - b_1\begin{vmatrix}a_2&c_2\\a_3&c_3\end{vmatrix} + c_1\begin{vmatrix}a_2&b_2\\a_3&b_3\end{vmatrix}$ you already use, with the minors and alternating signs simply given the formal names $C_{11}, C_{12}, C_{13}$ .

Figure

A three by three grid showing the checkerboard sign pattern plus minus plus minus plus minus plus minus plus used to convert a minor into a cofactor

Derivation

Derivation ACofactor expansion of a 3x3 determinant along any row

The goal is to show, starting only from the standard $3\times3$ expansion formula a student already knows, that the same determinant can be written compactly as a sum of entries times their signed minors along any chosen row or column, which is exactly what makes cofactor expansion a genuine computational shortcut rather than new arithmetic in disguise.

Step 1. Write the standard row-1 expansion for $A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$ , which is the definition of a $3\times3$ determinant learned earlier in this chapter:

|A| = a_1\begin{vmatrix}b_2&c_2\\b_3&c_3\end{vmatrix} - b_1\begin{vmatrix}a_2&c_2\\a_3&c_3\end{vmatrix} + c_1\begin{vmatrix}a_2&b_2\\a_3&b_3\end{vmatrix}.

Step 2. Identify each $2\times2$ determinant on the right as a minor of row 1. Deleting row 1 and column 1 leaves $M_{11} = \begin{vmatrix}b_2&c_2\\b_3&c_3\end{vmatrix}$ ; deleting row 1 and column 2 leaves $M_{12} = \begin{vmatrix}a_2&c_2\\a_3&c_3\end{vmatrix}$ ; deleting row 1 and column 3 leaves $M_{13} = \begin{vmatrix}a_2&b_2\\a_3&b_3\end{vmatrix}$ . The formula in Step 1 is therefore $|A| = a_1 M_{11} - b_1 M_{12} + c_1 M_{13}$ , with the signs already fixed as $+,-,+$ .

Step 3. Compare that fixed sign pattern $+,-,+$ against $(-1)^{i+j}$ for row 1: at $(1,1)$ , $i+j=2$ , so $(-1)^2=+1$ ; at $(1,2)$ , $i+j=3$ , so $(-1)^3=-1$ ; at $(1,3)$ , $i+j=4$ , so $(-1)^4=+1$ . The signs match exactly, so $M_{11}$ scaled by $+1$ is $C_{11}$ , $M_{12}$ scaled by $-1$ is $C_{12}$ , and $M_{13}$ scaled by $+1$ is $C_{13}$ . Substituting turns Step 2's formula into $|A| = a_1 C_{11} + b_1 C_{12} + c_1 C_{13}$ — every minor has become a cofactor and every sign is now automatic.

Step 4. Nothing in this argument used the fact that row 1 was chosen first; the same $2\times2$ blocks and the same $(-1)^{i+j}$ signs arise if the expansion is organised along row 2, row 3, or any column instead, because deleting a given row and column always leaves the same $2\times2$ minor regardless of which row supplied the multiplying entry. So the general statement holds for any row or column $k$ : $|A| = a_{k1}C_{k1} + a_{k2}C_{k2} + a_{k3}C_{k3}$ (or the analogous sum down a column), and this is the cofactor-expansion formula used for the rest of the chapter and for building the adjoint.

Worked example

For $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{bmatrix}$ , evaluate $|A|$ by cofactor expansion along the first column, then cross-check with the row-1 expansion.

Step 1. Find $C_{11}$ : delete row 1, column 1, giving $M_{11} = \begin{vmatrix}4 & 5\\0 & 6\end{vmatrix} = (4)(6)-(5)(0) = 24$ , so $C_{11} = (-1)^{1+1}(24) = +24$ .

Step 2. Find $C_{21}$ : delete row 2, column 1, giving $M_{21} = \begin{vmatrix}2 & 3\\0 & 6\end{vmatrix} = (2)(6)-(3)(0) = 12$ , so $C_{21} = (-1)^{2+1}(12) = -12$ .

Step 3. Find $C_{31}$ : delete row 3, column 1, giving $M_{31} = \begin{vmatrix}2 & 3\\4 & 5\end{vmatrix} = (2)(5)-(3)(4) = -2$ , so $C_{31} = (-1)^{3+1}(-2) = -2$ .

Step 4. Expand down column 1 using entries $a_{11}=1,\ a_{21}=0,\ a_{31}=1$ : $|A| = a_{11}C_{11} + a_{21}C_{21} + a_{31}C_{31} = 1(24) + 0(-12) + 1(-2) = 24 - 2 = 22$ .

Step 5. Cross-check along row 1: $|A| = 1\begin{vmatrix}4&5\\0&6\end{vmatrix} - 2\begin{vmatrix}0&5\\1&6\end{vmatrix} + 3\begin{vmatrix}0&4\\1&0\end{vmatrix} = 1(24) - 2(-5) + 3(-4) = 24 + 10 - 12 = 22$ . Both expansions agree: $|A| = 22$ .

Competitive stretch

The zero entry $a_{21}=0$ suggests expanding along row 2 instead, since that term drops out for free: $|A| = -a_{21}\!\begin{vmatrix}2&3\\0&6\end{vmatrix} + a_{22}\!\begin{vmatrix}1&3\\1&6\end{vmatrix} - a_{23}\!\begin{vmatrix}1&2\\1&0\end{vmatrix} = 0 + 4(6-3) - 5(0-2) = 12 + 10 = 22$ . Same value, fewer terms — choosing the row or column with the most zeros is the standard time-saving trick for larger determinants.

Common traps

Confusing a minor with a cofactor — a minor never carries a sign; only the cofactor is signed.
Getting the sign pattern backwards — it starts with $+$ at position $(1,1)$ and alternates in both directions like a checkerboard, not row-by-row constant.
Deleting the wrong row or column — always strike out the row and column the chosen entry itself occupies, never a different pair.
Forgetting that $(-1)^{i+j}$ depends only on the entry's position, not on its numeric value — a large entry and a zero entry in the same cell get the same sign.

Ask the tutor: Give me a specific matrix entry and I will check whether you have found its minor and cofactor correctly.

Why do we even need cofactors if minors already give us the numbers?

Because the sign matters and the raw minor alone does not carry it. Two entries can have identical minors but opposite cofactors purely because of where they sit in the matrix, and dropping or misapplying that sign changes the determinant's value entirely. Cofactors package the minor together with its correct sign so that expansion along any row or column, and later the construction of the adjoint, works with one consistent rule instead of a separate sign check every time.

Does it matter which row or column I expand along?

No — every valid expansion, along any row or any column, gives the same final determinant value, as the worked example's cross-check between column 1 and row 1 shows. What differs is only the amount of arithmetic: picking the row or column with the most zeros, as in the competitive stretch, removes whole terms before you even compute them, so it is worth a quick scan of the matrix before choosing which line to expand along.

How is this different from the plain 3 by 3 determinant formula I already learned?

It is not different arithmetic, only a different name for the same pieces. The $2\times2$ determinants inside the familiar row-1 expansion are exactly the minors $M_{11}, M_{12}, M_{13}$ , and the alternating $+,-,+$ signs you were already applying are exactly the cofactor signs $(-1)^{i+j}$ . Learning minors and cofactors formally is what lets that same idea generalise to any row or column, and to the adjoint of a matrix in the next topic, where the sign can no longer be left implicit.