Q.Find the minor of element $6$ in the determinant $\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$.

Concept: Minor of an element — the determinant of the submatrix obtained by deleting the row and column containing that element.

Step 1: The element $6$ is in row 2, column 3.

Step 2: Delete row 2 and column 3. The remaining $2 \times 2$ matrix is:

Step 3: The minor is the determinant of this submatrix:

Concept and Intuition

In matrix algebra, the minor of an element is a way to "zoom in" on the contribution of that single entry to the larger determinant. Think of it as the determinant of what remains when you remove the entire row and column that the element sits in. This is a foundational idea for computing larger determinants by expansion (Laplace expansion) and for defining cofactors and adjugates.

For a $3\times3$ determinant, each minor is simply the determinant of a $2\times2$ matrix. The key is to correctly identify which row and column to delete — a common mistake is to delete the wrong ones.

Step-by-Step Solution

1. Locate the element. The element $6$ is in the second row and third column of the determinant:

So its position is $(i,j) = (2,3)$.

2. Delete the row and column of that element. Remove row 2 (the row containing $4,5,6$) and column 3 (the column containing $3,6,9$). What remains is a $2\times2$ matrix formed from the intersection of rows 1 and 3 with columns 1 and 2:

3. Compute the determinant of this $2\times2$ matrix. For a $2\times2$ matrix $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$, the determinant is $ad - bc$. Here:

Method: Finding the Minor of a Specific Element

This method finds the minor of one named element in a determinant — a building block used constantly in cofactor expansion and the adjoint method.

Steps

Step 1: Locate the element's row and column

Identify exactly which row $i$ and column $j$ the given element sits in — write down $(i,j)$ explicitly before doing anything else.

Step 2: Delete that entire row and that entire column

Cross out every entry in row $i$ and every entry in column $j$. What remains is a smaller determinant, one size down (for a $3\times3$, always a $2\times2$).

Step 3: Write out the remaining entries in their original relative order

Keep the surviving rows and columns in the same order they appeared in the original matrix — don't rearrange them.

Step 4: Evaluate the smaller determinant

Apply the appropriate formula ($ad-bc$ for a $2\times2$ remainder) to get the minor's numeric value.

Step 5: Remember — no sign is attached at this stage

A minor is just the determinant of the submatrix; the $(-1)^{i+j}$ sign only gets attached when the minor becomes a cofactor, in a separate step.

Getting the row/column deletion right is the only place this method can go wrong — always double check by counting: for an $n\times n$ determinant, the minor of any single element must be $(n-1)\times(n-1)$.

Common Mistakes

Mistake 1: Deleting the wrong row or column

Why it's wrong: the minor of element $6$ requires deleting row 2 and column 3 specifically — accidentally deleting the wrong row (e.g. row 3, because $6$ "looks like" it's near the bottom) leaves an entirely different $2\times2$ submatrix and a wrong answer. Correct approach: write down the element's exact position $(i,j)$ first, then delete precisely row $i$ and column $j$ — never estimate by eye.

Mistake 2: Confusing the minor with the cofactor

Why it's wrong: the minor is just the determinant of the submatrix — no sign is attached — but a student who has just learned cofactors may reflexively multiply by $(-1)^{i+j}$ anyway, changing $-6$ into $+6$. Correct approach: the minor is asked for here, not the cofactor, so stop after evaluating the submatrix's determinant — do not attach a sign unless the cofactor is explicitly requested.