Minor of a Matrix Element – From Intuition to Precision
Imagine you have a large matrix — say a 3×3 grid of numbers. You pick one specific element, like the number in the second row and third column. Now ask: if I wanted to ignore that element and everything in its row and column, what smaller matrix remains? That smaller matrix is the key.
The minor of an element is simply the determinant of that leftover submatrix. It captures the "influence" of that element when you strip away its entire row and column.
The Intuition First
Think of a matrix as a system of equations. Each row is an equation, each column a variable. When you focus on one element, you are asking: what happens to the rest of the system if I remove the equation and the variable that this element belongs to? The minor tells you the "size" (determinant) of that reduced system.
For a 2×2 matrix, the minor of any element is just the single number diagonally opposite it — because removing one row and one column leaves a 1×1 matrix, whose determinant is that number itself.
For a 3×3 matrix, the minor of an element is the determinant of a 2×2 matrix formed by the four numbers that are not in the same row or column as the chosen element.
The Precise Statement
Let A be an n×n square matrix. Let aij be the element in the i-th row and j-th column.
Mij=det(matrix obtained by deleting row i and column j from A)
Mij is called the minor of the element aij.
A Concrete Example
Take the matrix:
A=123456789
Find the minor of the element a23=8 (row 2, column 3).
Delete row 2 and column 3. What remains?
(1346)
The minor is the determinant of this 2×2 matrix:
M23=(1)(6)−(4)(3)=6−12=−6
So the minor of 8 is −6.
Tip
The minor is not the element itself — it is the determinant of the submatrix left after removing that element's row and column. For a 1×1 matrix, the minor of the single element is 1 (the determinant of an empty matrix is defined as 1), but that's a special case.
Why Minors Matter
Minors are the building blocks of cofactors, which in turn are used to compute determinants of large matrices (Laplace expansion) and to find inverses. Every time you expand a determinant along a row or column, you are summing products of elements and their minors (with appropriate signs). …
Each minor is found by deleting that element's own row and column and evaluating what's left, then the cofactor attaches the sign (−1)i+j; a standard property of determinants is that a row's elements times a different row's cofactors always sum to zero. …
Compute each minor Mij, attach the sign (−1)i+j to get the cofactor Aij; then the cross sum (row-1 elements times row-3 cofactors) equals 0, confirming the standard property.
Mij=minor (delete row i,col j),Aij=(−1)i+jMij.
Given A=261−30554−7, so a11=2,a12=−3,a13=5.
Q.Assertion (A) : Minor of element a13 in the matrix 0122216−13 is 1221. Reason (R) : Minor of an element aij of a matrix is the determinant obtained by deleting its jth row and ith column in which the element lies. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
›Reveal solutionSolution
Deleting row 1 and column 3 gives 1221, so A is true; R reverses the row/column indices and is false.
The minor Mij of element aij is the determinant obtained by deleting the ithrow and jthcolumn of the matrix.
Element a13 sits in row 1, column 3 (i=1,j=3); its value is 6.
Delete row 1 and column 3 of 0122216−13: the remaining rows are (1,2) and (2,1), giving 1221. This matches the Assertion, so A is true. …