Q.If 2x43x92x1+3=0, then the value of x is :
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Minor of a Matrix Element
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.
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). …
Taking x common from the second row leaves a numeric determinant equal to 3, so the equation becomes $3 …
Factor x from row 2; the remaining determinant is 3, so 3x+3=0⇒x=−1.
A common factor in any row (or column) can be taken outside the determinant; expand a 3×3 determinant along a convenient row.
- 2x43x92x1=x214319211 (taking x from row 2). …
- CBSE 2024Set 465/RQPS/41 markMCQQ.The value of Δ=427929275593 is : (A) 0 (B) 1 (C) −3 (D) −15
›Reveal solutionSolution
Cofactor expansion along row 1 yields 0, so Δ=0.
a1b1c1a2b2c2a3b3c3=a1(b2c3−b3c2)−a2(b1c3−b3c1)+a3(b1c2−b2c1)
- Δ=427929275593.
- 42(7⋅3−9⋅5)=42(21−45)=42(−24)=−1008. …
- CBSE 2024Set 465/S/RQPS/41 markMCQQ.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 ith row and jth column 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. …
- CBSE 2023Set 465/EF1GH/41 markMCQQ.If 2x43x92x1+3=0, then the value of x is :(a) −1(b) 0(c) 1(d) 3
›Reveal solutionSolution
Factor x from row 2; the remaining determinant is 3, so 3x+3=0⇒x=−1.
A common factor in any row (or column) can be taken outside the determinant; expand a 3×3 determinant along a convenient row.
- 2x43x92x1=x214319211 (taking x from row 2). …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.