Class 12 Maths · Original EZT practice
Determinants (Class 12 Chapter 4): summary and practice
A determinant turns a square matrix into a single number - and that number tells you the area of a triangle, whether an inverse exists, and whether a system of equations has a unique solution. All questions are original EZT tutor practice following the chapter's own section order.
8 original worked examples, 8 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. The determinant of a matrix
For 2 x 2 the rule is ad - bc. For 3 x 3, expand along any row or column using the sign pattern + - +, - + -, + - +.
2. Area of a triangle
The area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is half the absolute value of the determinant built from the coordinates - the absolute value matters because area is never negative.
3. Minors and cofactors
The minor of an entry is the determinant of what remains after deleting its row and column. The cofactor adds a sign: Cij = (-1)^(i+j) Mij.
4. Adjoint and inverse
The adjoint is the transpose of the cofactor matrix. When |A| is not zero, A^-1 = adj A / |A|; when |A| = 0 the matrix is singular and has no inverse.
5. Solving systems of equations
Write the system as AX = B. When |A| is not zero, the unique solution is X = A^-1 B; otherwise the system is inconsistent or has infinitely many solutions.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Evaluate a 2 x 2 determinant
Choose the method: Main diagonal minus the other diagonal.
EZT practice question
Find the determinant of A = [[1, 2], [3, 4]].
Show the worked solution
- Rule: ad - bc.
- 1 x 4 - 2 x 3 = 4 - 6.
Answer: |A| = -2.
Check: A negative determinant is perfectly fine - it is a signed quantity.
Watch for this: Do not compute ad + bc; the cross term is subtracted.
Now try it yourself
Find |[[2, 3], [4, 6]]|.
Check your answer
0.
2 x 6 - 3 x 4 = 12 - 12, so the matrix is singular.
Expand a 3 x 3 determinant
Choose the method: Pick a row, apply + - + signs, multiply each entry by its minor.
EZT practice question
Evaluate |[[2, 0, 1], [1, 3, 2], [1, 1, 1]]|.
Show the worked solution
- Expand along row one: 2 x (3 - 2) - 0 x (1 - 2) + 1 x (1 - 3).
- = 2 - 0 - 2.
Answer: 0 - the matrix is singular.
Check: Rows one and three are linked: row one minus row three gives [1, -1, 0], and the zero determinant matches the dependent structure.
Watch for this: Watch the middle sign: the pattern along a row is + - +, not + + +.
Now try it yourself
Evaluate |[[1, 2, 3], [0, 1, 4], [5, 6, 0]]|.
Check your answer
1.
1 x (0 - 24) - 2 x (0 - 20) + 3 x (0 - 5) = -24 + 40 - 15 = 1.
Find a triangle's area by determinant
Choose the method: Half the absolute value of the coordinate determinant.
EZT practice question
Find the area of the triangle with vertices (0, 0), (4, 0) and (0, 3).
Show the worked solution
- Area = half of |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|.
- = half of |0 + 4 x 3 + 0| = half of 12.
Answer: 6 square units.
Check: Base 4 and height 3 give half x 4 x 3 = 6 - same answer.
Watch for this: Never drop the absolute value; a negative determinant would give negative area.
Now try it yourself
Find the area of the triangle (1, 1), (5, 1), (3, 4).
Check your answer
6 square units.
Half of |1 x (1 - 4) + 5 x (4 - 1) + 3 x (1 - 1)| = half of |-3 + 15| = 6.
Tell minors from cofactors
Choose the method: Minor first, then attach the sign.
EZT practice question
For A = [[1, 2], [3, 4]], find the minor and cofactor of the entry 1.
Show the worked solution
- Delete row one and column one: the minor is |4| = 4.
- Cofactor adds the sign (-1)^(1+1) = +1, so the cofactor is also 4.
Answer: Minor = 4, cofactor = 4.
Check: For the entry 2 the minor is 3 but the cofactor is -3 - the sign flips on that position.
Watch for this: Do not report the minor when the question asks for the cofactor; the sign pattern is the whole point.
Now try it yourself
What is the cofactor of the entry 3 in the same matrix?
Check your answer
-2.
Minor is 2 and the position sign is negative.
Build an inverse with the adjoint
Choose the method: A^-1 = adj A / |A|, only when |A| is not zero.
EZT practice question
Find the inverse of A = [[1, 2], [3, 4]] using the adjoint.
Show the worked solution
- |A| = -2, not zero, so the inverse exists.
- Cofactors: 4, -3, -2, 1; the adjoint is the transpose [[4, -2], [-3, 1]].
- Divide by |A|: adj A / -2 = [[-2, 1], [3/2, -1/2]].
Answer: A^-1 = [[-2, 1], [3/2, -1/2]].
Check: Multiply A by this inverse and you get the identity matrix.
Watch for this: The adjoint is the TRANSPOSE of the cofactor matrix - forgetting the transpose is the classic slip.
Now try it yourself
Does [[2, 3], [4, 6]] have an inverse?
Check your answer
No.
Its determinant is 0, so it is singular.
Spot a singular matrix
Choose the method: Compute |A|; zero means no inverse.
EZT practice question
What is a singular matrix?
Show the worked solution
- A square matrix with determinant zero is singular.
- A non-zero determinant means non-singular, and only then does A^-1 exist.
Answer: Singular means |A| = 0; such a matrix has no inverse.
Check: [[2, 0, 1], [1, 3, 2], [1, 1, 1]] has determinant 0, so it is singular.
Watch for this: Do not try to invert before checking the determinant; the formula divides by |A|.
Now try it yourself
Is [[1, 2, 3], [0, 1, 4], [5, 6, 0]] singular?
Check your answer
No - its determinant is 1.
Non-zero determinant means non-singular.
Solve a system with the inverse
Choose the method: Write AX = B, then X = A^-1 B.
EZT practice question
Solve x + 2y = 5 and 3x + 4y = 11 using matrices.
Show the worked solution
- A = [[1, 2], [3, 4]], so A^-1 = [[-2, 1], [3/2, -1/2]].
- x = -2 x 5 + 1 x 11 = 1.
- y = 3/2 x 5 - 1/2 x 11 = 7.5 - 5.5 = 2.
Answer: x = 1 and y = 2.
Check: Substitute back: 1 + 4 = 5 and 3 + 8 = 11 - both hold.
Watch for this: X = A^-1 B, not B times A^-1; the order matters because multiplication is not commutative.
Now try it yourself
Solve x + y = 3 and x - y = 1.
Check your answer
x = 2, y = 1.
Adding the equations gives 2x = 4; subtracting gives 2y = 2.
Decide if a system has a unique solution
Choose the method: Look at |A| first.
EZT practice question
How does the determinant decide whether a system AX = B has a unique solution?
Show the worked solution
- If |A| is not zero, A^-1 exists and X = A^-1 B is the one unique solution.
- If |A| = 0, the system is either inconsistent or has infinitely many solutions - checked via adj A times B.
Answer: A non-zero determinant gives exactly one solution; a zero determinant means no unique solution.
Check: This is why the singular test comes before any solving.
Watch for this: Do not apply X = A^-1 B when |A| = 0; the formula is undefined there.
Now try it yourself
Does x + y = 3, x - y = 1 have a unique solution?
Check your answer
Yes.
|[[1, 1], [1, -1]]| = -2, which is not zero.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Determinant of [[a, b], [c, d]]?
Show answer and reason
ad - bc.
Main diagonal minus the cross diagonal.
2. When does A^-1 exist?
Show answer and reason
Only when |A| is not zero.
A^-1 = adj A / |A| divides by the determinant.
3. Cofactor formula?
Show answer and reason
Cij = (-1)^(i+j) times the minor Mij.
The sign pattern alternates like a chessboard.
4. Solution of AX = B?
Show answer and reason
X = A^-1 B, when |A| is not zero.
Multiplying both sides by A^-1 on the left isolates X.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Find |[[2, 3], [4, 6]]| and say whether the inverse exists.
- Find the area of the triangle with vertices (0, 0), (2, 0), (0, 5).
- Solve x + y = 3 and x - y = 1.
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Questions about this practice page
- When does the inverse of a matrix exist?
- Only when its determinant is non-zero. Then A^-1 = adj A / |A|. A matrix with determinant zero is singular and has no inverse.
- How are minors and cofactors different?
- The minor is the smaller determinant left after deleting the entry's row and column; the cofactor is that minor with the sign (-1)^(i+j) attached.
- How do I find the area of a triangle with a determinant?
- Put the three vertices into the coordinate formula and take half the absolute value of the result - the absolute value keeps the area positive.
- What is the adjoint of a matrix?
- The transpose of the cofactor matrix. Together with the determinant it builds the inverse: A^-1 = adj A / |A|.
- Are these the NCERT exercise questions for Determinants (Class 12)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.