NCERT Solutions Class 12 Maths Chapter 4 Exercise 4.2 — Properties of Determinants

Why Properties of Determinants are a Game-Changer

Expanding a 3x3 determinant with complex variables can quickly turn into an algebraic nightmare with dozens of terms. The properties of determinants allow us to transform these matrices without changing their determinant value, or by changing them in predictable, simple ways. By applying operations like adding rows or factoring out constants, we can create columns or rows filled with zeroes. Once a row or column has two zeroes, expanding the determinant becomes a one-step trivial calculation. This systematic approach is the core focus of cbse class 12 determinants ex 4 2. Remember, using properties to show values without direct expansion is a favorite 4-mark or 6-mark question in board exams.

6 Core Properties You Must Memorize

Property 1 (Reflection Property)

The value of the determinant remains unchanged if its rows and columns are interchanged. (Det(A) = Det(A^T))

Property 2 (Sign-Change Property)

If any two rows (or columns) of a determinant are interchanged, then the sign of the determinant changes.

Property 3 (Identical Property)

If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

Property 4 (Scalar Multiplication Property)

If each element of a row (or column) of a determinant is multiplied by a constant k, then its value gets multiplied by k.

Property 5 (Sum Property)

If some or all elements of a row or column are expressed as the sum of two or more terms, then the determinant can be expressed as the sum of two or more determinants.

Property 6 (Invariance Property)

If to each element of any row or column of a determinant, the equimultiples of corresponding elements of other rows or columns are added, the value of the determinant remains the same. Example: Ri -> Ri + kRj.

A Systematic 4-Step Approach to Solve Exercise 4.2 Problems

1. Step 1: Look for Symmetry & Common Factors — Inspect the rows and columns. Look for cyclic variables (a, b, c) or numerical values that can be added together to produce a common term (like x + y + z or a + b + c) across an entire row or column.
2. Step 2: Apply Global Row or Column Additions — Apply operations like C1 -> C1 + C2 + C3 or R1 -> R1 + R2 + R3. This creates a uniform term in one row/column, which can then be factored out using Property 4 to yield a row or column of all 1s.
3. Step 3: Generate Zeroes — Once you have 1s in a row or column, use them to create zeroes. For example, if Column 1 has [1, 1, 1]^T, apply operations R2 -> R2 - R1 and R3 -> R3 - R1 to convert the column to [1, 0, 0]^T.
4. Step 4: Expand along the Row/Column with Maximum Zeroes — Expand along the row or column containing the two newly created zeroes. You will only need to compute a single 2x2 determinant, saving time and eliminating algebraic errors!

CBSE Board Exam Pitfalls & Success Strategies

1. Never perform conflicting simultaneous operations: Writing R1 -> R1 - R2 and R2 -> R2 - R1 in the exact same step is mathematically illegal and will earn you zero marks. You cannot use a row to modify another row while simultaneously modifying that row itself.
2. State your operations explicitly: In CBSE evaluation, marks are distributed step-by-step. Always write the exact operations (e.g., Applying C1 -> C1 - C2) on the right side of your proof.
3. Use the YoLearn Sketchpad: Drawing brackets and color-coding corresponding entries on our interactive whiteboard helps you visualize which elements subtract from where without mental fatigue.

Practice Questions with Solutions

• Q: Using properties of determinants, prove without expanding that: | x a x+a | | y b y+b | = 0 | z c z+c | A: Step 1: Let the given determinant be Delta. Step 2: Observe Column 3 (C3), which is a sum of two terms: (x+a), (y+b), and (z+c). We can apply Property 6 to modify Column 1 (C1). Step 3: Let's apply the column operation: C1 -> C1 + C2. This transforms Column 1 into: Row 1: x + a Row 2: y + b Row 3: z + c Step 4: Now, write the new determinant: Delta = | x+a a x+a | | y+b b y+b | | z+c c z+c | Step 5: Notice that Column 1 (C1) and Column 3 (C3) are completely identical. By Property 3, if any two columns of a determinant are identical, the value of the determinant is zero. Final answer: Therefore, Delta = 0. Hence proved.
• Q: Without expanding, prove that: | a-b b-c c-a | | b-c c-a a-b | = 0 | c-a a-b b-c | A: Step 1: Let the given determinant be Delta. Step 2: Observe the cyclic nature of the terms in each row. Let's add all rows to the first row to see if terms cancel out. Apply row operation: R1 -> R1 + R2 + R3. Step 3: Calculate the new elements for Row 1: Element 1: (a-b) + (b-c) + (c-a) = 0 Element 2: (b-c) + (c-a) + (a-b) = 0 Element 3: (c-a) + (a-b) + (b-c) = 0 Step 4: Rewrite the determinant with the new R1: Delta = | 0 0 0 | | b-c c-a a-b | | c-a a-b b-c | Step 5: Since all elements of the first row (R1) are zero, the entire value of the determinant is 0. Final answer: Therefore, Delta = 0. Hence proved.
• Q: Prove using properties of determinants: | 1 a a^2 | | 1 b b^2 | = (a-b)(b-c)(c-a) | 1 c c^2 | A: Step 1: Let the given determinant be Delta. Step 2: Since Column 1 consists of all 1s, we can easily create zeroes in this column using row operations. Apply operations: R2 -> R2 - R1 and R3 -> R3 - R1. Step 3: Apply these operations to find the new row values: Row 2 becomes: [0, b-a, b^2-a^2] Row 3 becomes: [0, c-a, c^2-a^2] Write the transformed determinant: Delta = | 1 a a^2 | | 0 b-a b^2-a^2 | | 0 c-a c^2-a^2 | Step 4: Take common factor (b-a) from Row 2 and (c-a) from Row 3: Delta = (b-a)(c-a) | 1 a a^2 | | 0 1 b+a | | 0 1 c+a | Step 5: Expand along Column 1 (C1): Delta = (b-a)(c-a) [1 (1(c+a) - 1(b+a))] Delta = (b-a)(c-a) (c + a - b - a) Delta = (b-a)(c-a)(c-b) Step 6: Adjust signs to match standard cyclic form: (b-a) = -(a-b) (c-b) = -(b-c) Multiplying the two negatives gives a positive, resulting in: (a-b)(b-c)(c-a). Final answer: Therefore, Delta = (a-b)(b-c)(c-a). Hence proved.
• Q: Prove the following identity using determinant properties: | x+4 2x 2x | | 2x x+4 2x | = (5x+4)(4-x)^2 | 2x 2x x+4 | A: Step 1: Let the given determinant be Delta. Step 2: Observe that adding all columns together gives the term (x+4 + 2x + 2x) = 5x+4 in each row. Apply column operation: C1 -> C1 + C2 + C3. Step 3: Rewrite the determinant with the modified C1: Delta = | 5x+4 2x 2x | | 5x+4 x+4 2x | | 5x+4 2x x+4 | Step 4: Take common factor (5x+4) out of Column 1: Delta = (5x+4) | 1 2x 2x | | 1 x+4 2x | | 1 2x x+4 | Step 5: Create zeroes in Column 1 by applying operations: R2 -> R2 - R1 and R3 -> R3 - R1. Delta = (5x+4) | 1 2x 2x | | 0 4-x 0 | | 0 0 4-x | Step 6: Expand along Column 1: Delta = (5x+4) [1 ((4-x)*(4-x) - 0)] Delta = (5x+4)(4-x)^2. Final answer: Delta = (5x+4)(4-x)^2. Hence proved.

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