Analogy Time: Think of a determinant as a "characteristic number" or a "score" for a matrix. Just like a person's fingerprint is unique to them, a determinant is a unique value that can tell us important things about a matrix, such as:
Visual Aid: Think of it as: (product going "down-right") - (product going "down-left").

(Self-correction: Cannot actually embed image, will describe visually)
Imagine arrows! An arrow from 'a' to 'd' (positive product), and an arrow from 'b' to 'c' (negative product).
| Sign Pattern for Cofactors (3x3) | ||
|---|---|---|
| + | - | + |
| - | + | - |
| + | - | + |
JEE Tip: While you can expand along any row or column, a smart move, especially for JEE, is to choose the row or column that contains the most zeros. This simplifies calculations immensely because any term multiplied by zero will vanish!
Step 1 (Positive diagonals):
$ (a cdot e cdot i) + (b cdot f cdot g) + (c cdot d cdot h) $
Step 2 (Negative diagonals):
$ - (c cdot e cdot g) - (a cdot f cdot h) - (b cdot d cdot i) $
CBSE vs. JEE Focus: For CBSE, mastering 2x2 and 3x3 determinants using cofactor expansion (and Sarrus' rule for 3x3 as a verification) is key. For JEE, this forms the absolute base. You'll need to be super fast and accurate with these calculations, as they will be components of more complex problems involving properties of determinants, adjoints, inverses, and solving systems of equations. Understanding the underlying mechanics of minors and cofactors is critical for higher-order determinants and theoretical questions.
Mastering the evaluation of determinants is fundamental for both JEE Main and board exams. While direct calculation is always an option, using mnemonics and shortcuts can significantly boost your speed and accuracy, especially under exam pressure. Here are some effective techniques:
When expanding a determinant using cofactors, remembering the alternating signs `(-1)^(i+j)` can be tricky. Here's a visual mnemonic:
+ - + ...
- + - ...
+ - + ...
...
+ - +
- + -
+ - +
For a 2x2 matrix A = [[a, b], [c, d]], the determinant is ad - bc. Remember it as:
a * d (positive product)
b * c (negative product)
This is a powerful visual shortcut specifically for 3x3 determinants. It's often quicker than cofactor expansion for these matrices.
Example: For A = [[a, b, c], [d, e, f], [g, h, i]]
| a b c | a b
| d e f | d e
| g h i | g h
det(A) = (aei + bfg + cdh) - (ceg + afh + bdi)
This method is highly recommended for 3x3 matrices in both CBSE and JEE, as it minimizes sign errors.
When evaluating higher-order determinants (3x3 or more) using cofactor expansion, always choose the row or column that contains the maximum number of zeros. This is a crucial shortcut:
Remembering key properties can often lead to instantaneous evaluation without any calculation:
Always check for these properties first, especially in multiple-choice questions, as they are designed to test your conceptual understanding.
By integrating these mnemonics and shortcuts into your practice, you'll find determinant evaluation less daunting and more efficient. Happy solving!
Evaluating determinants efficiently is crucial for both CBSE board exams and competitive exams like JEE Main. While direct expansion is always an option, leveraging properties of determinants can significantly reduce computation time and effort, especially for higher order matrices or those with complex entries.
This is the single most important strategy for evaluation.
By consistently applying these quick tips, you will not only solve determinant problems faster but also with greater accuracy. Practice is key to internalizing these strategies!
The determinant of a square matrix is not just a numerical value obtained through a formula; it carries significant intuitive meaning, especially in the context of linear transformations and systems of linear equations. Understanding this intuition goes beyond mere calculation and helps in grasping deeper mathematical concepts crucial for JEE.
This is arguably the most intuitive way to understand determinants:
Consider a 2x2 matrix $A = egin{bmatrix} a & b \ c & d end{bmatrix}$. If we view the columns (or rows) as two-dimensional vectors, say $mathbf{v_1} = egin{pmatrix} a \ c end{pmatrix}$ and $mathbf{v_2} = egin{pmatrix} b \ d end{pmatrix}$, then the absolute value of the determinant, $|ad-bc|$, represents the area of the parallelogram formed by these two vectors when originating from the same point.
Similarly, for a 3x3 matrix, if its columns (or rows) are considered as three-dimensional vectors, the absolute value of its determinant represents the volume of the parallelepiped (a 3D analogue of a parallelogram) formed by these three vectors.
The determinant provides crucial information about the solvability of a system of linear equations $AX=B$:
For both JEE and CBSE, understanding that a determinant of zero implies linear dependence of column/row vectors, leading to a "collapsed" space (zero area/volume), is fundamental. This directly impacts whether a matrix is invertible and the nature of solutions to systems of linear equations. It's not just a calculation, but a test for singularity and linear independence.
Keep these geometric and algebraic interpretations in mind, as they provide a powerful intuitive foundation for solving problems related to matrices and determinants.
Understanding the evaluation of determinants can sometimes feel like a dry, mechanical process. Analogies can help connect this abstract mathematical procedure to more intuitive, real-world scenarios, making the underlying logic clearer and more memorable.
Imagine a complex systemโit could be an economic model, a sports team's performance, or a scientific experiment. This system has multiple interacting components or factors. Each entry in a matrix (and subsequently, its determinant) represents a specific influence, contribution, or relationship within that system.
The determinant, when evaluated, acts like a single "Net Outcome Score" or a "System Health Indicator" that summarizes the overall effect of all these interacting parts. It provides a crucial piece of information about the system's nature, such as its stability, uniqueness of solutions, or potential for change.
Consider the matrix $$egin{pmatrix} a & b \ c & d end{pmatrix}$$. Its determinant is (ad - bc).
Analogy: Think of this as two competing forces or paths.
The analogy extends to more complex systems with more interacting components.
Analogy: Imagine a multi-player game or a complex machine with many gears. Each element (a_{ij}) in the determinant represents a specific player's contribution or a gear's output at a particular stage.
By thinking of determinant evaluation as calculating a "net outcome" or "system score," you can better appreciate how the individual elements combine in a specific, rule-bound way to yield a single, highly significant number that describes the entire system.
Keep practicing, and you'll soon master the art of determinant evaluation!
To effectively learn and master the "Evaluation of Determinants," a solid grasp of certain foundational mathematical concepts is crucial. These prerequisites ensure that you can focus on the new concepts related to determinants without being hindered by gaps in earlier learning. A strong foundation here will significantly ease your journey through Matrices and Determinants, which is a high-scoring unit for both JEE Main and board exams.
Here are the essential prerequisites:
For CBSE Board Exams, a strong grasp of arithmetic and basic algebra is sufficient. For JEE Main, a more robust and quick command over algebraic manipulations and simplification will be an added advantage, as questions might involve more complex expressions or require simplification to reach the final answer.
Make sure you revisit these foundational concepts if you feel any gaps. A little review now can save significant confusion later!
Navigating the evaluation of determinants can be tricky, and even seasoned students often fall into common traps during exams. Being aware of these pitfalls can significantly boost your accuracy and efficiency. Let's explore the typical mistakes to avoid:
Properties of determinants are powerful tools, but their incorrect application can lead to erroneous results:
By being mindful of these common traps and practicing meticulously, you can significantly improve your accuracy and speed in evaluating determinants.
For CBSE Board Examinations, the evaluation of determinants is a fundamental skill, but the emphasis often shifts from brute-force calculation to the smart application of properties. A strong understanding of determinant properties is crucial for scoring well in this section.
While JEE Main might test more complex manipulations and a deeper conceptual understanding of determinants, CBSE largely focuses on the systematic and correct application of properties and formulas. Proving identities using properties is a cornerstone for CBSE, whereas JEE might integrate determinants into more abstract problems or matrix operations.
To evaluate a 3x3 determinant or prove an identity, try to make two zeros in any one row or column using row/column operations. Once two zeros are created, expand the determinant along that row or column. This significantly reduces calculation errors and effort.
CBSE Tip: Practice a wide variety of "proving without expansion" type questions. These not only solidify your understanding of properties but are also highly recurrent in board exams.
For JEE Main, evaluating determinants is not merely about expanding them; it's about efficiently using properties to simplify and arrive at the solution quickly. This section highlights the key areas to focus on for competitive exams.
| 1 | 1 | 1 |
| a | b | c |
| aยฒ | bยฒ | cยฒ |
Tip for JEE: When faced with a 3x3 determinant or higher, resist the urge to expand immediately. Always spend a minute or two looking for opportunities to apply properties. Try to create two zeros in a row or column, or check if any row/column operations lead to common factors or make two rows/columns identical/proportional.
Mastering determinant evaluation for JEE is less about calculation and more about strategic thinking and applying properties judiciously. Practice various types of problems to build intuition.
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Students often incorrectly apply the property of scaling a determinant, confusing it with scalar multiplication of an entire matrix. They might factor out a constant 'k' and multiply the determinant by k^n (where 'n' is the order) even when 'k' is common to only one row or column, or conversely, factor 'k' out from a matrix-wide scalar multiple and only multiply by 'k'.
Arises from confusing det(kA) = k^n det(A) (for an 'n x n' matrix A) with the property that multiplying a single row or column of a determinant by 'k' multiplies its value by 'k'.
A constant 'k' can be factored out from only one row or one column at a time, multiplying the determinant by 'k'. The rule det(kA) = k^n det(A) applies when the entire matrix of order 'n' is scaled by 'k'.
Given D = | 2a b |
| 2c d |. Incorrectly writing D = 2^2 | a b |
| c d | (confusing with det(kA)) or D = 2 | a b |
| c d | implicitly factoring '2' from both rows/columns when '2' is only common to C1.
For D = | 2a b |
| 2c d |, factor '2' from C1: D = 2 | a b |
| c d |. The value is 2(ad - bc). This correctly reflects that scaling a single column by '2' scales the determinant by '2'.
| 1 | 0 |
| 0 | 1 |
| 1 | 1 |
| 1 | 1 |
| 1 | 0 |
| 0 | 1 |
| 1 | 1 |
| 1 | 1 |
| 1+1 | 0+1 |
| 0+1 | 1+1 |
| 2 | 1 |
| 1 | 2 |
| 1 | 2 | 3 |
| 0 | 1 | 4 |
| 5 | 6 | 0 |
| 1 | 2 | 3 |
| 0 | 1 | 4 |
| 5 | 6 | 0 |
(-1)^(i+j)) when expanding a determinant along a row or column using cofactors. This common oversight leads to an incorrect final value for the determinant, even if the minors are calculated correctly. M_ij) and a cofactor (C_ij). The cofactor C_ij is defined as (-1)^(i+j) * M_ij. Students sometimes directly use the minor M_ij in the expansion formula instead of the cofactor, thereby missing the crucial sign factor. (-1)^(i+j) sign factor. For a 3x3 determinant, the sign pattern for cofactors is a 'chessboard pattern':+ - +- + -+ - + |A|. Expanding along the first row, a student might incorrectly write:|A| = a_11 * M_11 + a_12 * M_12 + a_13 * M_13M_12 is correctly calculated, but the coefficient for a_12 * M_12 should be negative, not positive.|A|, the correct expansion along the first row is:|A| = a_11 * C_11 + a_12 * C_12 + a_13 * C_13|A| = a_11 * (+1)M_11 + a_12 * (-1)M_12 + a_13 * (+1)M_13|A| = a_11 * M_11 - a_12 * M_12 + a_13 * M_13det(A) = Σ a_ij * C_ij = Σ a_ij * (-1)^(i+j) * M_ij before starting the expansion.Before evaluating a determinant whose elements originate from physical measurements or quantities, always:
This ensures the mathematical operations within the determinant are performed on values that are consistent in their physical representation. This is more relevant for problems beyond core determinant evaluation, where determinants are applied as tools.
Consider a hypothetical problem where a determinant's value represents a physical quantity, and elements are formed from lengths given in mixed units:
Given: l1 = 5 cm, l2 = 0.1 m
Determinant = | l1 1 |
| l2 2 |
Student calculates (wrongly, without conversion):
| 5 1 | = (5 * 2) - (0.1 * 1) = 10 - 0.1 = 9.9
| 0.1 2 |
Here, l1 is in cm and l2 is in m. Direct calculation without conversion leads to an inconsistent result if the underlying problem expects consistent units.
Given: l1 = 5 cm, l2 = 0.1 m
Method 1: Convert l1 to meters:
l1 = 5 cm = 0.05 m
Determinant = | 0.05 1 | = (0.05 * 2) - (0.1 * 1) = 0.1 - 0.1 = 0
| 0.1 2 |
Method 2: Convert l2 to centimeters:
l2 = 0.1 m = 10 cm
Determinant = | 5 1 | = (5 * 2) - (10 * 1) = 10 - 10 = 0
| 10 2 |
The correct value of the determinant is 0, which significantly differs from 9.9 obtained by ignoring unit consistency. This highlights how crucial consistent units are when dealing with quantities derived from physical measurements.
(-1)^(i+j) is crucial for cofactors but not for minors.a_ij (at row i, column j) in a determinant expansion is given by (-1)^(i+j). A simpler way to remember this for the first row expansion (which is most common) is the checkerboard pattern of signs: + - +. For a 3x3 determinant expanded along the first row, the terms are a11 * C11 + a12 * C12 + a13 * C13, where C_ij is the cofactor. This translates to +a11 * M11 - a12 * M12 + a13 * M13 (where M_ij is the minor). | 1+ฮต 1 1 |
| 1 1+ฮต 1 |
| 1 1 1+ฮต |
| 1 1 1 |
| 1 1 1 |
| 1 1 1 |
| 1+ฮต 1 1 |
| 1 1+ฮต 1 |
| 1 1 1+ฮต |
| ฮต -ฮต 0 |
| 0 ฮต -ฮต |
| 1 1 1+ฮต |
det(kA) = k^n * det(A). This means the scalar 'k' is raised to the power of the order of the matrix 'n'. If you are simply multiplying a value `k` to an existing determinant `|A|`, it literally means `k` times the value of `|A|`, not transforming the matrix inside. A = [[a, b], [c, d]]. Student attempts to find det(2A) as: det(2A) = |[[2a, 2b], [2c, 2d]]| = (2a)(2d) - (2b)(2c) = 4ad - 4bc = 4(ad - bc). This is correct as a calculation, but the common conceptual mistake is to assume det(2A) = 2 * det(A) directly.A = [[a, b], [c, d]] and det(A) = ad - bc. det(2A), using the property det(kA) = k^n * det(A): det(2A) = 2^2 * det(A) = 4 * (ad - bc). k^n.(-1)^(i+j) for element a_ij).Evaluate: | 1 2 3 |
| 4 5 6 |
| 7 8 9 |
Incorrect Step: Expanding along R1, a common error could be: 1(5*9 - 6*8) + 2(4*9 - 6*7) + 3(4*8 - 5*7) (Mistake: Used '+' for the second term instead of '-').1(45 - 48) + 2(36 - 42) + 3(32 - 35)1(-3) + 2(-6) + 3(-3)-3 - 12 - 9 = -24 (Incorrect final value due to sign error).
Evaluate: | 1 2 3 |
| 4 5 6 |
| 7 8 9 |
Correct Approach: Expanding along R1 with correct signs:
1 * (5*9 - 6*8) - 2 * (4*9 - 6*7) + 3 * (4*8 - 5*7)1 * (45 - 48) - 2 * (36 - 42) + 3 * (32 - 35)1 * (-3) - 2 * (-6) + 3 * (-3)-3 + 12 - 99 - 9 = 0 (Correct final value).
+ - + pattern above the row/column you are expanding along before starting.aij in the cofactor expansion is determined by (-1)^(i+j). This creates an alternating pattern of signs:+ - +- + -+ - +aij is multiplied by (-1)^(i+j) before multiplying by its minor Mij to get the cofactor Cij. So, Cij = (-1)^(i+j) * Mij. A = 1 2 3 4 5 6 7 8 9
det(A) = 1 * (5*9 - 6*8) + 2 * (4*9 - 6*7) + 3 * (4*8 - 5*7)= 1 * (45 - 48) + 2 * (36 - 42) + 3 * (32 - 35)= 1 * (-3) + 2 * (-6) + 3 * (-3)= -3 - 12 - 9 = -24A = 1 2 3 4 5 6 7 8 9
det(A) = 1 * (5*9 - 6*8) - 2 * (4*9 - 6*7) + 3 * (4*8 - 5*7)= 1 * (45 - 48) - 2 * (36 - 42) + 3 * (32 - 35)= 1 * (-3) - 2 * (-6) + 3 * (-3)= -3 + 12 - 9 = 0| 100 cm | 3 |
| 2 m | 4 |
| 1 m | 3 |
| 2 m | 4 |
Consider evaluating the determinant of matrix A = [[a, b, c], [d, e, f], [g, h, i]] by expanding along the first row:
det(A) = a(ei - fh) + b(di - fg) + c(dh - ge)Here, the student incorrectly used a `+` sign before the `b` term, treating `b`'s contribution as positive instead of negative.
Using the correct sign convention for the same matrix A = [[a, b, c], [d, e, f], [g, h, i]] expanded along the first row:
det(A) = a(ei - fh) - b(di - fg) + c(dh - ge)The `b` term (at position (1,2)) has a cofactor sign of `(-1)^(1+2) = -1`, hence the negative sign. For JEE, this understanding is fundamental, and a sign error can lead to a completely wrong answer.
(-1)(i+j) rule for the element at position (i,j). This seemingly minor error can lead to a completely incorrect final value for the determinant. + - + / - + - / + - +).aij is (-1)(i+j), where 'i' is the row number and 'j' is the column number. For CBSE 12th exams and JEE, it's crucial to apply this rule consistently. Methodically apply this pattern or the formula for each term during expansion. A good practice is to mentally (or physically) draw the sign pattern for the chosen row/column before starting the expansion. Evaluate: | 1 2 3 || 4 5 6 || 7 8 9 |
Incorrect Expansion (along R1, common mistake at a12, assuming + for 2):
1(5*9 - 8*6) + 2(4*9 - 7*6) + 3(4*8 - 7*5)
= 1(45 - 48) + 2(36 - 42) + 3(32 - 35)
= 1(-3) + 2(-6) + 3(-3)
= -3 - 12 - 9 = -24The sign for the second term (element 2) should have been negative.
Correct Expansion (along R1):
1(5*9 - 8*6) - 2(4*9 - 7*6) + 3(4*8 - 7*5)
= 1(45 - 48) - 2(36 - 42) + 3(32 - 35)
= 1(-3) - 2(-6) + 3(-3)
= -3 + 12 - 9 = 0The correct sign pattern for R1 is + - +. This leads to the correct determinant value.
+ - + chessboard pattern for the chosen row/column before starting the expansion.+ or -) explicitly, then the element, then its minor. E.g., + a11(M11) - a12(M12) + a13(M13).(-1)(i+j) rule before proceeding with the minor calculations. This is a quick and effective check.Ri → kRi (or Ci → kCi) multiplies the determinant's value by k. To preserve the original determinant value, one must multiply the new determinant by 1/k. Similarly, when a common factor k is present in a specific row or column, it can be factored out *once* from that row/column. A common error, especially in JEE Advanced, is to confuse this with scalar multiplication of an entire matrix (where k is factored from every element, resulting in kn for an n x n matrix), leading to incorrect determinant evaluation. k A means every element of matrix A is multiplied by k) and incorrectly apply this to determinants. Lack of attention to detail, insufficient practice with determinant-specific properties, and rushing through steps under exam pressure also contribute to this oversight. Ri → kRi (or Ci → kCi) is performed on a determinant Δ to get Δ', then Δ' = kΔ. Therefore, Δ = (1/k)Δ'. Always remember to account for this factor k (or 1/k) when changing rows/columns by scalar multiplication.k can be taken out from *only one* row or *only one* column at a time. For an n x n matrix, if k is a common factor to *all* elements of the matrix, then det(kA) = kn det(A).Consider the determinant: Δ = | 6 12 |
| 2 5 |
The actual value is (6×5) - (12×2) = 30 - 24 = 6.
Wrong Approach: A student performs R1 → (1/6)R1 to simplify the first row, getting:Δ' = | 1 2 |
| 2 5 |
And then incorrectly assumes Δ = Δ'. They evaluate Δ' = (1×5) - (2×2) = 5 - 4 = 1, concluding that the original determinant Δ is 1.
Using the same determinant: Δ = | 6 12 |
| 2 5 |
Actual value is 6.
Correct Approach: If the operation R1 → (1/6)R1 is performed, the new determinant Δ' will be (1/6) times the original determinant Δ.
So, Δ' = (1/6)Δ.
The transformed determinant is:Δ' = | 1 2 |
| 2 5 |
Evaluating Δ' = (1×5) - (2×2) = 5 - 4 = 1.
Now, using the relationship Δ' = (1/6)Δ, we can find the original determinant:1 = (1/6)ΔΔ = 6 × 1 = 6.
This matches the actual value, correctly accounting for the scalar factor.
kn if kA is considered).det(A) = a_21 * C_21 + a_22 * C_22 + a_23 * C_23 = a_21 * (-1)^(2+1) * M_21 + a_22 * (-1)^(2+2) * M_22 + a_23 * (-1)^(2+3) * M_23 = -a_21 * M_21 + a_22 * M_22 - a_23 * M_23| 1 | 2 | 3 |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
| 1 | 2 | 3 |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | + | - | + |
| Row 2 | - | + | - |
| Row 3 | + | - | + |
det(kA) = k * det(A) instead of the correct property det(kA) = kn * det(A) for an n x n matrix. Additionally, students might incorrectly take a common factor k from a determinant if it's present in only one element, or conversely, assume multiplying a determinant by k means multiplying all its elements by k. kA means every element is multiplied by k, the rules for determinants are distinct. The property det(kA) = kn det(A) is often misremembered or not fully grasped, leading to errors in simplification and evaluation. k is multiplied to a determinant, it signifies that k has been multiplied to one specific row or one specific column of the original matrix.k can be taken out from a determinant only if it is common to all elements of a specific row or a specific column.n x n matrix A, if you multiply the entire matrix by a scalar k, then det(kA) = kn * det(A). This is because multiplying matrix A by k is equivalent to multiplying each of its n rows (or columns) by k.A = [[a, b], [c, d]]. A student might incorrectly write:det(2A) = 2 * det(A)
A = [[1, 2], [3, 4]], then det(A) = (1*4) - (2*3) = -2.2 * det(A) = 2 * (-2) = -4.det([[2, 4], [3, 5]]), a student might mistakenly try to factor out '2' from the determinant as 2 * det([[1, 2], [3, 5]]) without checking if '2' is common to the entire row/column, even if this specific example is correct for a row operation.A = [[1, 2], [3, 4]], where det(A) = -2.det(2A), first calculate 2A:2A = [[2*1, 2*2], [2*3, 2*4]] = [[2, 4], [6, 8]]
det(2A) = (2*8) - (4*6) = 16 - 24 = -8.det(kA) = k2 * det(A).det(2A) = 22 * det(A) = 4 * (-2) = -8. This matches the direct calculation.det([[2, 4], [3, 5]]), you can correctly factor 2 from the first row:det([[2, 4], [3, 5]]) = 2 * det([[1, 2], [3, 5]])
kA) and determinant properties (like det(kA)) follow distinct rules. Don't mix them up.n x n matrix A, det(kA) = kn det(A). Memorize this and understand its derivation through row/column operations.k means multiplying only one row or one column by k.A student encountering a 3x3 matrix A with det(A) = 5 might incorrectly calculate:
det(2A) = 2 * det(A) = 2 * 5 = 10Using the correct property for the same 3x3 matrix A with det(A) = 5:
det(2A) = 23 * det(A) = 8 * 5 = 40If A = [[a, b], [c, d]], then det(A) = ad - bc.
If kA = [[ka, kb], [kc, kd]], then det(kA) = (ka)(kd) - (kb)(kc) = kยฒ(ad - bc) = kยฒ det(A).This hands-on verification solidifies the understanding that 'k' is raised to the power 'n'.| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | + | - | + |
| Row 2 | - | + | - |
| Row 3 | + | - | + |
lim xโ0 det(A(x))) is involved, apply the limit only to the *final, simplified expression* of the determinant.limxโ0 (1/xยฒ) det([[1+2x, 1-x], [1+x, 1-2x]]).x:1+2x โ 1, 1-x โ 1, 1+x โ 1, 1-2x โ 1.det([[1, 1], [1, 1]]) = 0.limxโ0 (1/xยฒ) * 0 = 0. This result is incorrect.limxโ0 (1/xยฒ) det([[1+2x, 1-x], [1+x, 1-2x]]).det([[1+2x, 1-x], [1+x, 1-2x]])= (1+2x)(1-2x) - (1-x)(1+x)= (1 - 4xยฒ) - (1 - xยฒ)= 1 - 4xยฒ - 1 + xยฒ= -3xยฒ.limxโ0 (1/xยฒ) * (-3xยฒ) = limxโ0 (-3) = -3.-3, not 0, highlighting the danger of premature approximation.det(kA) for a matrix A), leading to k^n * det(A) instead of k * det(A) when 'k' is factored from a single row or column. Another mistake is misunderstanding that row/column operations of the type R_i → R_i + kR_j do not change the determinant's value. det(kA) = k^n det(A) (where kA means every element of matrix A is multiplied by k) with the process of factoring out 'k' from a single row or column of a determinant, which only multiplies the determinant by 'k'. Lack of careful practice with elementary operations and their specific effects on determinants also contributes. R_i → R_i + kR_j (or C_i → C_i + kC_j) leave the value of the determinant unchanged. This is a crucial property for simplification.Let A = |a b|
|c d|
det(A) = ad - bc.
Student's Incorrect Thought Process:
If R1 → 2*R1, then the new determinant B = |2a 2b|
| c d|
Student might incorrectly assume det(B) = 2^2 * det(A) = 4(ad-bc),
confusing it with det(2A) for a 2x2 matrix A.
Let A = |a b|
|c d|
det(A) = ad - bc.
Correct Scalar Multiplication:
If B is formed by R1 → 2*R1 from A:
B = |2a 2b|
| c d|
det(B) = (2a)d - (2b)c = 2ad - 2bc = 2(ad - bc) = 2 * det(A).
Here, '2' is factored out from only one row.
Correct Row Operation:
If C is formed by R1 → R1 + kR2 from A:
C = |a+kc b+kd|
| c d |
det(C) = (a+kc)d - (b+kd)c = ad + kcd - bc - kdc = ad - bc = det(A).
The determinant value remains unchanged.
det(kA) = k^n det(A) (when 'k' multiplies the entire matrix A of order 'n') and taking a common factor 'k' from a single row or column, which makes the determinant 'k' times the original.n x n matrix, multiplying one row or column by k multiplies the determinant by k, but multiplying the entire matrix by k multiplies the determinant by k^n. det(kA) = k^n det(A) (where A is an n x n matrix) with the property that extracting a common factor k from a single row or column leaves the determinant as k * (original determinant). This misapplication of scaling factors leads to significant errors in final determinant values. n x n matrix A is multiplied by a scalar k to form kA, then det(kA) = k^n det(A).A is multiplied by a scalar k, the new determinant is k * det(A). This is crucial for simplifying determinants by taking out common factors from rows/columns.Consider a 3x3 matrix A. If we transform it by multiplying the first row by 2 (R1 → 2R1), a common mistake is to assume the new determinant is 2^3 * det(A) = 8 * det(A).
For the same 3x3 matrix A:
R1 → 2R1, then the determinant of the new matrix is 2 * det(A).A is multiplied by 2 (i.e., every element of A is multiplied by 2, forming 2A), then det(2A) = 2^3 * det(A) = 8 * det(A).CBSE vs JEE: Both exams test these properties. JEE often embeds them in more complex problems involving multiple transformations or variable factors, requiring a deeper conceptual grasp.
Consider evaluating:
| 1 2 3 |
| 4 5 6 |
| 7 8 9 |
Wrong approach: Direct expansion (1(45-48) - 2(36-42) + 3(32-35)), leading to more terms and arithmetic. While simple for this specific determinant, the principle holds for more complex ones.
Using row/column operations for the same determinant:
| 1 2 3 |
| 4 5 6 |
| 7 8 9 |
Correct approach: Apply Rโ โ Rโ - Rโ and Rโ โ Rโ - Rโ
(Alternatively, Rโ โ Rโ - 4Rโ and Rโ โ Rโ - 7Rโ to create zeros in the first column)
Applying Rโ โ Rโ - Rโ and Rโ โ Rโ - Rโ:
| 1 2 3 |
| 3 3 3 |
| 3 3 3 |
Now, since Rโ and Rโ are identical, the determinant is 0. (Or Rโ โ Rโ - Rโ, which makes a row of zeros). This avoids complex arithmetic entirely.
JEE Tip: For determinants with elements in A.P., like the example, the determinant is often zero. Always check for such patterns or opportunities to make rows/columns identical or proportional.
(a+h) โ a for small h, or simplifying complex algebraic terms erroneously due to haste). This approach is fundamentally incorrect for evaluating determinants, which strictly require exact algebraic manipulations. Such 'approximations' inevitably lead to an exactly wrong answer. This also includes misjudging when rows/columns are truly proportional or identical without rigorous verification. Ri โ Ri + kRj, Ci โ Ci + kCj, etc.) to introduce zeros, which simplifies the expansion process. These operations preserve the determinant's value.| 1 2 3 |A common mistake is to (incorrectly) assume that for a 'small'
| 4 5 6 |
| x+h y+h z+h |
h, the third row can be 'approximated' by [x, y, z], thus replacing the original third row to simplify calculations.Incorrect: | 1 2 3 | -> | 1 2 3 |
| 4 5 6 | | 4 5 6 |
| x+h y+h z+h | | x y z | (This changes the value of the determinant.)
| 1 2 3 |The correct approach is to use the property that if a row/column is a sum of two terms, the determinant can be expressed as a sum of two determinants:
| 4 5 6 |
| x+h y+h z+h |
| 1 2 3 | + | 1 2 3 |Now, the second determinant can be simplified by taking 'h' common from the third row:
| 4 5 6 | | 4 5 6 |
| x y z | | h h h |
| 1 2 3 | + h * | 1 2 3 |This is the exact and correct way to handle such expressions, yielding an exact determinant value, not an approximation. The key is to apply determinant properties rigorously.
| 4 5 6 | | 4 5 6 |
| x y z | | 1 1 1 |
Cij = (-1)i+j Mij is often confused, leading to incorrect application of the (-1)i+j part.+ - + / - + - / + - +) leads to errors.det(A) = ∑ aij Cij, where Cij = (-1)i+j Mij is the cofactor. + - +
- + -
+ - +
a11M11 - a12M12 + a13M13. A = ∣ 1 2 3 ∣
∣ 4 5 6 ∣
∣ 7 8 9 ∣ 1 × (5×9 - 6×8) + 2 × (4×9 - 6×7) + 3 × (4×8 - 5×7)= 1(45-48) + 2(36-42) + 3(32-35)= 1(-3) + 2(-6) + 3(-3) = -3 - 12 - 9 = -24A:1 × (5×9 - 6×8) - 2 × (4×9 - 6×7) + 3 × (4×8 - 5×7)= 1(45-48) - 2(36-42) + 3(32-35)= 1(-3) - 2(-6) + 3(-3) = -3 + 12 - 9 = 0+ - + / - + - / + - +) next to your determinant.(-1)i+j to determine the sign for each cofactor.| 1 | 2 | 3 |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
| 1 | 2 | 3 |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
| 5 | 6 |
| 8 | 9 |
| 4 | 6 |
| 7 | 9 |
| 4 | 5 |
| 7 | 8 |
| + | - | + |
| - | + | - |
| + | - | + |
(-1)^(i+j)) when expanding a determinant along a row or column, which is crucial for defining cofactors. This leads to an incorrect final value for the determinant. +, -, +, ...) associated with each element's position are common contributing factors. aij in a determinant has an associated cofactor Cij = (-1)^(i+j) * Mij, where Mij is the minor. The sign (-1)^(i+j) dictates whether the minor is added or subtracted. For a 3x3 determinant, the sign pattern is:| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | + | - | + |
| Row 2 | - | + | - |
| Row 3 | + | - | + |
R1), the formula is det(A) = a11C11 + a12C12 + a13C13. A = | 1 2 3 | | 4 5 6 | | 7 8 9 |R1 as:det(A) = 1 * (5*9 - 6*8) + 2 * (4*9 - 6*7) + 3 * (4*8 - 5*7)- sign for the second term a12 is incorrectly taken as +).A and expanding along R1 correctly:det(A) = 1 * (5*9 - 6*8) - 2 * (4*9 - 6*7) + 3 * (4*8 - 5*7)= 1 * (45 - 48) - 2 * (36 - 42) + 3 * (32 - 35)= 1 * (-3) - 2 * (-6) + 3 * (-3)= -3 + 12 - 9= 0+,-,+) above the elements of the chosen row/column.| 2 | 4 | 6 |
|---|---|---|
| 1 | 3 | 5 |
| 7 | 9 | 11 |
| 2 | 4 | 6 |
|---|---|---|
| 1 | 3 | 5 |
| 7 | 9 | 11 |
| 1 | 2 | 3 |
|---|---|---|
| 1 | 3 | 5 |
| 7 | 9 | 11 |
+ - +
- + -
+ - +
| 1 | 2 | 3 |
|---|---|---|
| 4 | 5 | 6 |
| 7 | 8 | 9 |
| 1 | 2 | 3 |
|---|---|---|
| 4 | 5 | 6 |
| 7 | 8 | 9 |
Problem: Find the area of a triangle with vertices A(0,0), B(1 m, 50 cm), C(2 m, 3 m).
Incorrect approach: Directly forming the matrix with mixed units for coordinates:
Area = 0.5 * |det(
0 0 1
1 50 1
2 3 1
)|This will yield a numerically correct determinant but a physically meaningless area due to mixing meters and centimeters without conversion.
Correct approach: Convert all coordinates to a consistent unit, e.g., meters, before forming the matrix.
Area = 0.5 * |det(
0 0 1
1 0.5 1
2 3 1
)|Evaluating this determinant will correctly give the area in square meters (m2), providing a physically accurate result.
| + | - | + |
|---|---|---|
| - | + | - |
| + | - | + |
| 1 | 2 | 3 |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
Cij for an element aij is given by (-1)i+j Mij, where Mij is the minor.+ - +
- + -
+ - +
Evaluate:| 1 2 3 || 4 5 6 || 7 8 9 |
Incorrect Expansion along R1:
Determinant = 1(5×9 - 8×6) + 2(4×9 - 7×6) + 3(4×8 - 7×5)
= 1(45 - 48) + 2(36 - 42) + 3(32 - 35)
= 1(-3) + 2(-6) + 3(-3)
= -3 - 12 - 9
= -24
(Mistake: The sign for the second term (element '2') should be negative, not positive.)
Evaluate:| 1 2 3 || 4 5 6 || 7 8 9 |
Correct Expansion along R1:
Determinant = 1 × (-1)1+1 × (5×9 - 8×6) - 2 × (-1)1+2 × (4×9 - 7×6) + 3 × (-1)1+3 × (4×8 - 7×5)
= 1(45 - 48) - 2(36 - 42) + 3(32 - 35)
= 1(-3) - 2(-6) + 3(-3)
= -3 + 12 - 9
= 0
+ - +Always adhere to this pattern when choosing which element to expand and what sign to associate with its minor.
- + -
+ - +
det(kA). They might incorrectly take a common factor k from all elements and write k * det(A), or apply row/column operations like Ri → kRi without understanding its impact on the determinant value. n in det(kA) = kn det(A) for an n x n matrix. k is a common factor to all elements of a single row or a single column, it can be taken out of the determinant.n x n matrix A, det(kA) = kn det(A). This is a crucial JEE property.Ri → Ri + kRj (or Ci → Ci + kCj) does not change the value of the determinant.Ri → kRi (or Ci → kCi) multiplies the determinant by k. To keep the determinant value unchanged, one must divide by k outside the determinant.Consider A = [[2, 4], [6, 8]].
Incorrect: det(A) = 2 * |[1, 2], [3, 4]| (taking 2 common from all elements).
Incorrect: If 2A = [[4, 8], [12, 16]], then det(2A) = 2 * det(A).Consider A = [[2, 4], [6, 8]].
Correct Factorization:
det(A) = |[2, 4],
[6, 8]|
= 2 * |[1, 2],
[6, 8]| (taking 2 common from R1 only)
= 2 * 2 * |[1, 2],
[3, 4]| (taking 2 common from R2 after R1 was factored)
= 4 * (1*4 - 2*3) = 4 * (-2) = -8.
Verification: Direct expansion: det(A) = (2*8) - (4*6) = 16 - 24 = -8.
Correct det(kA):
Let k=2. For a 2x2 matrix A, det(2A) = 22 * det(A) = 4 * det(A).
If A = [[2, 4], [6, 8]], then det(A) = -8.
So, det(2A) = 4 * (-8) = -32.
Let's check by calculating 2A: 2A = [[4, 8], [12, 16]].
det(2A) = (4*16) - (8*12) = 64 - 96 = -32. This matches!
| 1 | 2 | 3 |
|---|---|---|
| 0 | 4 | 1 |
| -1 | 0 | 2 |
| 1 | 2 | 3 |
|---|---|---|
| 0 | 4 | 1 |
| -1 | 0 | 2 |
| 4 | 1 |
| 0 | 2 |
| 0 | 1 |
| -1 | 2 |
| 0 | 4 |
| -1 | 0 |
(-1)i+j factor for each element's cofactor.+ - + / - + - / + - +) for elements.(-1)i+j. For CBSE and JEE, meticulously applying this sign is crucial. The pattern for a 3x3 matrix is:+ - +- + -+ - + | a | b | c |
| d | e | f |
| g | h | i |
|A| = a(ei - fh) + b(di - fg) + c(dh - eg)b term is incorrectly taken as positive instead of negative.| a | b | c |
| d | e | f |
| g | h | i |
|A| = a(ei - fh) - b(di - fg) + c(dh - eg)b term, which corresponds to the (1,2) position.| Cofactor Sign Pattern | ||
|---|---|---|
| + | - | + |
| - | + | - |
| + | - | + |
| 1 | 2 | 3 |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
| 5 | 6 |
| 8 | 9 |
| 4 | 6 |
| 7 | 9 |
| 4 | 5 |
| 7 | 8 |
| + | - | + |
| - | + | - |
| + | - | + |
Always apply these alternating signs correctly when expanding along any row or column.
+ - +
- + -
+ - +
+ - + - + - + - + | 1 2 3 | | 4 5 6 | | 7 8 9 | | 1 2 3 | | 4 5 6 | | 7 8 9 | Consider the determinant:
| 1 2 3 |A student might perform the operation R1 → 2R1, obtaining:
| 4 5 6 |
| 7 8 9 |
| 2 4 6 |
| 4 5 6 |
| 7 8 9 |
And then incorrectly assume that the determinant of the new matrix is the same as the original one, i.e., Det(new matrix) = Det(original matrix). This is incorrect.
Using the same original determinant:
| 1 2 3 |Let its value be D.
| 4 5 6 |
| 7 8 9 |
✗ Wrong Approach:
Evaluate D = | sin(0.01) cos(0.01) |
| -cos(0.01) sin(0.01) |
Student's Approximation: Using sin(x) ≈ x and cos(x) ≈ 1 for small x:
Elements become: sin(0.01) ≈ 0.01, cos(0.01) ≈ 1
D ≈ | 0.01 1 |
| -1 0.01 |
D ≈ (0.01)(0.01) - (1)(-1) = 0.0001 + 1 = 1.0001
✓ Correct Approach:
Evaluate D = | sin(0.01) cos(0.01) |
| -cos(0.01) sin(0.01) |
Exact Evaluation:
D = (sin(0.01))(sin(0.01)) - (cos(0.01))(-cos(0.01))
D = sin²(0.01) + cos²(0.01)
D = 1
Comparison: The approximated value (1.0001) is close but not exact. In JEE Advanced, options are often numerically very close, making the exact answer (1) critical. This highlights how an identity (sin²x + cos²x = 1) can be missed by premature approximation.
+ - +
- + -
+ - +
| 1 2 3 |
| 4 5 6 |
| 7 8 9 |
| 1 2 3 |
| 4 5 6 |
| 7 8 9 |
det(A) = ∑j=1n aij Cijdet(A) = a11M11 + a12M12 + a13M13 (Incorrect sign for `a12`)det(A) = a11(+M11) + a12(-M12) + a13(+M13)det(A) = a11C11 + a12C12 + a13C13 where `C11 = M11`, `C12 = -M12`, `C13 = M13`.| + | - | + |
|---|---|---|
| - | + | - |
| + | - | + |
Ri → kRi (multiplying a row by a scalar k), they often forget that this operation scales the determinant by k. This means the new determinant is k times the original determinant. They also sometimes incorrectly 'factor out' a scalar k that isn't common to all elements of a single row or column, or apply scalar factoring as if it were a matrix operation (where det(kA) = kndet(A) for an n x n matrix, not simply k*det(A) as many wrongly assume during intermediate steps). Ri → Ri + kRj does not change the determinant, and incorrectly extend this 'no change' rule to scalar multiplication operations. Hasty application of formulas without deep conceptual understanding, especially under exam pressure, further contributes to this error. Ri → kRi (or Ci → kCi), the determinant of the resulting matrix is k times the determinant of the original matrix. To maintain the original determinant's value, you must multiply the new determinant by 1/k, effectively writing: Det(A) = (1/k) * Det(Matrix after Ri → kRi).Ri → Ri + kRj (or Ci → Ci + kCj) leaves the determinant value unchanged.k can only be taken out if it is common to all elements of a single row or a single column.Let D = 1 2 3 4
The actual value of D = (1*4) - (2*3) = 4 - 6 = -2.
Student's Incorrect Step: The student performs R1 → 2R1 to simplify, getting D' = .2 4 3 4
Incorrect Assumption: The student then incorrectly assumes that D' still equals D, or forgets to account for the scalar multiplication, proceeding to evaluate D' = (2*4) - (4*3) = 8 - 12 = -4 and incorrectly stating that the original determinant D = -4.
Let D = 1 2 3 4
The actual value of D = (1*4) - (2*3) = 4 - 6 = -2.
Correct Approach (to keep original D value): To use the operation R1 → 2R1 and maintain the original determinant's value, one must write:
D = (1/2) * (2*1) (2*2) 3 4
D = (1/2) * 2 4 3 4
Now, evaluate the determinant on the RHS:
D = (1/2) * [(2*4) - (4*3)] = (1/2) * [8 - 12] = (1/2) * [-4] = -2.
This correctly yields the original determinant value, showing the necessary compensation for the scalar operation.
Ri → kRi, immediately write down the compensating factor 1/k outside the determinant to avoid forgetting it.Consider the determinant:
$$ egin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 end{vmatrix} $$
Incorrect expansion along the first row (e.g., applying wrong sign for a12):
$$ Delta = 1(5cdot9 - 6cdot8) + 2(4cdot9 - 6cdot7) + 3(4cdot8 - 5cdot7) $$
Here, the sign for the element '2' (a12) should be negative, not positive.
Using the same determinant:
$$ egin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 end{vmatrix} $$
Correct expansion along the first row:
$$ Delta = 1 cdot ext{Cofactor}(a_{11}) + 2 cdot ext{Cofactor}(a_{12}) + 3 cdot ext{Cofactor}(a_{13}) $$
$$ Delta = 1 cdot ((-1)^{1+1} egin{vmatrix} 5 & 6 \ 8 & 9 end{vmatrix}) + 2 cdot ((-1)^{1+2} egin{vmatrix} 4 & 6 \ 7 & 9 end{vmatrix}) + 3 cdot ((-1)^{1+3} egin{vmatrix} 4 & 5 \ 7 & 8 end{vmatrix}) $$
$$ Delta = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) $$
$$ Delta = 1(-3) - 2(-6) + 3(-3) $$
$$ Delta = -3 + 12 - 9 = 0 $$
A = | 2 ฮฉ 1000 mฮฉ |A student might incorrectly calculate the determinant as:
| 5 ฮฉ 1 kฮฉ |
A' = | 2 ฮฉ 1 ฮฉ |Now, calculate the determinant:
| 5 ฮฉ 1000 ฮฉ |
| + | - | + |
| - | + | - |
| + | - | + |
Always prioritize simplification using elementary row (R) or column (C) operations before expanding. The primary goal is to create as many zeros as possible in a single row or column. This drastically reduces the number of terms to be evaluated during expansion, often simplifying the determinant to a very straightforward form or even zero.
Consider the determinant: D = | x+a x+b x+c | | y+a y+b y+c | | z+a z+b z+c |
Directly expanding this 3x3 determinant involves calculating 6 terms, each being a product of three binomials. This is highly error-prone and extremely time-consuming, making it an impractical approach in JEE Main.
For the determinant from the 'Wrong Example':
Apply column operations C2 → C2 - C1 and C3 → C3 - C1.
This transforms the determinant to:D = | x+a b-a c-a | | y+a b-a c-a | | z+a b-a c-a |
Since the second and third columns are now identical, the value of the determinant is zero. This powerful simplification avoids all complex calculations, arriving at the answer quickly and accurately.
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