| Apples | Bananas | Cherries | |
|---|---|---|---|
| Store A | 50 | 30 | 20 |
| Store B | 40 | 35 | 25 |
| Math | Physics | Chemistry | |
|---|---|---|---|
| Rahul | 85 | 78 | 92 |
| Priya | 90 | 81 | 88 |
Mastering the fundamental concepts of matrices and their algebraic operations is crucial for success in both CBSE Board Exams and JEE Main. These quick tips will help you navigate common pitfalls and strengthen your understanding.
Keep these quick tips handy as you practice problems. A strong grasp of these algebraic rules forms the foundation for more advanced topics in matrices and determinants.
Welcome to the intuitive world of Matrices and their Algebra! Understanding matrices isn't just about crunching numbers; it's about seeing them as powerful tools for organizing and manipulating data. Think of them as sophisticated spreadsheets or structured lists.
Just like you add, subtract, or multiply regular numbers to solve problems, you need operations for matrices to manipulate the structured data they hold. These operations allow us to combine, scale, and transform the information efficiently.
This is where it gets interesting and powerful. Matrix multiplication is not element-wise multiplication. It represents a more complex combination or transformation of data.
i-th row and j-th column of the product matrix, you take the i-th row of the first matrix and the j-th column of the second matrix, multiply their corresponding elements, and sum them up.By grasping these intuitive ideas, you'll find matrix algebra much less intimidating and appreciate its elegant way of handling complex data structures and operations, which forms the backbone of many advanced mathematical and computational applications.
This is perhaps one of the most intuitive applications. In 2D and 3D computer graphics, matrices are used extensively to perform transformations like translation (moving an object), scaling (resizing), rotation (turning), and reflection. Every point in a graphic object can be represented as a vector, and a transformation matrix can be multiplied with this vector to change its position, size, or orientation. Modern video games, animated movies, and CAD software heavily rely on matrix algebra for rendering realistic visuals.
Matrices play a crucial role in securing information. In cryptography, messages can be converted into numerical form (e.g., ASCII values) and then arranged into a matrix. This message matrix is then encrypted by multiplying it with an invertible encoding matrix (the key). To decrypt the message, the receiver uses the inverse of the encoding matrix. This process, often involving large matrices, makes messages very difficult to break without the correct key.
Matrices are used to model complex economic systems. For instance, the Leontief Input-Output Model uses matrices to analyze the interdependencies between different sectors of an economy. It helps economists understand how production in one sector affects others, aiding in policy-making and resource allocation. Matrices are also used in game theory to represent payoffs in strategic interactions.
In the age of big data, matrices are foundational. Datasets are often represented as matrices, where rows might be data points and columns are features. Matrix operations are at the heart of algorithms for:
JEE and CBSE Relevance: While direct questions asking for "real-world applications" are rare in JEE Main and Advanced, understanding these uses can provide a deeper conceptual grasp of why matrices and their operations are structured the way they are. For CBSE, some word problems might implicitly involve scenarios that can be modeled using matrices, encouraging practical thinking.
Matrices, therefore, are not just theoretical constructs but essential tools that underpin many modern technologies and analytical methods. Mastering matrix algebra equips you with a powerful problem-solving capability.
Analogies serve as powerful tools in mathematics, simplifying complex concepts by relating them to familiar real-world scenarios. For 'Matrices and Algebra of Matrices,' understanding these analogies can provide an intuitive grasp, especially useful for initial learning and conceptual clarity in both CBSE Board exams and JEE Main preparation.
This is the most complex operation and often requires a more abstract analogy:
Wood Metal
A [ 2 3 ]
B [ 4 1 ]
SupplierX SupplierY
Wood [ 10 12 ]
Metal[ 5 6 ]
SupplierX SupplierY
A [ (2*10 + 3*5) = 35 (2*12 + 3*6) = 42 ]
B [ (4*10 + 1*5) = 45 (4*12 + 1*6) = 54 ]
By relating these operations to everyday scenarios, students can build a stronger conceptual foundation for mastering matrices in their JEE and board exam preparations.
JEE Relevance: Although seemingly basic, these are the most frequent sources of calculation errors in competitive exams. Speed and accuracy in arithmetic save time and ensure correctness.
Motivation: Mastering these foundational concepts will make your journey through matrices much smoother. Many students struggle with matrices not due to the matrix concepts themselves, but because of weaknesses in basic arithmetic and algebra. Spend a little time shoring up these areas if you feel less confident.
Navigating matrices requires precision. Students often fall into predictable traps during exams. Being aware of these common pitfalls can significantly improve accuracy and prevent loss of marks.
This is arguably the most frequent mistake. Unlike scalar multiplication, matrix multiplication is generally not commutative. That is, AB ≠ BA, even if both products are defined and of the same order.
(A+B)^2 or (A-B)^2 require careful expansion as (A+B)^2 = A^2 + AB + BA + B^2, not A^2 + 2AB + B^2 unless AB = BA.In scalar algebra, if ab = 0 then a=0 or b=0. Similarly, if ax = ay (and a ≠ 0), then x = y. These laws generally do not hold true for matrices.
AB = O (the zero matrix), it is not necessary that A = O or B = O. Both A and B can be non-zero matrices.AX = AY (where A is a non-zero matrix), it is not necessary that X = Y unless matrix A is invertible.Basic operations on matrices have strict dimension requirements. Overlooking these leads to undefined operations.
AB to be defined, the number of columns in matrix A must be equal to the number of rows in matrix B. If A is m × n and B is p × q, then n must equal p. The resulting matrix AB will be of order m × q.Students sometimes treat the scalar 0 interchangeably with the zero matrix O. While A + O = A and A + (-A) = O are true, in equations involving multiplication, a scalar 0 is different from O.
X = A + 0, it is incorrect. It should be X = A + O (if 0 represents the zero matrix). The context usually clarifies, but be mindful of the distinction.
By internalizing these common traps, you can approach matrix problems with greater confidence and accuracy. Double-check assumptions related to commutativity and cancellation laws, and always verify dimension compatibility.
Mastering matrices and their algebra is fundamental for both JEE and Board exams. This section encapsulates the most critical concepts and properties you must remember.
Key Takeaways: Matrices and Algebra of Matrices
| Exam Focus | Key Aspects |
|---|---|
| CBSE Boards | Focus on basic operations (addition, scalar multiplication, matrix multiplication), properties of transpose, and expressing a matrix as sum of symmetric and skew-symmetric parts. |
| JEE Main | Emphasis on implications of non-commutative matrix multiplication, properties like $(AB)^T = B^T A^T$, identification of idempotent/nilpotent/involutory matrices, and applying these properties in complex problems. |
Remember these fundamental concepts and properties thoroughly. They are the building blocks for solving more advanced problems in Matrices and Determinants!
A systematic problem-solving approach is crucial for tackling matrix and algebra of matrices questions effectively in both JEE and board exams. Understanding the underlying concepts is key, but applying them methodically under exam conditions requires a structured strategy.
Follow these steps to efficiently solve problems involving matrix operations:
Example: Given matrices $A = egin{bmatrix} 1 & 2 \ 3 & 4 end{bmatrix}$ and $B = egin{bmatrix} -1 & 0 \ 1 & 2 end{bmatrix}$, find $C$ such that $C = 2A + B^T$.
Mastering this systematic approach will build confidence and significantly improve your speed and accuracy in matrix-related problems.
Example (CBSE-style):
If A = [ [2, 3], [1, 4] ] and B = [ [1, 0], [2, 5] ], find (A + B)T and verify that (A + B)T = AT + BT.
This type of question directly tests matrix addition, transpose definition, and a key property, which is very typical for CBSE board exams.
Mastering these fundamental concepts and practicing a variety of problems from your NCERT textbook and previous year's CBSE papers will ensure a strong score in this section.
This section focuses on the critical areas within "Matrices and algebra of matrices" that are frequently tested in the JEE Main examination. While basic operations are covered, JEE often delves deeper into properties, special types of matrices, and conceptual applications.
| Aspect | CBSE Board Exams | JEE Main |
|---|---|---|
| Operations | Focus on basic addition, scalar and matrix multiplication. | Includes advanced scenarios of multiplication, powers of matrices. |
| Matrix Types | Basic understanding of diagonal, symmetric, skew-symmetric. | Deep conceptual understanding and application of Idempotent, Nilpotent, Involutory matrices. |
| Properties | Basic properties of addition and transpose. | Emphasis on non-commutativity of multiplication, (AB)' = B'A', and decomposition of matrices. Trace properties are also relevant. |
| Question Style | Direct calculation, procedural questions. | Conceptual, multi-step, often involving properties and special matrix types. |
JEE Tip: Always double-check matrix dimensions before performing multiplication. Be extremely careful with the order of multiplication (AB ≠ BA) as it's a common trap.
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AX = B
X = BA-1 <-- INCORRECT!This incorrectly assumes that post-multiplying by A-1 on the left side of the equation (which would be (AX)A-1) results in X, and also that BA-1 is equivalent to A-1B.AX = B
A-1(AX) = A-1B <-- Pre-multiplying by A-1
(A-1A)X = A-1B
IX = A-1B
X = A-1B <-- CORRECT!Similarly, for XA = B, we would post-multiply by A-1 to get X = BA-1.Let A =
| 1 | 2 |
| 3 | 4 |
Wrong Calculation:
2A =
| 2 * 1 | 2 |
| 2 * 3 | 4 |
| 2 | 2 |
| 6 | 4 |
Here, the scalar '2' was incorrectly applied only to the first column, leaving other elements unchanged.
Let A =
| 1 | 2 |
| 3 | 4 |
Correct Calculation:
2A =
| 2 * 1 | 2 * 2 |
| 2 * 3 | 2 * 4 |
| 2 | 4 |
| 6 | 8 |
Each element of matrix A is correctly multiplied by the scalar '2', adhering to the definition of scalar multiplication.
| 2 | 5 |
| 1 | 3 |
| 100 | 200 |
| 300 | 400 |
| 2 | 5 |
| 1 | 3 |
| 100 | 200 |
| 300 | 400 |
| 2+100 | 5+200 |
| 1+300 | 3+400 |
| 102 | 205 |
| 301 | 403 |
A common minor error in Matrices and Algebra of Matrices, particularly for JEE Main, is making sign mistakes while calculating cofactors. Students often correctly find the minor of an element but forget to apply the alternating sign factor, (-1)^(i+j), where i and j are the row and column indices of the element.
(-1)^(i+j) rule or the checkerboard sign pattern (+ - +, - + -, etc.) is sometimes momentarily forgotten under exam pressure.i+j (e.g., thinking 1+2 is even instead of odd) can lead to an incorrect sign.Always explicitly remember and apply the sign factor (-1)^(i+j) for each cofactor. Visualizing the checkerboard sign pattern can be very helpful:
+ - +
- + -
+ - +
For CBSE, this is fundamental to finding the adjoint and inverse. For JEE, it's a critical step within larger problems involving determinants or matrix inverses, where a single sign error can lead to a completely wrong answer.
Given matrix A = egin{pmatrix} 2 & 3 \ 4 & 5 end{pmatrix}. Find the cofactor C_12 (cofactor of element 3).
Incorrect approach: Student calculates minor M_12 = 4 and wrongly assumes C_12 = M_12 = 4 (forgetting the sign).
Given matrix A = egin{pmatrix} 2 & 3 \ 4 & 5 end{pmatrix}. Find the cofactor C_12 (cofactor of element 3).
Correct approach:
i=1, j=2.M_12: This is the determinant of the matrix obtained by deleting row 1 and column 2, which is det(4) = 4.C_12 = (-1)^(1+2) * M_12 = (-1)^3 * 4 = -1 * 4 = -4.(-1)^(i+j) in an intermediate step, especially for complex matrices.i+j is even or odd.Given matrix $A = egin{pmatrix} cos(0.01) & sin(0.01) \ -sin(0.01) & cos(0.01) end{pmatrix}$.
Incorrect Approximation: Approximating $A$ as $A_{approx} = egin{pmatrix} 1 & 0.01 \ -0.01 & 1 end{pmatrix}$ by using $cos x approx 1$ and $sin x approx x$ for small $x$. If the problem asks for $A^{100}$, using $A_{approx}^{100}$ will give a significantly different and wrong result compared to the exact calculation.
For the same matrix $A = egin{pmatrix} cos(0.01) & sin(0.01) \ -sin(0.01) & cos(0.01) end{pmatrix}$, if we need to calculate $A^{100}$, recognize that $A$ is a 2D rotation matrix $R( heta)$ where $ heta = 0.01$ radians.
Correct Approach: The property of rotation matrices states $R( heta)^n = R(n heta)$. Therefore, $A^{100} = R(100 imes 0.01) = R(1) = egin{pmatrix} cos(1) & sin(1) \ -sin(1) & cos(1) end{pmatrix}$. This method maintains exactness and uses matrix properties directly, avoiding premature scalar approximations.
| 1 | 2 |
| 3 | 4 |
| 5 | 6 |
| 7 | 8 |
| 1 | 2 |
| 3 | 4 |
| 5 | 6 |
| 7 | 8 |
| (1*5+2*7) | (1*6+2*8) |
| (3*5+4*7) | (3*6+4*8) |
| 19 | 22 |
| 43 | 50 |
| 5 | 6 |
| 7 | 8 |
| 1 | 2 |
| 3 | 4 |
| (5*1+6*3) | (5*2+6*4) |
| (7*1+8*3) | (7*2+8*4) |
| 23 | 34 |
| 31 | 46 |
3A = [[3 * 0.33, 3 * 0.4], [3 * 0.14, 3 * 0.3]]
= [[0.99, 1.2], [0.42, 0.9]] (Introduces rounding errors, e.g., 0.99 instead of 1)
3A = [[3 * (1/3), 3 * (2/5)], [3 * (1/7), 3 * (3/10)]]
= [[1, 6/5], [3/7, 9/10]] (Maintains exact fractional values)
| Col 1 | Col 2 | Col 3 | |
|---|---|---|---|
| Row 1 | + | - | + |
| Row 2 | - | + | - |
| Row 3 | + | - | + |
| Item 1 |
|---|
| Item 2 |
| 10 |
| 15 |
| Item 1 |
|---|
| Item 2 |
| 150 |
| 200 |
| 10 + 150 |
| 15 + 200 |
| 160 |
| 215 |
| 150 |
| 200 |
| 11250 |
| 15000 |
| 10 |
| 15 |
| 10000 |
| 15000 |
| 10000 + 11250 |
| 15000 + 15000 |
| 21250 |
| 30000 |
| 2 | 3 |
| 1 | 4 |
| 6 | 3 |
| 1 | 4 |
| 6 | 9 |
| 1 | 12 |
| 2 | 3 |
| 1 | 4 |
| 3 × 2 | 3 × 3 |
| 3 × 1 | 3 × 4 |
| 6 | 9 |
| 3 | 12 |
(A+B)Β² = AΒ² + 2AB + BΒ²This is wrong because it assumes AB = BA, which is not generally true for matrices.
(A+B)Β² = (A+B)(A+B)This is the correct expansion, maintaining the order of multiplication at all steps.
= A(A+B) + B(A+B)
= AΒ² + AB + BA + BΒ²
det(2A) = 2 * det(A) = 2 * 5 = 10det(2A) = 23 * det(A) (since the order n=3)det(2A) = 8 * 5 = 40| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | + | - | + |
| Row 2 | - | + | - |
| Row 3 | + | - | + |
// Incorrect Element-wise Multiplication
Wrong A * B = [[1*5, 2*6], [3*7, 4*8]]
= [[5, 12], [21, 32]]
// Incorrect Commutativity (e.g., calculating BA instead of AB)
C11 = (1 * 5) + (2 * 7) = 5 + 14 = 19
C12 = (1 * 6) + (2 * 8) = 6 + 16 = 22
C21 = (3 * 5) + (4 * 7) = 15 + 28 = 43
C22 = (3 * 6) + (4 * 8) = 18 + 32 = 50
Therefore, AB = [[19, 22], [43, 50]]
A student incorrectly attempts to multiply Matrix P (order 3x2) with Matrix Q (order 3x4). They might mistakenly assume a 3x4 or 3x2 result, or try to multiply them without checking compatibility. The fundamental error here is not recognizing that the inner "units" (dimensions) 2 and 3 do not match, making the product PQ undefined.
Another common mistake is when given A (2x3) and B (3x2), students might correctly find AB (2x2) but then incorrectly assume BA will also be 2x2 (it's 3x3) or even 2x3. They fail to 'convert' the dimensions correctly for the reversed product.
Consider Matrix A of order 2x3 and Matrix B of order 3x4.
(-1)^(i+j), where 'i' is the row number and 'j' is the column number. (-1)^(i+j) factor.i+j).(-1)^(i+j) for each element's minor when calculating its cofactor. For a 3x3 matrix, the sign pattern is:+ - +- + -+ - +(-1)^(i+j). Ensure meticulous attention to detail at each step of expansion. For JEE Advanced, speed combined with accuracy is crucial; practicing systematic approaches is key. A = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]. Expanding along the first row, a common mistake for the element a_12 = 2 is to take its cofactor as +(4*9 - 6*7), ignoring the negative sign from (-1)^(1+2) = -1.A = [[1, 2, 3], [4, 5, 6], [7, 8, 9]], the cofactor of a_12 (element 2) is:C_12 = (-1)^(1+2) * det([[4, 6], [7, 9]])C_12 = -1 * (4*9 - 6*7)C_12 = -1 * (36 - 42)C_12 = -1 * (-6) = 61*C_11 + 2*C_12 + 3*C_13, where C_12 correctly includes the sign.(I + A)^2 β I + 2A(I + A)^2 = (I + A)(I + A) = I^2 + IA + AI + A^2 = I + A + A + A^2 = I + 2A + A^2| 1 | 0 |
| 0 | 0 |
| 0 | 1 |
| 0 | 0 |
| 0 | 1 |
| 0 | 0 |
| 0 | 0 |
| 0 | 0 |
| 1 | 0 |
| 0 | 0 |
| 0 | 1 |
| 0 | 0 |
| 0 | 0 |
| 0 | 0 |
| 0 | 0 |
| 0 | 0 |
| 1 | 1 |
| 0 | 0 |
| 1 | 1 |
| 0 | 0 |
| 1 | 1 |
| 0 | 0 |
(A + B)Β² = AΒ² + 2AB + BΒ²(A - B)(A + B) = AΒ² - BΒ²(A + B)Β² = (A + B)(A + B) = A(A + B) + B(A + B) = AΒ² + AB + BA + BΒ²(A - B)(A + B) = A(A + B) - B(A + B) = AΒ² + AB - BA - BΒ²+ - +- + -+ - +| 2 | -1 | 3 |
| 1 | 0 | -2 |
| 4 | 5 | 1 |
(A+B)Β² = AΒ² + 2AB + BΒ²(A+B)Β² = (A+B)(A+B) = A(A+B) + B(A+B) = AΒ² + AB + BA + BΒ²| 1 | 2 |
| 3 | 4 |
| 5 |
| 6 |
| 1 | 2 |
| 3 | 4 |
| 5 |
| 6 |
| (1*5 + 2*6) |
| (3*5 + 4*6) |
| 17 |
| 39 |
A prevalent error is assuming matrix multiplication is commutative (AB = BA), treating matrices like scalars. However, matrix multiplication is generally non-commutative.
This mistake stems from over-generalizing scalar algebra properties (a × b = b × a) to matrices. A lack of deep understanding of matrix-specific algebraic rules and insufficient practice contribute to this conceptual gap. Students fail to recognize that the order of operations significantly impacts the result in matrix algebra.
Always remember that matrix multiplication is NOT generally commutative (AB ≠ BA). The order of multiplication is critical. Commutativity occurs only in specific cases (e.g., a matrix and its inverse or identity matrix). For JEE Main, adhere strictly to matrix multiplication rules, understanding that changing the order often changes the outcome.
Incorrectly using (AB)-1 = A-1B-1 instead of the correct B-1A-1, or carelessly interchanging the order of matrices in algebraic manipulations (e.g., transforming AX = B into XA = B without proper pre-multiplication by an inverse).
Consider matrices:
A = [[1, 2],
[3, 4]]
B = [[0, 1],
[1, 0]]
Calculating their products:
AB =
[[2, 1],
[4, 3]]
BA =
[[3, 4],
[1, 2]]
Since AB ≠ BA, this demonstrates that matrix multiplication is non-commutative.
Let A and B be two matrices.
Incorrect expansion:
(A+B)2 = A2 + 2AB + B2
Let A and B be two matrices.
Correct expansion:
(A+B)2 = (A+B)(A+B)
= A(A+B) + B(A+B)
= A.A + A.B + B.A + B.B
= A2 + AB + BA + B2
| 2 | -1 |
| 3 | 4 |
| -1 | 5 |
| 0 | -2 |
| 2 - (-1) | -1 - 5 |
| 3 - 0 | 4 - (-2) |
| 2 - 1 | -6 |
| 3 | 4 - 2 |
| 1 | -6 |
| 3 | 2 |
| 2 - (-1) | -1 - 5 |
| 3 - 0 | |
| 4 - (-2) |
| 2 + 1 | -6 |
| 3 | 4 + 2 |
| 3 | -6 |
| 3 | 6 |
CBSE & JEE Tip: This dimensional compatibility is analogous to ensuring 'units' are consistent before calculations in physics problems. Treat matrix orders as crucial 'units' that dictate permissible operations.
| + | - | + |
|---|---|---|
| - | + | - |
| + | - | + |
| 1 | 2 |
| 3 | 4 |
| 0 | 1 |
| 1 | 0 |
| 2 | 1 |
| 4 | 3 |
| 3 | 4 |
| 1 | 2 |
(A + B)2 = A2 + 2AB + B2(A - B)(A + B) = A2 - B2(A + B)2 = (A + B)(A + B) = A(A + B) + B(A + B) = A2 + AB + BA + B2(A - B)(A + B) = A(A + B) - B(A + B) = A2 + AB - BA - B2Given matrices A and B. Incorrectly assuming:
(A + B)Β² = AΒ² + 2AB + BΒ²
or
(A - B)(A + B) = AΒ² - BΒ²
Given matrices A and B. The correct expansions are:
(A + B)Β² = (A + B)(A + B) = AΒ² + AB + BA + BΒ²
(A - B)(A + B) = A(A + B) - B(A + B) = AΒ² + AB - BA - BΒ²
These simplified forms (like AΒ² + 2AB + BΒ² or AΒ² - BΒ²) are only valid if and only if AB = BA.
+ - +).
- + -
+ - +
A = [[1, 2, 3], [4, 5, 6]] (2x3)
B = [[7, 8], [9, 10], [11, 12]] (3x2)
Wrong: A + B is attempted. This is incorrect because A and B do not have the same order.
A = [[1, 2], [3, 4], [5, 6]] (3x2)
B = [[7, 8, 9], [10, 11, 12], [13, 14, 15]] (3x3)
Wrong: A Γ B is attempted. This is incorrect because columns of A (2) ≠ rows of B (3).
A = [[1, 2], [3, 4]] (2x2)
B = [[5, 6], [7, 8]] (2x2)
Correct: A + B is possible because both are 2x2 matrices.
A + B = [[1+5, 2+6], [3+7, 4+8]] = [[6, 8], [10, 12]] (2x2)
A = [[1, 2, 3], [4, 5, 6]] (2x3)
B = [[7, 8], [9, 10], [11, 12]] (3x2)
Correct: A Γ B is possible because columns of A (3) = rows of B (3).
Resulting matrix will be of order 2x2. Calculation proceeds as per matrix multiplication rules.
Students frequently err by writing:
(AB)T = ATBTIf A is 2x3 and B is 3x2, then AB is 2x2, so (AB)T is 2x2. However, AT is 3x2 and BT is 2x3. The product ATBT would be 3x3. This dimensional mismatch (2x2 ≠ 3x3) clearly shows the formula ATBT is incorrect, and often undefined.
Applying the Reversal Law for A (2x3) and B (3x2):
(AB)T = BTATHere, BT is 2x3 and AT is 3x2. Their product BTAT results in a 2x2 matrix. This dimension (2x2) correctly matches that of (AB)T. Dimensional consistency is a key indicator of the correct formula.
Cij in the product matrix C, take the ith row of matrix A and the jth column of matrix B. Multiply their corresponding elements and then sum these products. This is often referred to as the 'row-by-column' rule. A = [[1, 2], [3, 4]] and B = [[5, 6], [7, 8]]. A x B = [[1*5, 2*6], [3*7, 4*8]] = [[5, 12], [21, 32]]A = [[1, 2], [3, 4]] and B = [[5, 6], [7, 8]]. C11 = (1 * 5) + (2 * 7) = 5 + 14 = 19C12 = (1 * 6) + (2 * 8) = 6 + 16 = 22C21 = (3 * 5) + (4 * 7) = 15 + 28 = 43C22 = (3 * 6) + (4 * 8) = 18 + 32 = 50A x B = [[19, 22], [43, 50]]. This is a crucial concept for both CBSE and JEE.If A and B are matrices, then (A+B)2 is always equal to A2 + 2AB + B2.For any two square matrices A and B of the same order:
(A+B)2 = (A+B)(A+B)
= AΒ·A + AΒ·B + BΒ·A + BΒ·B
= A2 + AB + BA + B2.
Only if A and B commute (i.e., AB = BA), can this be further simplified to A2 + 2AB + B2.A student might incorrectly expand:
(A + B)Β² = AΒ² + 2AB + BΒ²
This is wrong because it assumes AB = BA, which allows combining AB + BA into 2AB.
The correct expansion of (A + B)Β² is:
(A + B)Β² = (A + B)(A + B)
= A(A + B) + B(A + B)
= AΒ² + AB + BA + BΒ²
Since AB ≠ BA in general, this cannot be simplified further to AΒ² + 2AB + BΒ².
Similarly, (A - B)(A + B) = AΒ² + AB - BA - BΒ² ≠ AΒ² - BΒ².
| Col 1 | Col 2 | Col 3 | |
|---|---|---|---|
| Row 1 | + | - | + |
| Row 2 | - | + | - |
| Row 3 | + | - | + |
The correct approach involves a two-step process:
Consider a problem where a system involves two forces: F1 = 10 N and F2 = 5 kN.
Incorrect Approach:
Students might form a matrix of forces directly as:
M = [[10],
[5]]
If this matrix is then used, for example, to calculate the sum of forces (if a vector sum was intended conceptually) or multiplied by another matrix representing physical properties, the numerical result will be inconsistent and physically meaningless because '10' represents Newtons while '5' represents kilonewtons. A sum of these values (e.g., 10+5=15) would incorrectly mix units, leading to a wrong answer for total force.
For the same forces F1 = 10 N and F2 = 5 kN:
Correct Approach:
Step 1: Ensure Unit Consistency.
Convert all forces to a single unit, for instance, Newtons:
F1 = 10 N
F2 = 5 kN = 5 * 1000 N = 5000 N
Step 2: Form the Matrix with Consistent Numerical Values.
M = [[10],
[5000]]
Now, any matrix operation performed on M will use numerically consistent values. For example, if we were to conceptually sum these forces, the result would be 5010 N, which is physically correct. Similarly, if M is part of a larger matrix equation, all elements are now in the same base unit, preventing errors.
If A and B are matrices:
1. (A+B)2 = A2 + 2AB + B2 (Incorrect!)
2. (AB)-1 = A-1B-1 (Incorrect!)
3. (AB)T = ATBT (Incorrect!)For matrices A and B:
1. (A+B)2 = A2 + AB + BA + B2
2. (AB)-1 = B-1A-1 (provided A, B are invertible)
3. (AB)T = BTAT
Example:
Let A = [[1, 2], [3, 4]] and B = [[0, 1], [1, 0]]
AB = [[1*0+2*1, 1*1+2*0], [3*0+4*1, 3*1+4*0]] = [[2, 1], [4, 3]]
BA = [[0*1+1*3, 0*2+1*4], [1*1+0*3, 1*2+0*4]] = [[3, 4], [1, 2]]
Clearly, AB β BA.Given matrices A and B, students might incorrectly simplify expressions like:
(A + B)Β² = AΒ² + 2AB + BΒ²
This is wrong because (A + B)Β² = (A + B)(A + B) = AΒ² + AB + BA + BΒ².
Since AB β BA generally, 2AB β AB + BA.
Let A = [[1, 2], [0, 1]] and B = [[1, 0], [1, 1]]
Calculate AB:
AB = [[(1*1 + 2*1), (1*0 + 2*1)],
[(0*1 + 1*1), (0*0 + 1*1)]]
= [[3, 2],
[1, 1]]
Calculate BA:
BA = [[(1*1 + 0*0), (1*2 + 0*1)],
[(1*1 + 1*0), (1*2 + 1*1)]]
= [[1, 2],
[1, 3]]
As seen, AB β BA. This example clearly demonstrates the non-commutative nature of matrix multiplication.
(A + B)Β² = AΒ² + 2AB + BΒ² (Incorrect)(A - B)(A + B) = AΒ² - BΒ² (Incorrect)(A + B)Β² = (A + B)(A + B) = A(A + B) + B(A + B) = AΒ² + AB + BA + BΒ²(A - B)(A + B) = A(A + B) - B(A + B) = AΒ² + AB - BA - BΒ²AB and BA are kept separate because they are generally not equal. For CBSE and JEE Main, understanding this distinction is fundamental.(AB)X = C
(AB)-1(AB)X = (AB)-1C
IX = B-1A-1C
X = B-1A-1C
[1 2](2x2) and B =
[3 4]
[5 6 7](2x3).
[8 9 10]
[1 2](2x2) and B =
[3 4]
[5 6 7](2x3).
[8 9 10]
+ - +
- + -
+ - +
Wrong: For any two square matrices A and B of the same order, (A+B)Β² = AΒ² + 2AB + BΒ².
Correct: For any two square matrices A and B of the same order, (A+B)Β² = (A+B)(A+B) = A(A+B) + B(A+B) = AΒ² + AB + BA + BΒ².
JEE Tip: This identity simplifies to AΒ² + 2AB + BΒ² ONLY IF AB = BA. Always perform the full expansion unless commutativity is guaranteed.
Given matrices A and B, a student might incorrectly assume:
(A + B)Β² = AΒ² + 2AB + BΒ²
This expansion is only valid if AB = BA, which is a special case, not a general rule.
Let A = [[1, 2], [3, 4]] and B = [[5, 6], [7, 8]].
AB = [[1*5 + 2*7, 1*6 + 2*8], [3*5 + 4*7, 3*6 + 4*8]] = [[5+14, 6+16], [15+28, 18+32]] = [[19, 22], [43, 50]]
BA = [[5*1 + 6*3, 5*2 + 6*4], [7*1 + 8*3, 7*2 + 8*4]] = [[5+18, 10+24], [7+24, 14+32]] = [[23, 34], [31, 46]]
Clearly, AB β BA.
Therefore, the correct expansion for (A+B)Β² is (A+B)(A+B) = A(A+B) + B(A+B) = AΒ² + AB + BA + BΒ².
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