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Matrices - Operation on matrices: Addition, multiplication and multiplication with a scalar

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Matrix Addition: Two matrices AA and BB can be added if and only if they are of the same order (dimensions). Visualize this as placing one grid over another of identical size; the resulting sum C=A+BC = A + B is formed by adding elements at the same corresponding positions, such that cij=aij+bijc_{ij} = a_{ij} + b_{ij}.

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Scalar Multiplication: This operation involves multiplying every single entry of a matrix AA by a constant number kk. Geometrically, this represents a uniform scaling of the matrix data, where a matrix grid expands or contracts based on the value of kk. For example, k[aij]=[k⋅aij]k[a_{ij}] = [k \cdot a_{ij}].

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Matrix Multiplication Compatibility: Multiplication of two matrices AA and BB is only defined if the number of columns in the first matrix AA is equal to the number of rows in the second matrix BB. If AA is an m×nm \times n matrix and BB is an n×pn \times p matrix, the resulting matrix CC will have the dimensions m×pm \times p.

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The Row-Column Rule: To find the element in the ii-th row and jj-th column of the product ABAB, you compute the dot product of the ii-th row of AA and the jj-th column of BB. Visualize the row of the first matrix sliding horizontally across and 'pairing up' with the vertical column of the second matrix, multiplying the pairs and summing them up.

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Non-Commutative Property: Unlike real numbers, matrix multiplication is generally not commutative (AB≠BAAB \neq BA). Changing the order of multiplication often changes the result or makes the operation impossible due to dimension mismatch. Visually, the interaction between rows and columns changes entirely when the matrices are swapped.

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Additive Identity and Inverse: The Null Matrix OO (a matrix where every entry is zero) acts as the additive identity, meaning A+O=AA + O = A. For every matrix AA, there exists an additive inverse −A-A (where every element's sign is flipped) such that A+(−A)=OA + (-A) = O.

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Multiplicative Identity: The Identity matrix II is a square matrix with 11s along the main diagonal (from top-left to bottom-right) and 00s everywhere else. It acts as the number '1' in matrix algebra, satisfying the property AI=IA=AAI = IA = A.

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Distributive and Associative Laws: Matrix multiplication is associative, meaning A(BC)=(AB)CA(BC) = (AB)C, and distributive over addition, meaning A(B+C)=AB+ACA(B + C) = AB + AC. This allows for complex algebraic manipulation of matrix equations similar to standard algebra, provided the order of multiplication is maintained.

📐Formulae

A+B=[aij+bij]A + B = [a_{ij} + b_{ij}] (Addition of matrices of the same order)

kA=[k⋅aij]k A = [k \cdot a_{ij}] (Scalar multiplication)

cij=∑k=1naikbkjc_{ij} = \sum_{k=1}^{n} a_{ik} b_{kj} (General term for matrix multiplication product C=ABC = AB)

A+B=B+AA + B = B + A (Commutative Law of Addition)

(A+B)+C=A+(B+C)(A + B) + C = A + (B + C) (Associative Law of Addition)

k(A+B)=kA+kBk(A + B) = kA + kB (Distributive Law for Scalars)

A(BC)=(AB)CA(BC) = (AB)C (Associative Law of Multiplication)

A(B+C)=AB+ACA(B + C) = AB + AC (Distributive Law of Multiplication)

💡Examples

Problem 1:

Given A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} and B=[0−123]B = \begin{bmatrix} 0 & -1 \\ 2 & 3 \end{bmatrix}, find 2A+3B2A + 3B.

Solution:

Step 1: Calculate 2A2A. Multiply every element of AA by 22: 2A=[2(1)2(2)2(3)2(4)]=[2468]2A = \begin{bmatrix} 2(1) & 2(2) \\ 2(3) & 2(4) \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix} Step 2: Calculate 3B3B. Multiply every element of BB by 33: 3B=[3(0)3(−1)3(2)3(3)]=[0−369]3B = \begin{bmatrix} 3(0) & 3(-1) \\ 3(2) & 3(3) \end{bmatrix} = \begin{bmatrix} 0 & -3 \\ 6 & 9 \end{bmatrix} Step 3: Add the resulting matrices element-wise: 2A+3B=[2+04+(−3)6+68+9]=[211217]2A + 3B = \begin{bmatrix} 2+0 & 4+(-3) \\ 6+6 & 8+9 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 12 & 17 \end{bmatrix}

Explanation:

This problem demonstrates combining scalar multiplication and matrix addition. We scale each matrix individually before adding corresponding entries.

Problem 2:

Find the product ABAB if A=[2132]A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} and B=[1−104]B = \begin{bmatrix} 1 & -1 \\ 0 & 4 \end{bmatrix}.

Solution:

Step 1: Calculate element c11c_{11} (Row 1 of A×A \times Col 1 of BB): (2)(1)+(1)(0)=2+0=2(2)(1) + (1)(0) = 2 + 0 = 2 Step 2: Calculate element c12c_{12} (Row 1 of A×A \times Col 2 of BB): (2)(−1)+(1)(4)=−2+4=2(2)(-1) + (1)(4) = -2 + 4 = 2 Step 3: Calculate element c21c_{21} (Row 2 of A×A \times Col 1 of BB): (3)(1)+(2)(0)=3+0=3(3)(1) + (2)(0) = 3 + 0 = 3 Step 4: Calculate element c22c_{22} (Row 2 of A×A \times Col 2 of BB): (3)(−1)+(2)(4)=−3+8=5(3)(-1) + (2)(4) = -3 + 8 = 5 Result: AB=[2235]AB = \begin{bmatrix} 2 & 2 \\ 3 & 5 \end{bmatrix}

Explanation:

This uses the Row-Column rule. We multiply elements of the rows of the first matrix by elements of the columns of the second matrix and sum them to find each entry of the product.