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Matrices - Simple properties of addition, multiplication and scalar multiplication

Grade 12CBSE

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

🔑Concepts

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Matrix Order and Notation: A matrix AA of order m×nm \times n is a rectangular array of numbers arranged in mm horizontal rows and nn vertical columns. The element aija_{ij} represents the entry located at the intersection of the ithi^{th} row and jthj^{th} column. Visually, imagine a grid where the first index tells you how far down to go and the second index tells you how far across to go.

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Matrix Addition: Addition is only defined for matrices of the same order. If A=[aij]A = [a_{ij}] and B=[bij]B = [b_{ij}], then A+B=[aij+bij]A + B = [a_{ij} + b_{ij}]. Visually, this is like stacking two transparent grids of the same size and adding the numbers that overlap at each position. It follows the commutative law A+B=B+AA + B = B + A and the associative law (A+B)+C=A+(B+C)(A + B) + C = A + (B + C).

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Scalar Multiplication: Multiplying a matrix AA by a scalar kk (a real number) results in a matrix where every element aija_{ij} is replaced by k⋅aijk \cdot a_{ij}. Geometrically, this acts as a scaling factor that expands or contracts every value in the matrix grid uniformly.

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Existence of Additive Identity and Inverse: The zero matrix OO (where every element is 0) serves as the additive identity such that A+O=O+A=AA + O = O + A = A. For every matrix AA, there exists an additive inverse −A-A (where every element is −aij-a_{ij}) such that A+(−A)=OA + (-A) = O.

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Matrix Multiplication Rule: Two matrices AA and BB can be multiplied to form ABAB if and only if the number of columns in AA is equal to the number of rows in BB. If AA is of order m×nm \times n and BB is of order n×pn \times p, the resulting matrix CC will be of order m×pm \times p. Visually, you take the ithi^{th} row of AA and 'dot product' it with the jthj^{th} column of BB to find the element cijc_{ij}.

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Properties of Matrix Multiplication: Multiplication is non-commutative in general, meaning ABAB is usually not equal to BABA. However, it is associative A(BC)=(AB)CA(BC) = (AB)C and distributive over addition A(B+C)=AB+ACA(B + C) = AB + AC. The Identity matrix II (a square matrix with 1s on the main diagonal and 0s elsewhere) acts as the multiplicative identity such that AI=IA=AAI = IA = A.

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Properties of Scalar Multiplication: Scalar multiplication distributes over matrix addition, expressed as k(A+B)=kA+kBk(A + B) = kA + kB, and over the sum of scalars, expressed as (k+l)A=kA+lA(k + l)A = kA + lA. This ensures that linear combinations of matrices behave predictably during algebraic manipulation.

📐Formulae

Addition: A+B=[aij+bij]A + B = [a_{ij} + b_{ij}], where A,B∈Mm×nA, B \in M_{m \times n}

Scalar Multiplication: kA=[k⋅aij]kA = [k \cdot a_{ij}]

Matrix Multiplication: cij=∑k=1naikbkjc_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}

Additive Identity: A+O=AA + O = A

Multiplicative Identity: AI=IA=AAI = IA = A

Distributive Property: A(B+C)=AB+ACA(B + C) = AB + AC and (A+B)C=AC+BC(A + B)C = AC + BC

Scalar Distribution: k(A+B)=kA+kBk(A + B) = kA + kB and (k+l)A=kA+lA(k + l)A = kA + lA

💡Examples

Problem 1:

Given matrices A=[2432]A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} and B=[13−25]B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}, find 3A−B3A - B.

Solution:

Step 1: Find 3A3A by multiplying every element of AA by 3. 3A=3[2432]=[3(2)3(4)3(3)3(2)]=[61296]3A = 3 \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 3(2) & 3(4) \\ 3(3) & 3(2) \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} Step 2: Subtract matrix BB from 3A3A element-wise. 3A−B=[61296]−[13−25]3A - B = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} - \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} =[6−112−39−(−2)6−5]=[59111]= \begin{bmatrix} 6 - 1 & 12 - 3 \\ 9 - (-2) & 6 - 5 \end{bmatrix} = \begin{bmatrix} 5 & 9 \\ 11 & 1 \end{bmatrix}

Explanation:

This problem demonstrates scalar multiplication followed by matrix subtraction. Note how the signs change when subtracting a negative number in the bottom-left element.

Problem 2:

If A=[1−230]A = \begin{bmatrix} 1 & -2 \\ 3 & 0 \end{bmatrix} and B=[−12]B = \begin{bmatrix} -1 \\ 2 \end{bmatrix}, calculate the product ABAB.

Solution:

Step 1: Check the orders. AA is 2×22 \times 2 and BB is 2×12 \times 1. Since columns of AA (2) = rows of BB (2), multiplication is possible. The result will be 2×12 \times 1. Step 2: Calculate each element using the row-by-column rule. For c11c_{11} (Row 1 of AA and Column 1 of BB): (1)(−1)+(−2)(2)=−1−4=−5(1)(-1) + (-2)(2) = -1 - 4 = -5 For c21c_{21} (Row 2 of AA and Column 1 of BB): (3)(−1)+(0)(2)=−3+0=−3(3)(-1) + (0)(2) = -3 + 0 = -3 Step 3: Write the final matrix. AB=[−5−3]AB = \begin{bmatrix} -5 \\ -3 \end{bmatrix}

Explanation:

This example shows the standard procedure for matrix multiplication. We sum the products of corresponding elements from the rows of the first matrix and the columns of the second.