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Algebra - Operations on Matrices: Addition, Multiplication, Scalar Multiplication

Grade 12ICSE

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

🔑Concepts

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Order and Equality: A matrix is a rectangular array of numbers arranged in mm rows and nn columns, denoted as m×nm \times n. Two matrices are equal only if they have the exact same dimensions and every corresponding element aija_{ij} is equal to bijb_{ij}. Visually, this means both grids must have the same shape and identical values in every cell.

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Matrix Addition and Subtraction: These operations are only possible for matrices of the same order. To add or subtract, you perform the operation on corresponding elements. Visually, imagine stacking two identical grids and adding the numbers that occupy the same position (aij±bija_{ij} \pm b_{ij}).

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Scalar Multiplication: Multiplying a matrix by a scalar kk involves multiplying every individual element of the matrix by that constant. Visually, this scales the entire content of the matrix grid by the factor kk. For example, k[abcd]=[kakbkckd]k \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} ka & kb \\ kc & kd \end{bmatrix}.

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Matrix Multiplication Rule: Two matrices AA and BB can be multiplied to form ABAB only if the number of columns in AA is equal to the number of rows in BB. If AA is m×nm \times n and BB is n×pn \times p, the resulting matrix CC will have the order m×pm \times p.

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The Multiplication Process: To find the element in the ii-th row and jj-th column of the product, you multiply the elements of the ii-th row of the first matrix by the corresponding elements of the jj-th column of the second matrix and sum them up. Visually, this follows a 'Row-by-Column' movement pattern.

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Commutativity and Associativity: While matrix addition is commutative (A+B=B+AA + B = B + A), matrix multiplication is generally NOT commutative (AB≠BAAB \neq BA). However, both addition and multiplication are associative, meaning (A+B)+C=A+(B+C)(A + B) + C = A + (B + C) and (AB)C=A(BC)(AB)C = A(BC).

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Identity and Zero Matrices: The Identity matrix II is a square matrix with 11s on the main diagonal (top-left to bottom-right) and 00s elsewhere; it acts as the multiplicative identity (AI=IA=AAI = IA = A). The Zero matrix OO contains only zeros and acts as the additive identity (A+O=AA + O = A).

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Distributive Property: Matrix multiplication distributes over matrix addition. This is expressed as A(B+C)=AB+ACA(B + C) = AB + AC and (A+B)C=AC+BC(A + B)C = AC + BC, provided the dimensions allow for these operations.

📐Formulae

(A±B)ij=aij±bij(A \pm B)_{ij} = a_{ij} \pm b_{ij}

(kA)ij=k⋅aij(kA)_{ij} = k \cdot a_{ij}

(AB)ij=∑k=1naikbkj(AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}

k(A+B)=kA+kBk(A + B) = kA + kB

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

AB≠BAAB \neq BA (General Non-commutativity of Multiplication)

A⋅I=I⋅A=AA \cdot I = I \cdot A = A

A+(−A)=OA + (-A) = O

💡Examples

Problem 1:

Given matrices A=[2134]A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} and B=[1−203]B = \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}, calculate 3A−2B3A - 2B.

Solution:

Step 1: Multiply matrix AA by scalar 33: 3A=3[2134]=[3(2)3(1)3(3)3(4)]=[63912]3A = 3 \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3(2) & 3(1) \\ 3(3) & 3(4) \end{bmatrix} = \begin{bmatrix} 6 & 3 \\ 9 & 12 \end{bmatrix}

Step 2: Multiply matrix BB by scalar 22: 2B=2[1−203]=[2(1)2(−2)2(0)2(3)]=[2−406]2B = 2 \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 2(1) & 2(-2) \\ 2(0) & 2(3) \end{bmatrix} = \begin{bmatrix} 2 & -4 \\ 0 & 6 \end{bmatrix}

Step 3: Subtract 2B2B from 3A3A: 3A−2B=[6−23−(−4)9−012−6]=[4796]3A - 2B = \begin{bmatrix} 6 - 2 & 3 - (-4) \\ 9 - 0 & 12 - 6 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ 9 & 6 \end{bmatrix}

Explanation:

This problem demonstrates scalar multiplication followed by matrix subtraction. Each element of the matrix is scaled first, and then corresponding elements are subtracted.

Problem 2:

Find the product ABAB if A=[1230]A = \begin{bmatrix} 1 & 2 \\ 3 & 0 \end{bmatrix} and B=[4125]B = \begin{bmatrix} 4 & 1 \\ 2 & 5 \end{bmatrix}.

Solution:

Step 1: Check dimensions. AA is 2×22 \times 2 and BB is 2×22 \times 2. Resulting matrix will be 2×22 \times 2.

Step 2: Calculate element c11c_{11} (Row 1 of AA ×\times Col 1 of BB): c11=(1×4)+(2×2)=4+4=8c_{11} = (1 \times 4) + (2 \times 2) = 4 + 4 = 8

Step 3: Calculate element c12c_{12} (Row 1 of AA ×\times Col 2 of BB): c12=(1×1)+(2×5)=1+10=11c_{12} = (1 \times 1) + (2 \times 5) = 1 + 10 = 11

Step 4: Calculate element c21c_{21} (Row 2 of AA ×\times Col 1 of BB): c21=(3×4)+(0×2)=12+0=12c_{21} = (3 \times 4) + (0 \times 2) = 12 + 0 = 12

Step 5: Calculate element c22c_{22} (Row 2 of AA ×\times Col 2 of BB): c22=(3×1)+(0×5)=3+0=3c_{22} = (3 \times 1) + (0 \times 5) = 3 + 0 = 3

Final Matrix AB=[811123]AB = \begin{bmatrix} 8 & 11 \\ 12 & 3 \end{bmatrix}

Explanation:

This shows the Row-by-Column multiplication method. Each entry in the resulting matrix is a sum of products of corresponding entries from the rows of the first matrix and columns of the second.