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Matrices - Properties of multiplication of matrices

Grade 12CBSE

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

🔑Concepts

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The product of two matrices AA and BB is defined if the number of columns in AA is equal to the number of rows in BB.

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Non-commutativity: In general, matrix multiplication is not commutative, meaning AB≠BAAB \neq BA, even if both ABAB and BABA are defined and have the same order.

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Associative Law: For any three matrices A,BA, B and CC, the property (AB)C=A(BC)(AB)C = A(BC) holds whenever both sides of the equality are defined.

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Distributive Law: For three matrices A,BA, B and CC, matrix multiplication is distributive over addition: A(B+C)=AB+ACA(B + C) = AB + AC and (A+B)C=AC+BC(A + B)C = AC + BC.

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Existence of Multiplicative Identity: For every square matrix AA of order nn, there exists an identity matrix II of the same order such that IA=AI=AIA = AI = A.

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Zero Product: The product of two non-zero matrices can be a zero matrix. This differs from real numbers where ab=0ab = 0 implies a=0a=0 or b=0b=0.

📐Formulae

(AB)C=A(BC)(AB)C = A(BC)

A(B+C)=AB+ACA(B + C) = AB + AC

(A+B)C=AC+BC(A + B)C = AC + BC

AI=IA=AAI = IA = A

AB≠BA (Generally)AB \neq BA \text{ (Generally)}

💡Examples

Problem 1:

Given A=[100−1]A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} and B=[0110]B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, verify that AB≠BAAB \neq BA.

Solution:

First, calculate ABAB: AB=AB =\begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}==\begin{bmatrix} (1)(0) + (0)(1) & (1)(1) + (0)(0) \ (0)(0) + (-1)(1) & (0)(1) + (-1)(0) \end{bmatrix}==\begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} Next, calculate BABA: BA=BA =\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix}==\begin{bmatrix} (0)(1) + (1)(0) & (0)(0) + (1)(-1) \ (1)(1) + (0)(0) & (1)(0) + (0)(-1) \end{bmatrix}==\begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} Since [01−10]≠[0−110]\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \neq \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, we have AB≠BAAB \neq BA.

Explanation:

This example demonstrates the non-commutative property of matrix multiplication. Even though both matrices are square and the products are defined, the resulting matrices are different.

Problem 2:

Find the product ABAB if A=[0−102]A = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix} and B=[3500]B = \begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix}.

Solution:

Multiply the matrices row by column: AB=AB =\begin{bmatrix} 0 & -1 \ 0 & 2 \end{bmatrix} \begin{bmatrix} 3 & 5 \ 0 & 0 \end{bmatrix}==\begin{bmatrix} (0)(3) + (-1)(0) & (0)(5) + (-1)(0) \ (0)(3) + (2)(0) & (0)(5) + (2)(0) \end{bmatrix}==\begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}

Explanation:

This shows that the product of two non-zero matrices AA and BB can result in a zero matrix OO. Unlike in scalar algebra, AB=OAB = O does not necessarily mean A=OA = O or B=OB = O.

Properties of multiplication of matrices Class 12 Notes & Examples