If I = 1 0 0 1 then find - I .

Given, I = 1 0 0 1 The negative matrix of I = - I Now by changing the signs of each element of matrix I We get, I = - 1 0 0 - 1 Therefore, the negative matrix of matrix I = - I = - 1 0 0 - 1

If A = cos α sin α - sin α cos α , then verify that A ' A = I

Given: A = cos α sin α - sin α cos α To get transpose interchange rows and columns. ⇒ A ' = cos α - sin α sin α cos α Therefore A ' A = cos α - sin α sin α cos α cos α sin α - sin α cos α = cos 2 α + sin 2 α cos α sin α - cos α sin α cos α sin α - cos α sin α cos 2 α + sin 2 α = 1 0 0 1 = I ∴ A ' A = I Hence, proved

Identify the elements of matrix A = 5 6 9 2 3 4 7 1 8

Matrix is a rectangular array of numbers, expressions or symbols arranged in rows and columns. The numbers, symbols or expressions in the matrix are called elements of matrix. For a matrix X = a b c d , each of a , b , c , d is an element of the matrix. Given, A = 5 6 9 2 3 4 7 1 8 So, the elements of A are 5 , 6 , 9 , 2 , 3 , 4 , 7 , 1 , 8

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Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Algebra>Exercise

Embibe Experts Solutions for Chapter: Algebra, Exercise 1: Exercise

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Algebra, Exercise 1: Exercise

Attempt the free practice questions on Chapter 3: Algebra, Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Crash Course (Based on Revised Syllabus-2023) solutions are prepared by Experienced Embibe Experts.