Complex Numbers
Introduction - Consider a simple quadratic equation x 2 + 1 = 0. There is no real number which satisfies this equation. So there was a need to find a system which could answer to this problem. Euler used the symbol 'i' to denote to solve the above equation. Complex ..
Introduction - Consider a simple quadratic equation x 2 + 1 = 0. There is no real number which satisfies this equation. So there was a need to find a system which could answer to this problem. Euler used the symbol 'i' to denote to solve the above equation. Complex ..Introduction
Consider a simple quadratic equation x 2 + 1 = 0. There is no real number which satisfies this equation. So there was a need to find a system which could answer to this problem. Euler used the symbol 'i' to denote to solve the above equation. Complex number h..
Consider a simple quadratic equation x 2 + 1 = 0. There is no real number which satisfies this equation. So there was a need to find a system which could answer to this problem. Euler used the symbol 'i' to denote to solve the above equation. Complex number h..Summary
The following are the steps to solve a system of linear equations using Cramer's rule. Step 1: Find the value of the determinant Step 2: If D 0, then the system has unique solution, given by Where D 1 , D 2 and D 3 are the determinants obtained from D by replacing respectively t..
The following are the steps to solve a system of linear equations using Cramer's rule. Step 1: Find the value of the determinant Step 2: If D 0, then the system has unique solution, given by Where D 1 , D 2 and D 3 are the determinants obtained from D by replacing respectively t..Proof:
r = (0, 1, 2, 3, . . . n) Hence the total number of subsets i..
r = (0, 1, 2, 3, . . . n) Hence the total number of subsets i..Proof:
We shall use PMI to prove tha..
We shall use PMI to prove tha..Proof:
Replacing q by ix, ..
Replacing q by ix, ..Proof:
Similarly, ..
Similarly, ..Proof:
From the definition of inverse of a matrix, we have (AB)(AB) - 1 = I or A - 1 (AB)(AB) - 1 = A - 1 I (Pre-multiplying both sides by A - 1 ) or (A - 1 A) B (AB) - 1 = A - 1 (Since A - 1 I = A - 1 ) or I B (AB) - 1 = A - 1 or B (AB) - 1 = A - 1 or (B - 1 B)(AB) - 1 =B - ..
Proof:
Let p(x) be a polynomial divided by (x-a). Let q(x) be the quotient and R be the remainder. By division algorithm, Dividend = (Divisor x quotient) + Remainder p(x) = q(x) . (x-a) + R Substitute x = a, p(a) = q(a) (a-a) + R p(a) = R (a - a = 0, 0 - q (a) = 0) Hence Remainder = p(..
Result
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