Linear Algebra Identities

In general, identiites is defined as the condition, that satisfies for all the values of the involved  variables. The linear algebra identities deals with one variable and two variable solutions. The identities are involved in certain condition, by using this condition we can find the values in the given equation. The identities are involved in two variable or variable with numbers. Linear algebra identities is useful to work the problem quick and easy. But the identities must be read or memorize.

Understanding Simplifying Trigonometric Identities is always challenging for me but thanks to all math help websites to help me out. 

The linear algebra identities as shown in below:

Problems using the linear algebra identities

Example 1:

Simplify, (3x + 7y)² 

Solution:

Using the linear algebra identities simplify the (3x + 7y)²

(3x + 7y)² using the identities

(a+b)² = a²+2ab+b².

Here 3x is substitute to a and  7y is substitute in b

(3x + 7y)²= (3x)² + 2(3x)(7y) + (7y)²

= 9x²+ 42xy + 49y²

(3x + 7y)²  = 9x²+ 42xy + 49y²

Example 2:

simplify, (11a – 7b)²

Solution:

using the linear algebra identities simplify the (11a – 7b)²

(11a – 7b)² using the identities

(a-b)² = a²-2ab+b².

Here 11a is substitute to a and 7b is substitute in b

= (11a)² – 2(11a) (7b) + (7b)²

= 121a² – 154ab + 49b².

(11a – 7b)² = 121a² – 154ab + 49b².

Example 3:

Simplify, (2p + 5q)(2p – 5q)

Solution:

using the linear algebra identities simplify the (2p + 5q)(2p – 5q)

(2p + 5q)(2p – 5q) using the identities

(a+b)(a-b) = (a²-b²).

Here substitute a for 2p and b for 5q.

= (2p)² – (5q)²

= 4p² – 25q²

(2p + 5q)(2p – 5q) =  4p² – 25q²

Problem solving linear algebra identities with three variable

Simplify the function using the three variable  linear algebra identities

a² + b² + c² ≡ (a + b + c)² – 2(ab + bc + ca)

(i) (2x + y + 2z)²

(ii) (x – 2y + z)²

(iii) (2p – 3q – r)²

(iv) (2a + 3b − 2c)²

Solution:
simplify problem in the given three variable linear algebra identities

(i) (2x + y + 2z)² = [(2x) + y + (2z)]² = (2x)² + y² + (2z)² + 2(2x)y + 2y(2z) + 2(2z)(2x)

= 4x² + y² + 4z² + 4xy + 4yz + 8zx.

(ii) (x – 2y + z)² = [x + (–2y) + z]²= x² + (–2y)² + z² + 2x(–2y) +2 (–2y)z + 2zx

= x² + 4y² + z² – 4xy – 4yz + 2zx.

(iii) (2p – 3q – r)² = [(2p) + (–3q) + (–r)]²

= (2p)² + (–3q)² + (–r)² + 2(2p) (–3q) + 2(–3q) (–r) + 2(–r)(2p).

= 4p² + 9q² + r² – 12pq + 6qr – 4rp.

(iv) (2a + 3b – 2c)² = [(2a) + (3b) + (–2c)]²

= (2a)² + (3b)² + (–2c)² + 2(2a)(3b) + 2(3b)(–2c) + 2(–2c)(2a)

= 4a² + 9b² + 4c² + 12ab – 12bc – 8ca.