LINEAR ALGEBRA

Linear Algebra



Linear Algebra

Answer 1)

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '9y' to each side of the equation.

5x + -9y + 9y = 0 + 9y

Combine like terms: -9y + 9y = 0

5x + 0 = 0 + 9y

5x = 0 + 9y

Remove the zero:

5x = 9y

Divide each side by '5'.

x = 1.8y

Simplifying

x = 1.8y

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-11y' to each side of the equation.

9x + 11y + -11y = 0 + -11y

Combine like terms: 11y + -11y = 0

9x + 0 = 0 + -11y

9x = 0 + -11y

Remove the zero:

9x = -11y

Divide each side by '9'.

x = -1.222222222y

Simplifying

x = -1.222222222y

Answer 2)

7x - 3y + 2z

Solving

7x + -3y + 2z = 0

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '3y' to each side of the equation.

7x + -3y + 3y + 2z = 0 + 3y

Combine like terms: -3y + 3y = 0

7x + 0 + 2z = 0 + 3y

7x + 2z = 0 + 3y

Remove the zero:

7x + 2z = 3y

Add '-2z' to each side of the equation.

7x + 2z + -2z = 3y + -2z

Combine like terms: 2z + -2z = 0

7x + 0 = 3y + -2z

7x = 3y + -2z

Divide each side by '7'.

x = 0.4285714286y + -0.2857142857z

Simplifying

x = 0.4285714286y + -0.2857142857z

5x+4y-z=0

Solving

5x + 4y + -1z = 0

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-4y' to each side of the equation.

5x + 4y + -4y + -1z = 0 + -4y

Combine like terms: 4y + -4y = 0

5x + 0 + -1z = 0 + -4y

5x + -1z = 0 + -4y

Remove the zero:

5x + -1z = -4y

Add 'z' to each side of the equation.

5x + -1z + z = -4y + z

Combine like terms: -1z + z = 0

5x + 0 = -4y + z

5x = -4y + z

Divide each side by '5'.

x = -0.8y + 0.2z

Simplifying

x = -0.8y + 0.2z

Answer 2)

I)

Let A be an m by n matrix. The space spanned by the rows of A is called the row space of A, denoted RS(A); it is a subspace of R n . The space spanned by the columns of A is called the column space of A, denoted CS(A); it is a subspace of R mThe collection { r1, r2, …, r m } consisting of the rows of A may not form a basis for RS(A), because the collection may not be linearly independent. However, a maximal linearly independent subset of { r1, r2, …, r m } does give a basis for the row space. Since the maximum number of linearly independent rows of A is equal to the rank of A.

II)

Consider a square matrix A = [ai j ] of order n × n. Its n components aii form the main diagonal, which runs from ...