Algebra questions

Question

Just need very simplified answers on how to do these questions if i am asked. Thank you.

Answer 1

(10).

The solution as shown below.

Answer 2

(9).

The solution as shown below.

Answer 3

(8).

The solution as shown below.

Answer 4

(7).

The solution as shown below.

Answer 5

(6).

Free Variables Theorem for Homogeneous System :

If a homogeneous linear system has " n " unknowns, and if the reduced row echelon form of its augmented matrix has " r " nonzero rows, then the system has " n - r " free variables.

Answer 6

(5).

(a). If a row-reduced matrix has a row of zeros on the left side, but the right hand side isn't zero then the system has no solution.

For example, the row-reduced matrix is

There is no solution here. You can write that as the null set Ø, the empty set {}, or no solution.

 

(b). If a row-reduced matrix has the same number of non-zero rows as variables then the system has unique or one solution.

For example, the row-reduced matrix is

When you convert the augmented matrix back into equation form, you get x = 3, y = 1, and z = 2.

 

(d). If a row-reduced matrix has more variables than non-zero rows then the system has unique or infinitely many solutions.

For example, the row-reduced matrix is

The first equation will be x + 3z = 4. Solving for x gives x = 4 - 3z.

The second equation will be y - 2z = 3. Solving for y gives y = 3 + 2z.

The z column is not cleared out (all zeros except for one number) so the other variables will be defined in terms of z. Therefore, z will be the parameter t and the solution is ...

x = 4 - 3t, y = 3 + 2t, z = t.

Answer 7

(4).

Use elementary row operations to rewrite the augmented matrix in row-echelon form.

We should operate from left to right by columns, using elementary row operations to obtain zeros in all entries directly below the leading 1 ' s.

For example, the augment matrix of the first column has no leading 1 in upper left corner.

To rewrite the augmented matrix in row-echelon form, Interchange R1 and R2 so first column has leading 1 in upper left corner. 

.

 

Answer 8

(3).

For example, the 3 × 4 matrix is .

Now, apply elementary row operations until you obtain zeros above each of the leading 1’s, as follows.

Perform operations on R2 so third column has zeros above its leading 1.

The matrix is now in reduced row-echelon form.

Answer 9

(2).

For example, the system of linear equations are .

Begin by rewriting the linear system and aligning the variables.

Next, use the coefficients and constant terms as the matrix entries. Include zeros for the coefficients of the missing variables.

The augmented matrix has four rows and five columns, so it is a 4 × 5 matrix.

Answer 10

(1).

For the examples of equations are

Equation 1 : 3x + y = 7.

The equation can be written in the form y = - 3x + 7 (y = mx + b), this function is linear.

Equation 2 : y = 2x + 8.

The equation can be written in the form y = mx + b, this function is linear.

Equation 3 : y = 3x² + 4x + 1.

The equation cannot be written in the form y = mx + b. So, this function is nonlinear.

Equation 4 : y = 3/x.

The equation cannot be written in the form y = mx + b. So, this function is nonlinear.

Equation 5 : y = 3x³ + 4x² + 4x + 1.

The equation cannot be written in the form y = mx + b. So, this function is nonlinear.

Equation 6 : y = 3x.

The equation can be written in the form y = mx + b, this function is linear.