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3 years ago

Solve the system by using a matrix equation 3x-5y=3 3x-4y=3

Mathematics

2 answers:

natka813 [3]3 years ago

6 0

Answer: B. (1,0)

Step-by-step explanation:

1. We will use the matrix method to solve this problem.

3 -5 3

3 -4 3

2. Apply to Row2 : Row2 - Row1.

3 -5 3

0 1 0

3. Simplify rows.

3 -5 3

0 1 0

Note: The matrix is now in row echelon form.

The steps below are for back substitution.

4. Apply to Row1 : Row1 + 5 Row2.

3 0 3

0 1 0

5. Simplify rows.

1 0 1

0 1 0

Note: The matrix is now in reduced row echelon form.

6. Therefore,

x=1

y=0

or (1,0).

babunello [35]3 years ago

3 0

Answer:

The correct answer is B. (1,0).

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Klio2033 [76]

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$\{T(v_1), T(v_2), T(v_3)\}$ is linearly dependent set.

Step-by-step explanation:

Given: $\{v_1,v_2,v_3\}$ is a linearly dependent set in set of real numbers R

To show: the set $\{T(v_1), T(v_2), T(v_3)\}$ is linearly dependent.

Solution:

If $\{v_1,v_2,v_3,...,v_n\}$ is a set of linearly dependent vectors then there exists atleast one $k_i:i=1,2,3,...,n$ such that $k_1v_1+k_2v_2+k_3v_3+...+k_nv_n=0$

Consider $k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0$

A linear transformation T: U→V satisfies the following properties:

1. $T(u_1+u_2)=T(u_1)+T(u_2)$

2. $T(au)=aT(u)$

Here, $u,u_1,u_2$ ∈ U

As T is a linear transformation,

$k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0\\T(k_1v_1)+T(k_2v_2)+T(k_3v_3)=0\\T(k_1v_1+k_2v_2+k_3v_3)=0\\$

As $\{v_1,v_2,v_3\}$ is a linearly dependent set,

$k_1v_1+k_2v_2+k_3v_3=0$ for some $k_i\neq 0:i=1,2,3$

So, for some $k_i\neq 0:i=1,2,3$

$k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0$

Therefore, set $\{T(v_1), T(v_2), T(v_3)\}$ is linearly dependent.

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