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2 years ago

A) (i) Write down the value of x.

Mathematics

1 answer:

Mazyrski [523]2 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

x = 65° ( vertically opposite angles )

y = 71° ( alternate angles )

The sum of the 3 angles in a triangle = 180° , then

x + y + z = 180° , that is

65° + 71° + z = 180°

136° + z = 180° ( subtract 136° from both sides )

z = 44°

So x = 65°, y = 71° , z = 44°

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Let T: set of real numbers R Superscript nℝnright arrow→set of real numbers R Superscript mℝm be a linear transformation, and

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Answer:

$\{T(v_1), T(v_2), T(v_3)\}$ 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 $k_1v_1+k_2v_2+k_3v_3+...+k_nv_n=0$

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

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

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

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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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