Math / Algebra

QuestionExample 5.2.1. Let $T\left(x_{1}, x_{2}, x_{3}\right)=\left(x_{1}-x_{2},-2 x_{1}+2 x_{2}+x_{3},-x_{1}+x_{2}+x_{3}\right)$ and $\vec{b}=(1,0,-1)$ . Determine whether $\vec{b} \in T\left(\mathbb{R}^{3}\right)$ .

Studdy Solution

STEP 1

What is this asking? Is the vector $(1, 0, -1)$ in the range of the transformation $T$ ?
In simpler terms, can we find some input vector that, when transformed by $T$ , produces $(1, 0, -1)$ ? Watch out! Don't mix up the domain and range!
We're looking for an input vector in $\mathbb{R}^3$ , not an output vector.

STEP 2

1. Set up the system of equations
2. Solve the system

STEP 3

We want to find out if there exist $x_1$ , $x_2$ , and $x_3$ such that $T(x_1, x_2, x_3) = (1, 0, -1)$ .

STEP 4

Remember, $T(x_1, x_2, x_3)$ is defined as $(x_1 - x_2, -2x_1 + 2x_2 + x_3, -x_1 + x_2 + x_3)$ .

STEP 5

So, we need to see if we can solve: $x_1 - x_2 = 1$ $-2x_1 + 2x_2 + x_3 = 0$ $-x_1 + x_2 + x_3 = -1$

STEP 6

From the first equation, we get $x_1 = 1 + x_2$ .
We did this so we can substitute it into the other equations and get rid of $x_1$ !

STEP 7

Substituting into the second equation, we have $-2(1 + x_2) + 2x_2 + x_3 = 0$ .

STEP 8

This simplifies to $-2 - 2x_2 + 2x_2 + x_3 = 0$ , which means $x_3 = 2$ .
Boom! We found $x_3$ !

STEP 9

Now, substitute $x_1 = 1 + x_2$ and $x_3 = 2$ into the third equation: $-(1 + x_2) + x_2 + 2 = -1$ .

STEP 10

This simplifies to $-1 - x_2 + x_2 + 2 = -1$ , which gives us $1 = -1$ .
Whoa, that's not right!

STEP 11

This contradiction tells us there's no solution to the system of equations.
There are no values of $x_1$ , $x_2$ , and $x_3$ that satisfy all three equations simultaneously.

STEP 12

Since there's no solution to the system, the vector $(1, 0, -1)$ is not in the range of the transformation $T$ .
So, $\vec{b} \notin T(\mathbb{R}^3)$ .

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QuestionExample 5.2.1. Let $T\left(x_{1}, x_{2}, x_{3}\right)=\left(x_{1}-x_{2},-2 x_{1}+2 x_{2}+x_{3},-x_{1}+x_{2}+x_{3}\right)$ and $\vec{b}=(1,0,-1)$ . Determine whether $\vec{b} \in T\left(\mathbb{R}^{3}\right)$ .

Studdy Solution

STEP 1

STEP 2

1. Set up the system of equations
2. Solve the system

STEP 3

We want to find out if there exist $x_1$ , $x_2$ , and $x_3$ such that $T(x_1, x_2, x_3) = (1, 0, -1)$ .

STEP 4

Remember, $T(x_1, x_2, x_3)$ is defined as $(x_1 - x_2, -2x_1 + 2x_2 + x_3, -x_1 + x_2 + x_3)$ .

STEP 5

So, we need to see if we can solve: $x_1 - x_2 = 1$ $-2x_1 + 2x_2 + x_3 = 0$ $-x_1 + x_2 + x_3 = -1$

STEP 6

From the first equation, we get $x_1 = 1 + x_2$ .
We did this so we can substitute it into the other equations and get rid of $x_1$ !

STEP 7

Substituting into the second equation, we have $-2(1 + x_2) + 2x_2 + x_3 = 0$ .

STEP 8

This simplifies to $-2 - 2x_2 + 2x_2 + x_3 = 0$ , which means $x_3 = 2$ .
Boom! We found $x_3$ !

STEP 9

Now, substitute $x_1 = 1 + x_2$ and $x_3 = 2$ into the third equation: $-(1 + x_2) + x_2 + 2 = -1$ .

STEP 10

This simplifies to $-1 - x_2 + x_2 + 2 = -1$ , which gives us $1 = -1$ .
Whoa, that's not right!

STEP 11

This contradiction tells us there's no solution to the system of equations.
There are no values of $x_1$ , $x_2$ , and $x_3$ that satisfy all three equations simultaneously.

STEP 12

Since there's no solution to the system, the vector $(1, 0, -1)$ is not in the range of the transformation $T$ .
So, $\vec{b} \notin T(\mathbb{R}^3)$ .

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Studdy Solution

STEP 1

STEP 2

1. Set up the system of equations2. Solve the system

STEP 3

We want to find out if there exist x1x_1x1​, x2x_2x2​, and x3x_3x3​ such that T(x1,x2,x3)=(1,0,−1)T(x_1, x_2, x_3) = (1, 0, -1)T(x1​,x2​,x3​)=(1,0,−1).

STEP 4

Remember, T(x1,x2,x3)T(x_1, x_2, x_3)T(x1​,x2​,x3​) is defined as (x1−x2,−2x1+2x2+x3,−x1+x2+x3)(x_1 - x_2, -2x_1 + 2x_2 + x_3, -x_1 + x_2 + x_3)(x1​−x2​,−2x1​+2x2​+x3​,−x1​+x2​+x3​).

STEP 5

So, we need to see if we can solve: x1−x2=1x_1 - x_2 = 1x1​−x2​=1 −2x1+2x2+x3=0-2x_1 + 2x_2 + x_3 = 0−2x1​+2x2​+x3​=0−x1+x2+x3=−1-x_1 + x_2 + x_3 = -1−x1​+x2​+x3​=−1

STEP 6

From the first equation, we get x1=1+x2x_1 = 1 + x_2x1​=1+x2​.We did this so we can substitute it into the other equations and get rid of x1x_1x1​!

STEP 7

Substituting into the second equation, we have −2(1+x2)+2x2+x3=0-2(1 + x_2) + 2x_2 + x_3 = 0−2(1+x2​)+2x2​+x3​=0.

STEP 8

This simplifies to −2−2x2+2x2+x3=0-2 - 2x_2 + 2x_2 + x_3 = 0−2−2x2​+2x2​+x3​=0, which means x3=2x_3 = 2x3​=2.Boom! We found x3x_3x3​!

STEP 9

Now, substitute x1=1+x2x_1 = 1 + x_2x1​=1+x2​ and x3=2x_3 = 2x3​=2 into the third equation: −(1+x2)+x2+2=−1-(1 + x_2) + x_2 + 2 = -1−(1+x2​)+x2​+2=−1.

STEP 10

This simplifies to −1−x2+x2+2=−1-1 - x_2 + x_2 + 2 = -1−1−x2​+x2​+2=−1, which gives us 1=−11 = -11=−1.Whoa, that's not right!

STEP 11

This contradiction tells us there's no solution to the system of equations.There are no values of x1x_1x1​, x2x_2x2​, and x3x_3x3​ that satisfy all three equations simultaneously.

STEP 12

Since there's no solution to the system, the vector (1,0,−1)(1, 0, -1)(1,0,−1) is *not* in the range of the transformation TTT.So, b⃗∉T(R3)\vec{b} \notin T(\mathbb{R}^3)b∈/T(R3).

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1. Set up the system of equations
2. Solve the system

We want to find out if there exist $x_1$ , $x_2$ , and $x_3$ such that $T(x_1, x_2, x_3) = (1, 0, -1)$ .

Remember, $T(x_1, x_2, x_3)$ is defined as $(x_1 - x_2, -2x_1 + 2x_2 + x_3, -x_1 + x_2 + x_3)$ .

So, we need to see if we can solve: $x_1 - x_2 = 1$ $-2x_1 + 2x_2 + x_3 = 0$ $-x_1 + x_2 + x_3 = -1$

From the first equation, we get $x_1 = 1 + x_2$ .
We did this so we can substitute it into the other equations and get rid of $x_1$ !

Substituting into the second equation, we have $-2(1 + x_2) + 2x_2 + x_3 = 0$ .

This simplifies to $-2 - 2x_2 + 2x_2 + x_3 = 0$ , which means $x_3 = 2$ .
Boom! We found $x_3$ !

Now, substitute $x_1 = 1 + x_2$ and $x_3 = 2$ into the third equation: $-(1 + x_2) + x_2 + 2 = -1$ .

This simplifies to $-1 - x_2 + x_2 + 2 = -1$ , which gives us $1 = -1$ .
Whoa, that's not right!

This contradiction tells us there's no solution to the system of equations.
There are no values of $x_1$ , $x_2$ , and $x_3$ that satisfy all three equations simultaneously.

Since there's no solution to the system, the vector $(1, 0, -1)$ is not in the range of the transformation $T$ .
So, $\vec{b} \notin T(\mathbb{R}^3)$ .