Solved on Feb 02, 2024

Find the transition matrix from basis $F$ to basis $E$ where $E=[(1,2)^{T},(3,7)^{T}]$ and $F=[(1,-2)^{T},(3,-5)^{T}]$ in $\mathbb{R}^{2}$ .

STEP 1

Assumptions
1. $E$ and $F$ are two ordered bases for $\mathbb{R}^{2}$ .
2. $E=\left[(1,2)^{T},(3,7)^{T}\right]$ and $F=\left[(1,-2)^{T},(3,-5)^{T}\right]$ .
3. We need to find the transition matrix from basis $F$ to basis $E$ .
4. A transition matrix converts coordinates from one basis to another.

STEP 2

Let's denote the transition matrix from basis $F$ to basis $E$ as $P$ . The columns of $P$ are the coordinates of the vectors of basis $E$ expressed in terms of basis $F$ .

STEP 3

To find the transition matrix $P$ , we need to solve the following system of equations for each vector in basis $E$ :
$P \begin{bmatrix} 1 \\ -2 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \quad \text{and} \quad P \begin{bmatrix} 3 \\ -5 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix}$

STEP 4

Let $P = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , where $a, b, c, d$ are the unknowns we need to solve for. We will use the given vectors of $F$ and $E$ to create a system of linear equations.

STEP 5

The first vector of $E$ is $(1,2)^T$ , and the first vector of $F$ is $(1,-2)^T$ . We can write the first equation as:
$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 \\ -2 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

STEP 6

Multiplying the matrices on the left side gives us two equations:
$a - 2b = 1 \\ c - 2d = 2$

STEP 7

The second vector of $E$ is $(3,7)^T$ , and the second vector of $F$ is $(3,-5)^T$ . We can write the second equation as:
$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 3 \\ -5 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix}$

STEP 8

Multiplying the matrices on the left side gives us two more equations:
$3a - 5b = 3 \\ 3c - 5d = 7$

STEP 9

Now we have a system of four equations with four unknowns:
$\begin{cases} a - 2b = 1 \\ c - 2d = 2 \\ 3a - 5b = 3 \\ 3c - 5d = 7 \end{cases}$

STEP 10

We can solve this system using various methods such as substitution, elimination, or matrix operations. Let's use substitution to solve for $a$ and $c$ first.

STEP 11

From the first equation, we can express $a$ in terms of $b$ :
$a = 1 + 2b$

STEP 12

Substitute the expression for $a$ into the third equation:
$3(1 + 2b) - 5b = 3$

STEP 13

Simplify and solve for $b$ :
$3 + 6b - 5b = 3 \\ b = 0$

STEP 14

Now that we have $b$ , we can find $a$ :
$a = 1 + 2(0) \\ a = 1$

STEP 15

Next, we solve for $c$ and $d$ . From the second equation, we can express $c$ in terms of $d$ :
$c = 2 + 2d$

STEP 16

Substitute the expression for $c$ into the fourth equation:
$3(2 + 2d) - 5d = 7$

STEP 17

Simplify and solve for $d$ :
$6 + 6d - 5d = 7 \\ d = 1$

STEP 18

Now that we have $d$ , we can find $c$ :
$c = 2 + 2(1) \\ c = 4$

STEP 19

Now we have all the elements of the transition matrix $P$ :
$P = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}$

STEP 20

However, we must verify that this matrix actually transforms the vectors of $F$ into the vectors of $E$ . We do this by multiplying $P$ by the vectors of $F$ and checking if we get the vectors of $E$ .

STEP 21

Verify the first vector:
$\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 - 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

STEP 22

Verify the second vector:
$\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ -5 \end{bmatrix} = \begin{bmatrix} 3 \\ 12 - 5 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix}$

STEP 23

Since both verifications are correct, we conclude that the transition matrix from $F$ to $E$ is:
$P = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}$
None of the given answer choices match this result, so it seems there might be an error in the provided options. The correct transition matrix is the one we've calculated.

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Find the transition matrix from basis FFF to basis EEE where E=[(1,2)T,(3,7)T]E=[(1,2)^{T},(3,7)^{T}]E=[(1,2)T,(3,7)T] and F=[(1,−2)T,(3,−5)T]F=[(1,-2)^{T},(3,-5)^{T}]F=[(1,−2)T,(3,−5)T] in R2\mathbb{R}^{2}R2.

STEP 1

STEP 2

Let's denote the transition matrix from basis FFF to basis EEE as PPP. The columns of PPP are the coordinates of the vectors of basis EEE expressed in terms of basis FFF.

STEP 3

STEP 4

Let P=[abcd]P = \begin{bmatrix} a & b \\ c & d \end{bmatrix}P=[ac​bd​], where a,b,c,da, b, c, da,b,c,d are the unknowns we need to solve for. We will use the given vectors of FFF and EEE to create a system of linear equations.

STEP 5

STEP 6

Multiplying the matrices on the left side gives us two equations:a−2b=1c−2d=2 a - 2b = 1 \\ c - 2d = 2 a−2b=1c−2d=2

STEP 7

STEP 8

Multiplying the matrices on the left side gives us two more equations:3a−5b=33c−5d=7 3a - 5b = 3 \\ 3c - 5d = 7 3a−5b=33c−5d=7

STEP 9

Now we have a system of four equations with four unknowns:{a−2b=1c−2d=23a−5b=33c−5d=7 \begin{cases} a - 2b = 1 \\ c - 2d = 2 \\ 3a - 5b = 3 \\ 3c - 5d = 7 \end{cases} ⎩⎨⎧​a−2b=1c−2d=23a−5b=33c−5d=7​

STEP 10

We can solve this system using various methods such as substitution, elimination, or matrix operations. Let's use substitution to solve for aaa and ccc first.

STEP 11

From the first equation, we can express aaa in terms of bbb:a=1+2b a = 1 + 2b a=1+2b

STEP 12

Substitute the expression for aaa into the third equation:3(1+2b)−5b=3 3(1 + 2b) - 5b = 3 3(1+2b)−5b=3

STEP 13

Simplify and solve for bbb:3+6b−5b=3b=0 3 + 6b - 5b = 3 \\ b = 0 3+6b−5b=3b=0

STEP 14

Now that we have bbb, we can find aaa:a=1+2(0)a=1 a = 1 + 2(0) \\ a = 1 a=1+2(0)a=1

STEP 15

Next, we solve for ccc and ddd. From the second equation, we can express ccc in terms of ddd:c=2+2d c = 2 + 2d c=2+2d

STEP 16

Substitute the expression for ccc into the fourth equation:3(2+2d)−5d=7 3(2 + 2d) - 5d = 7 3(2+2d)−5d=7

STEP 17

Simplify and solve for ddd:6+6d−5d=7d=1 6 + 6d - 5d = 7 \\ d = 1 6+6d−5d=7d=1

STEP 18

Now that we have ddd, we can find ccc:c=2+2(1)c=4 c = 2 + 2(1) \\ c = 4 c=2+2(1)c=4

STEP 19

Now we have all the elements of the transition matrix PPP:P=[abcd]=[1041] P = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} P=[ac​bd​]=[14​01​]

STEP 20

However, we must verify that this matrix actually transforms the vectors of FFF into the vectors of EEE. We do this by multiplying PPP by the vectors of FFF and checking if we get the vectors of EEE.

STEP 21

Verify the first vector:[1041][1−2]=[14−2]=[12] \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 - 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} [14​01​][1−2​]=[14−2​]=[12​]

STEP 22

Verify the second vector:[1041][3−5]=[312−5]=[37] \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ -5 \end{bmatrix} = \begin{bmatrix} 3 \\ 12 - 5 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix} [14​01​][3−5​]=[312−5​]=[37​]

STEP 23

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Find the transition matrix from basis $F$ to basis $E$ where $E=[(1,2)^{T},(3,7)^{T}]$ and $F=[(1,-2)^{T},(3,-5)^{T}]$ in $\mathbb{R}^{2}$ .

Let's denote the transition matrix from basis $F$ to basis $E$ as $P$ . The columns of $P$ are the coordinates of the vectors of basis $E$ expressed in terms of basis $F$ .

Let $P = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , where $a, b, c, d$ are the unknowns we need to solve for. We will use the given vectors of $F$ and $E$ to create a system of linear equations.

Multiplying the matrices on the left side gives us two equations:
$a - 2b = 1 \\ c - 2d = 2$

Multiplying the matrices on the left side gives us two more equations:
$3a - 5b = 3 \\ 3c - 5d = 7$

Now we have a system of four equations with four unknowns:
$\begin{cases} a - 2b = 1 \\ c - 2d = 2 \\ 3a - 5b = 3 \\ 3c - 5d = 7 \end{cases}$

We can solve this system using various methods such as substitution, elimination, or matrix operations. Let's use substitution to solve for $a$ and $c$ first.

From the first equation, we can express $a$ in terms of $b$ :
$a = 1 + 2b$

Substitute the expression for $a$ into the third equation:
$3(1 + 2b) - 5b = 3$

Simplify and solve for $b$ :
$3 + 6b - 5b = 3 \\ b = 0$

Now that we have $b$ , we can find $a$ :
$a = 1 + 2(0) \\ a = 1$

Next, we solve for $c$ and $d$ . From the second equation, we can express $c$ in terms of $d$ :
$c = 2 + 2d$

Substitute the expression for $c$ into the fourth equation:
$3(2 + 2d) - 5d = 7$

Simplify and solve for $d$ :
$6 + 6d - 5d = 7 \\ d = 1$

Now that we have $d$ , we can find $c$ :
$c = 2 + 2(1) \\ c = 4$

Now we have all the elements of the transition matrix $P$ :
$P = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}$

However, we must verify that this matrix actually transforms the vectors of $F$ into the vectors of $E$ . We do this by multiplying $P$ by the vectors of $F$ and checking if we get the vectors of $E$ .

Verify the first vector:
$\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 - 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

Verify the second vector:
$\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ -5 \end{bmatrix} = \begin{bmatrix} 3 \\ 12 - 5 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix}$