Solved on Feb 27, 2024

Solve the system $A \mathbf{x}=\mathbf{b}$ using the given factorization $A=P^{\top} L U$ , where $\mathbf{b}=\begin{bmatrix}1 \\ 1 \\ 7\end{bmatrix}$ and $\mathbf{x}=\begin{bmatrix}2 \\ 3 \\ 2\end{bmatrix}$ .

STEP 1

Assumptions
1. The matrix $A$ is given and has a factorization $A=P^{\top} L U$ .
2. The permutation matrix $P$ , lower triangular matrix $L$ , and upper triangular matrix $U$ are provided.
3. The vector $\mathbf{b}$ is given.
4. The system of equations is $A \mathbf{x}=\mathbf{b}$ .
5. The identity matrix is denoted by $I$ .
6. The property $P P^{\top} = I$ is given and will be used in the solution.

STEP 2

First, we need to apply the permutation matrix $P$ to the vector $\mathbf{b}$ to get $P \mathbf{b}$ .
$P \mathbf{b} = \left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{l} 1 \\ 1 \\ 7 \end{array}\right]$

STEP 3

Perform the matrix multiplication to find $P \mathbf{b}$ .
$P \mathbf{b} = \left[\begin{array}{l} 1 \\ 1 \\ 7 \end{array}\right]$

STEP 4

Now, we need to solve the system $L U \mathbf{x} = P \mathbf{b}$ , which is equivalent to the original system $A \mathbf{x} = \mathbf{b}$ .

STEP 5

First, let's solve the system $L \mathbf{y} = P \mathbf{b}$ for $\mathbf{y}$ , where $\mathbf{y} = U \mathbf{x}$ .

STEP 6

Write down the system $L \mathbf{y} = P \mathbf{b}$ using the given matrices $L$ and $P \mathbf{b}$ .
$\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \frac{1}{2} & -\frac{1}{2} & 1 \end{array}\right] \left[\begin{array}{l} y_1 \\ y_2 \\ y_3 \end{array}\right] = \left[\begin{array}{l} 1 \\ 1 \\ 7 \end{array}\right]$

STEP 7

Since $L$ is a lower triangular matrix, we can solve for $\mathbf{y}$ using forward substitution.

STEP 8

From the first row, we directly get $y_1 = 1$ .

STEP 9

From the second row, we directly get $y_2 = 1$ .

STEP 10

From the third row, we have $\frac{1}{2} y_1 - \frac{1}{2} y_2 + y_3 = 7$ .

STEP 11

Substitute $y_1$ and $y_2$ into the third equation.
$\frac{1}{2} \cdot 1 - \frac{1}{2} \cdot 1 + y_3 = 7$

STEP 12

Simplify the equation to solve for $y_3$ .
$y_3 = 7 + \frac{1}{2} - \frac{1}{2}$

STEP 13

Calculate the value of $y_3$ .
$y_3 = 7$

STEP 14

Now we have the vector $\mathbf{y} = \left[\begin{array}{l} 1 \\ 1 \\ 7 \end{array}\right]$ .

STEP 15

Next, we need to solve the system $U \mathbf{x} = \mathbf{y}$ for $\mathbf{x}$ .

STEP 16

Write down the system $U \mathbf{x} = \mathbf{y}$ using the given matrix $U$ and vector $\mathbf{y}$ .
$\left[\begin{array}{rrr} 2 & 3 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & -\frac{7}{2} \end{array}\right] \left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{l} 1 \\ 1 \\ 7 \end{array}\right]$

STEP 17

Since $U$ is an upper triangular matrix, we can solve for $\mathbf{x}$ using back substitution.

STEP 18

From the third row, we get $-\frac{7}{2} x_3 = 7$ .

STEP 19

Solve for $x_3$ .
$x_3 = -2$

STEP 20

From the second row, we have $x_2 - x_3 = 1$ .

STEP 21

Substitute $x_3$ into the second equation.
$x_2 - (-2) = 1$

STEP 22

Solve for $x_2$ .
$x_2 = 1 - 2 = -1$

STEP 23

From the first row, we have $2 x_1 + 3 x_2 + 2 x_3 = 1$ .

STEP 24

Substitute $x_2$ and $x_3$ into the first equation.
$2 x_1 + 3 \cdot (-1) + 2 \cdot (-2) = 1$

STEP 25

Solve for $x_1$ .
$2 x_1 - 3 - 4 = 1$

STEP 26

Simplify the equation to find $x_1$ .
$2 x_1 = 1 + 3 + 4$

STEP 27

Calculate the value of $x_1$ .
$x_1 = \frac{8}{2} = 4$

STEP 28

Now we have the solution vector $\mathbf{x} = \left[\begin{array}{l} 4 \\ -1 \\ -2 \end{array}\right]$ .
The solution to the system $A \mathbf{x} = \mathbf{b}$ is $\mathbf{x} = \left[\begin{array}{l} 4 \\ -1 \\ -2 \end{array}\right]$ .

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Solve the system Ax=bA \mathbf{x}=\mathbf{b}Ax=b using the given factorization A=P⊤LUA=P^{\top} L UA=P⊤LU, where b=[117]\mathbf{b}=\begin{bmatrix}1 \\ 1 \\ 7\end{bmatrix}b=⎣⎡​117​⎦⎤​ and x=[232]\mathbf{x}=\begin{bmatrix}2 \\ 3 \\ 2\end{bmatrix}x=⎣⎡​232​⎦⎤​.

STEP 1

STEP 2

STEP 3

Perform the matrix multiplication to find PbP \mathbf{b}Pb.Pb=[117]P \mathbf{b} = \left[\begin{array}{l} 1 \\ 1 \\ 7 \end{array}\right]Pb=⎣⎡​117​⎦⎤​

STEP 4

Now, we need to solve the system LUx=PbL U \mathbf{x} = P \mathbf{b}LUx=Pb, which is equivalent to the original system Ax=bA \mathbf{x} = \mathbf{b}Ax=b.

STEP 5

First, let's solve the system Ly=PbL \mathbf{y} = P \mathbf{b}Ly=Pb for y\mathbf{y}y, where y=Ux\mathbf{y} = U \mathbf{x}y=Ux.

STEP 6

STEP 7

Since LLL is a lower triangular matrix, we can solve for y\mathbf{y}y using forward substitution.

STEP 8

From the first row, we directly get y1=1y_1 = 1y1​=1.

STEP 9

From the second row, we directly get y2=1y_2 = 1y2​=1.

STEP 10

From the third row, we have 12y1−12y2+y3=7\frac{1}{2} y_1 - \frac{1}{2} y_2 + y_3 = 721​y1​−21​y2​+y3​=7.

STEP 11

Substitute y1y_1y1​ and y2y_2y2​ into the third equation.12⋅1−12⋅1+y3=7\frac{1}{2} \cdot 1 - \frac{1}{2} \cdot 1 + y_3 = 721​⋅1−21​⋅1+y3​=7

STEP 12

Simplify the equation to solve for y3y_3y3​.y3=7+12−12y_3 = 7 + \frac{1}{2} - \frac{1}{2}y3​=7+21​−21​

STEP 13

Calculate the value of y3y_3y3​.y3=7y_3 = 7y3​=7

STEP 14

Now we have the vector y=[117]\mathbf{y} = \left[\begin{array}{l} 1 \\ 1 \\ 7 \end{array}\right]y=⎣⎡​117​⎦⎤​.

STEP 15

Next, we need to solve the system Ux=yU \mathbf{x} = \mathbf{y}Ux=y for x\mathbf{x}x.

STEP 16

STEP 17

Since UUU is an upper triangular matrix, we can solve for x\mathbf{x}x using back substitution.

STEP 18

From the third row, we get −72x3=7-\frac{7}{2} x_3 = 7−27​x3​=7.

STEP 19

Solve for x3x_3x3​.x3=−2x_3 = -2x3​=−2

STEP 20

From the second row, we have x2−x3=1x_2 - x_3 = 1x2​−x3​=1.

STEP 21

Substitute x3x_3x3​ into the second equation.x2−(−2)=1x_2 - (-2) = 1x2​−(−2)=1

STEP 22

Solve for x2x_2x2​.x2=1−2=−1x_2 = 1 - 2 = -1x2​=1−2=−1

STEP 23

From the first row, we have 2x1+3x2+2x3=12 x_1 + 3 x_2 + 2 x_3 = 12x1​+3x2​+2x3​=1.

STEP 24

Substitute x2x_2x2​ and x3x_3x3​ into the first equation.2x1+3⋅(−1)+2⋅(−2)=12 x_1 + 3 \cdot (-1) + 2 \cdot (-2) = 12x1​+3⋅(−1)+2⋅(−2)=1

STEP 25

Solve for x1x_1x1​.2x1−3−4=12 x_1 - 3 - 4 = 12x1​−3−4=1

STEP 26

Simplify the equation to find x1x_1x1​.2x1=1+3+42 x_1 = 1 + 3 + 42x1​=1+3+4

STEP 27

Calculate the value of x1x_1x1​.x1=82=4x_1 = \frac{8}{2} = 4x1​=28​=4

STEP 28

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Solve the system $A \mathbf{x}=\mathbf{b}$ using the given factorization $A=P^{\top} L U$ , where $\mathbf{b}=\begin{bmatrix}1 \\ 1 \\ 7\end{bmatrix}$ and $\mathbf{x}=\begin{bmatrix}2 \\ 3 \\ 2\end{bmatrix}$ .

Perform the matrix multiplication to find $P \mathbf{b}$ .
$P \mathbf{b} = \left[\begin{array}{l} 1 \\ 1 \\ 7 \end{array}\right]$

Now, we need to solve the system $L U \mathbf{x} = P \mathbf{b}$ , which is equivalent to the original system $A \mathbf{x} = \mathbf{b}$ .

First, let's solve the system $L \mathbf{y} = P \mathbf{b}$ for $\mathbf{y}$ , where $\mathbf{y} = U \mathbf{x}$ .

Since $L$ is a lower triangular matrix, we can solve for $\mathbf{y}$ using forward substitution.

From the first row, we directly get $y_1 = 1$ .

From the second row, we directly get $y_2 = 1$ .

From the third row, we have $\frac{1}{2} y_1 - \frac{1}{2} y_2 + y_3 = 7$ .

Substitute $y_1$ and $y_2$ into the third equation.
$\frac{1}{2} \cdot 1 - \frac{1}{2} \cdot 1 + y_3 = 7$

Simplify the equation to solve for $y_3$ .
$y_3 = 7 + \frac{1}{2} - \frac{1}{2}$

Calculate the value of $y_3$ .
$y_3 = 7$

Now we have the vector $\mathbf{y} = \left[\begin{array}{l} 1 \\ 1 \\ 7 \end{array}\right]$ .

Next, we need to solve the system $U \mathbf{x} = \mathbf{y}$ for $\mathbf{x}$ .

Since $U$ is an upper triangular matrix, we can solve for $\mathbf{x}$ using back substitution.

From the third row, we get $-\frac{7}{2} x_3 = 7$ .

Solve for $x_3$ .
$x_3 = -2$

From the second row, we have $x_2 - x_3 = 1$ .

Substitute $x_3$ into the second equation.
$x_2 - (-2) = 1$

Solve for $x_2$ .
$x_2 = 1 - 2 = -1$

From the first row, we have $2 x_1 + 3 x_2 + 2 x_3 = 1$ .

Substitute $x_2$ and $x_3$ into the first equation.
$2 x_1 + 3 \cdot (-1) + 2 \cdot (-2) = 1$

Solve for $x_1$ .
$2 x_1 - 3 - 4 = 1$

Simplify the equation to find $x_1$ .
$2 x_1 = 1 + 3 + 4$

Calculate the value of $x_1$ .
$x_1 = \frac{8}{2} = 4$