Math

QuestionCreate the augmented matrix from AA and bb for Ax=bA \mathbf{x}=\mathbf{b}, then solve for x\mathbf{x}.

Studdy Solution

STEP 1

Assumptions1. The matrix AA is a3x3 matrix with elements A=[143333]A=\left[\begin{array}{rrr} 1 & & -4 \\ -3 & -3 &3 \\ 3 & & \end{array}\right]. The vector bb is a3x1 vector with elements b=[438]b=\left[\begin{array}{r} 4 \\ 3 \\ -8 \end{array}\right]3. The matrix equation is Ax=bA \mathbf{x}=\mathbf{b}, where x\mathbf{x} is the vector we want to solve for.

STEP 2

First, we need to write the augmented matrix for the linear system that corresponds to the matrix equation Ax=bA \mathbf{x}=\mathbf{b}. The augmented matrix is formed by appending the matrix AA and the vector bb.
[Ab]=[1244228][A|b] = \left[\begin{array}{rrr|r} 1 &2 & -4 &4 \\ - & - & & \\ &2 &2 & -8\end{array}\right]

STEP 3

Now, we need to solve the system of equations represented by the augmented matrix. We can do this by performing row operations to transform the matrix into row echelon form or reduced row echelon form.
Let's start by swapping row1 and row3 to bring the largest absolute value in the first column to the top.
[Ab]=[3228333312][A|b] = \left[\begin{array}{rrr|r} 3 &2 &2 & -8 \\ -3 & -3 &3 &3 \\ 1 &2 & - &\end{array}\right]

STEP 4

Next, we add row1 to row2 to eliminate the -3 in the second row, first column.
[Ab]=[3228011244][A|b] = \left[\begin{array}{rrr|r} 3 &2 &2 & -8 \\ 0 & -1 & & - \\ 1 &2 & -4 &4\end{array}\right]

STEP 5

Now, we multiply row2 by -1 to make the leading coefficient in the second row positive.
[Ab]=[322801551244][A|b] = \left[\begin{array}{rrr|r} 3 &2 &2 & -8 \\ 0 &1 & -5 &5 \\ 1 &2 & -4 &4\end{array}\right]

STEP 6

Next, we subtract row3 from row1 to eliminate the1 in the first row, third column.
[Ab]=[2061201551244][A|b] = \left[\begin{array}{rrr|r} 2 &0 &6 & -12 \\ 0 &1 & -5 &5 \\ 1 &2 & -4 &4\end{array}\right]

STEP 7

Now, we divide row1 by2 to make the leading coefficient in the first row equal to1.
[Ab]=[103601551244][A|b] = \left[\begin{array}{rrr|r} 1 &0 &3 & -6 \\ 0 &1 & -5 &5 \\ 1 &2 & -4 &4\end{array}\right]

STEP 8

Next, we subtract row1 from row3 to eliminate the1 in the third row, first column.
[Ab]=[1036015502710][A|b] = \left[\begin{array}{rrr|r} 1 &0 &3 & -6 \\ 0 &1 & -5 &5 \\ 0 &2 & -7 &10\end{array}\right]

STEP 9

Now, we subtract2 times row2 from row3 to eliminate the2 in the third row, second column.
[Ab]=[36553][A|b] = \left[\begin{array}{rrr|r} & &3 & -6 \\ & & -5 &5 \\ & &3 &\end{array}\right]

STEP 10

Finally, we divide row3 by3 to make the leading coefficient in the third row equal to.
[Ab]=[036055000][A|b] = \left[\begin{array}{rrr|r} &0 &3 & -6 \\ 0 & & -5 &5 \\ 0 &0 & &0\end{array}\right]

STEP 11

Now, we can write the solution as a vector by reading off the values from the rightmost column of the augmented matrix.x=[650]\mathbf{x} = \left[\begin{array}{r} -6 \\ 5 \\ 0\end{array}\right]So, the solution to the system of equations is x=[650]\mathbf{x} = \left[\begin{array}{r} -6 \\ 5 \\ 0\end{array}\right].

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