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To solve for the vector [tex]\( x \)[/tex] such that [tex]\( T(x) = A x = b \)[/tex], where
[tex]\[ A = \begin{pmatrix} 1 & -5 & -6 \\ -3 & 6 & 0 \end{pmatrix} \][/tex]
and
[tex]\[ b = \begin{pmatrix} -3 \\ 0 \end{pmatrix}, \][/tex]
we need to determine [tex]\( x \)[/tex] such that the equation [tex]\( A x = b \)[/tex] holds true.
### Step 1: Formulate the Equation
We are looking for [tex]\( x \)[/tex] in the form:
[tex]\[ x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \][/tex]
such that:
[tex]\[ \begin{pmatrix} 1 & -5 & -6 \\ -3 & 6 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} -3 \\ 0 \end{pmatrix}. \][/tex]
### Step 2: Set Up the System of Linear Equations
This matrix equation translates to the following system of linear equations:
[tex]\[ \begin{cases} 1 \cdot x_1 - 5 \cdot x_2 - 6 \cdot x_3 = -3 \\ -3 \cdot x_1 + 6 \cdot x_2 = 0 \end{cases} \][/tex]
### Step 3: Solve the Second Equation
From the second equation:
[tex]\[ -3x_1 + 6x_2 = 0 \][/tex]
we can solve for one of the variables in terms of the other:
[tex]\[ -3x_1 = -6x_2 \][/tex]
[tex]\[ x_1 = 2x_2 \][/tex]
### Step 4: Substitute [tex]\( x_1 = 2 x_2 \)[/tex] into the First Equation
Next, substitute [tex]\( x_1 = 2 x_2 \)[/tex] into the first equation:
[tex]\[ 1 \cdot (2 x_2) - 5 \cdot x_2 - 6 \cdot x_3 = -3 \][/tex]
[tex]\[ 2 x_2 - 5 x_2 - 6 x_3 = -3 \][/tex]
[tex]\[ -3 x_2 - 6 x_3 = -3 \][/tex]
### Step 5: Solve for [tex]\( x_2 \)[/tex]
Divide the equation by -3:
[tex]\[ x_2 + 2 x_3 = 1 \][/tex]
[tex]\[ x_2 = 1 - 2 x_3 \][/tex]
### Step 6: Express [tex]\( x_1, x_2, x_3 \)[/tex]
We've determined:
[tex]\[ x_1 = 2 x_2 = 2 (1 - 2 x_3) = 2 - 4 x_3, \][/tex]
therefore:
[tex]\[ x_1 = 2 - 4 x_3 \][/tex]
[tex]\[ x_2 = 1 - 2 x_3 \][/tex]
[tex]\[ x_3 = x_3 \][/tex]
The solution vector [tex]\( x \)[/tex] can be expressed as:
[tex]\[ x = \begin{pmatrix} 2 - 4 x_3 \\ 1 - 2 x_3 \\ x_3 \end{pmatrix}. \][/tex]
### Step 7: Determine Specific Solution
We use the specific instance when [tex]\( x_3 = 0 \)[/tex] to find a particular solution:
[tex]\[ x_3 = 0 \][/tex]
[tex]\[ x_1 = 2 \][/tex]
[tex]\[ x_2 = 1 \][/tex]
So, one particular solution is:
[tex]\[ x = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}. \][/tex]
### Step 8: General Solution and Uniqueness
The general solution is:
[tex]\[ x = \begin{pmatrix} 2 - 4 x_3 \\ 1 - 2 x_3 \\ x_3 \end{pmatrix}. \][/tex]
This implies that there are infinitely many solutions due to the free variable [tex]\( x_3 \)[/tex]. Therefore, the solution is not unique.
### Conclusion
The vector [tex]\( x \)[/tex] whose image under [tex]\( T \)[/tex] is [tex]\( b \)[/tex] is given by:
[tex]\[ x = \begin{pmatrix} 2 - 4 x_3 \\ 1 - 2 x_3 \\ x_3 \end{pmatrix}, \][/tex]
where [tex]\( x_3 \)[/tex] is any real number. Consequently, the solution is not unique. In the given example, a solution can also be represented by [tex]\( x = \begin{pmatrix} 0.09523809523809512 \\ 0.047619047619047616 \\ 0.47619047619047583 \end{pmatrix} \)[/tex].
[tex]\[ A = \begin{pmatrix} 1 & -5 & -6 \\ -3 & 6 & 0 \end{pmatrix} \][/tex]
and
[tex]\[ b = \begin{pmatrix} -3 \\ 0 \end{pmatrix}, \][/tex]
we need to determine [tex]\( x \)[/tex] such that the equation [tex]\( A x = b \)[/tex] holds true.
### Step 1: Formulate the Equation
We are looking for [tex]\( x \)[/tex] in the form:
[tex]\[ x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \][/tex]
such that:
[tex]\[ \begin{pmatrix} 1 & -5 & -6 \\ -3 & 6 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} -3 \\ 0 \end{pmatrix}. \][/tex]
### Step 2: Set Up the System of Linear Equations
This matrix equation translates to the following system of linear equations:
[tex]\[ \begin{cases} 1 \cdot x_1 - 5 \cdot x_2 - 6 \cdot x_3 = -3 \\ -3 \cdot x_1 + 6 \cdot x_2 = 0 \end{cases} \][/tex]
### Step 3: Solve the Second Equation
From the second equation:
[tex]\[ -3x_1 + 6x_2 = 0 \][/tex]
we can solve for one of the variables in terms of the other:
[tex]\[ -3x_1 = -6x_2 \][/tex]
[tex]\[ x_1 = 2x_2 \][/tex]
### Step 4: Substitute [tex]\( x_1 = 2 x_2 \)[/tex] into the First Equation
Next, substitute [tex]\( x_1 = 2 x_2 \)[/tex] into the first equation:
[tex]\[ 1 \cdot (2 x_2) - 5 \cdot x_2 - 6 \cdot x_3 = -3 \][/tex]
[tex]\[ 2 x_2 - 5 x_2 - 6 x_3 = -3 \][/tex]
[tex]\[ -3 x_2 - 6 x_3 = -3 \][/tex]
### Step 5: Solve for [tex]\( x_2 \)[/tex]
Divide the equation by -3:
[tex]\[ x_2 + 2 x_3 = 1 \][/tex]
[tex]\[ x_2 = 1 - 2 x_3 \][/tex]
### Step 6: Express [tex]\( x_1, x_2, x_3 \)[/tex]
We've determined:
[tex]\[ x_1 = 2 x_2 = 2 (1 - 2 x_3) = 2 - 4 x_3, \][/tex]
therefore:
[tex]\[ x_1 = 2 - 4 x_3 \][/tex]
[tex]\[ x_2 = 1 - 2 x_3 \][/tex]
[tex]\[ x_3 = x_3 \][/tex]
The solution vector [tex]\( x \)[/tex] can be expressed as:
[tex]\[ x = \begin{pmatrix} 2 - 4 x_3 \\ 1 - 2 x_3 \\ x_3 \end{pmatrix}. \][/tex]
### Step 7: Determine Specific Solution
We use the specific instance when [tex]\( x_3 = 0 \)[/tex] to find a particular solution:
[tex]\[ x_3 = 0 \][/tex]
[tex]\[ x_1 = 2 \][/tex]
[tex]\[ x_2 = 1 \][/tex]
So, one particular solution is:
[tex]\[ x = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}. \][/tex]
### Step 8: General Solution and Uniqueness
The general solution is:
[tex]\[ x = \begin{pmatrix} 2 - 4 x_3 \\ 1 - 2 x_3 \\ x_3 \end{pmatrix}. \][/tex]
This implies that there are infinitely many solutions due to the free variable [tex]\( x_3 \)[/tex]. Therefore, the solution is not unique.
### Conclusion
The vector [tex]\( x \)[/tex] whose image under [tex]\( T \)[/tex] is [tex]\( b \)[/tex] is given by:
[tex]\[ x = \begin{pmatrix} 2 - 4 x_3 \\ 1 - 2 x_3 \\ x_3 \end{pmatrix}, \][/tex]
where [tex]\( x_3 \)[/tex] is any real number. Consequently, the solution is not unique. In the given example, a solution can also be represented by [tex]\( x = \begin{pmatrix} 0.09523809523809512 \\ 0.047619047619047616 \\ 0.47619047619047583 \end{pmatrix} \)[/tex].
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