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Sagot :
To solve the system of equations using the matrix tool, we start by writing the system in the form of [tex]\[A\mathbf{x} = \mathbf{b}\][/tex] where [tex]\(A\)[/tex] is the coefficient matrix, [tex]\(\mathbf{x}\)[/tex] is the vector of variables, and [tex]\(\mathbf{b}\)[/tex] is the constants vector.
The given system of linear equations is:
[tex]\[ \begin{array}{rcl} 4x + 11y &=& -5 \\ 4x + 8y &=& -8 \end{array} \][/tex]
First, identify the coefficient matrix [tex]\(A\)[/tex], the variables vector [tex]\(\mathbf{x}\)[/tex], and the constants vector [tex]\(\mathbf{b}\)[/tex]:
[tex]\[ \begin{pmatrix} 4 & 11 \\ 4 & 8 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -5 \\ -8 \end{pmatrix} \][/tex]
Here,
[tex]\[ A = \begin{pmatrix} 4 & 11 \\ 4 & 8 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} -5 \\ -8 \end{pmatrix} \][/tex]
We can use the matrix method to find [tex]\(\mathbf{x} = A^{-1}\mathbf{b}\)[/tex], where [tex]\(A^{-1}\)[/tex] is the inverse of the coefficient matrix [tex]\(A\)[/tex].
To find the inverse of [tex]\(A\)[/tex], first calculate the determinant of [tex]\(A\)[/tex]:
[tex]\[ \text{det}(A) = (4 \cdot 8) - (4 \cdot 11) = 32 - 44 = -12 \][/tex]
Since the determinant is non-zero, the inverse of [tex]\(A\)[/tex] exists. Now, calculate the inverse of the matrix [tex]\(A\)[/tex]:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \frac{1}{-12} \begin{pmatrix} 8 & -11 \\ -4 & 4 \end{pmatrix} = \begin{pmatrix} -\frac{2}{3} & \frac{11}{12} \\ \frac{1}{3} & -\frac{1}{3} \end{pmatrix} \][/tex]
Now, multiply [tex]\(A^{-1}\)[/tex] by [tex]\(\mathbf{b}\)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} -\frac{2}{3} & \frac{11}{12} \\ \frac{1}{3} & -\frac{1}{3} \end{pmatrix} \begin{pmatrix} -5 \\ -8 \end{pmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ \begin{pmatrix} -\frac{2}{3} & \frac{11}{12} \\ \frac{1}{3} & -\frac{1}{3} \end{pmatrix} \begin{pmatrix} -5 \\ -8 \end{pmatrix} = \begin{pmatrix} (-\frac{2}{3}) \cdot (-5) + \frac{11}{12} \cdot (-8) \\ (\frac{1}{3}) \cdot (-5) + (-\frac{1}{3}) \cdot (-8) \end{pmatrix} \][/tex]
Calculate each element step-by-step:
[tex]\[ \begin{pmatrix} \frac{10}{3} - \frac{88}{12} \\ -\frac{5}{3} + \frac{8}{3} \end{pmatrix} = \begin{pmatrix} \frac{10}{3} - \frac{22}{3} \\ \frac{3}{3} \end{pmatrix} = \begin{pmatrix} -\frac{12}{3} \\ 1 \end{pmatrix} = \begin{pmatrix} -4 \\ 1 \end{pmatrix} \][/tex]
Thus, the solution to the given system of equations is the ordered pair [tex]\((-4, 1)\)[/tex].
Therefore, the solution is:
[tex]\[ \boxed{(-4, 1)} \][/tex]
The given system of linear equations is:
[tex]\[ \begin{array}{rcl} 4x + 11y &=& -5 \\ 4x + 8y &=& -8 \end{array} \][/tex]
First, identify the coefficient matrix [tex]\(A\)[/tex], the variables vector [tex]\(\mathbf{x}\)[/tex], and the constants vector [tex]\(\mathbf{b}\)[/tex]:
[tex]\[ \begin{pmatrix} 4 & 11 \\ 4 & 8 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -5 \\ -8 \end{pmatrix} \][/tex]
Here,
[tex]\[ A = \begin{pmatrix} 4 & 11 \\ 4 & 8 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} -5 \\ -8 \end{pmatrix} \][/tex]
We can use the matrix method to find [tex]\(\mathbf{x} = A^{-1}\mathbf{b}\)[/tex], where [tex]\(A^{-1}\)[/tex] is the inverse of the coefficient matrix [tex]\(A\)[/tex].
To find the inverse of [tex]\(A\)[/tex], first calculate the determinant of [tex]\(A\)[/tex]:
[tex]\[ \text{det}(A) = (4 \cdot 8) - (4 \cdot 11) = 32 - 44 = -12 \][/tex]
Since the determinant is non-zero, the inverse of [tex]\(A\)[/tex] exists. Now, calculate the inverse of the matrix [tex]\(A\)[/tex]:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \frac{1}{-12} \begin{pmatrix} 8 & -11 \\ -4 & 4 \end{pmatrix} = \begin{pmatrix} -\frac{2}{3} & \frac{11}{12} \\ \frac{1}{3} & -\frac{1}{3} \end{pmatrix} \][/tex]
Now, multiply [tex]\(A^{-1}\)[/tex] by [tex]\(\mathbf{b}\)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} -\frac{2}{3} & \frac{11}{12} \\ \frac{1}{3} & -\frac{1}{3} \end{pmatrix} \begin{pmatrix} -5 \\ -8 \end{pmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ \begin{pmatrix} -\frac{2}{3} & \frac{11}{12} \\ \frac{1}{3} & -\frac{1}{3} \end{pmatrix} \begin{pmatrix} -5 \\ -8 \end{pmatrix} = \begin{pmatrix} (-\frac{2}{3}) \cdot (-5) + \frac{11}{12} \cdot (-8) \\ (\frac{1}{3}) \cdot (-5) + (-\frac{1}{3}) \cdot (-8) \end{pmatrix} \][/tex]
Calculate each element step-by-step:
[tex]\[ \begin{pmatrix} \frac{10}{3} - \frac{88}{12} \\ -\frac{5}{3} + \frac{8}{3} \end{pmatrix} = \begin{pmatrix} \frac{10}{3} - \frac{22}{3} \\ \frac{3}{3} \end{pmatrix} = \begin{pmatrix} -\frac{12}{3} \\ 1 \end{pmatrix} = \begin{pmatrix} -4 \\ 1 \end{pmatrix} \][/tex]
Thus, the solution to the given system of equations is the ordered pair [tex]\((-4, 1)\)[/tex].
Therefore, the solution is:
[tex]\[ \boxed{(-4, 1)} \][/tex]
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