Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
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]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
