How do you solve the system #a+5b=1, 7a-2b=44# using matrices?

Answer 1

#a=6, b=-1#

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Answer 2

To solve the system of equations (a + 5b = 1) and (7a - 2b = 44) using matrices, we can represent the system in matrix form (Ax = B), where:

[ A = \begin{bmatrix} 1 & 5 \ 7 & -2 \end{bmatrix} ] [ x = \begin{bmatrix} a \ b \end{bmatrix} ] [ B = \begin{bmatrix} 1 \ 44 \end{bmatrix} ]

To solve for (x), we use the matrix equation (x = A^{-1}B), where (A^{-1}) is the inverse of matrix (A).

First, we need to find the inverse of matrix (A): [ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} -2 & -5 \ -7 & 1 \end{bmatrix} ] Where (\text{det}(A)) is the determinant of matrix (A).

[ \text{det}(A) = (1 \times (-2)) - (5 \times 7) = -2 - 35 = -37 ]

So, (A^{-1}) becomes: [ A^{-1} = \frac{1}{-37} \begin{bmatrix} -2 & -5 \ -7 & 1 \end{bmatrix} ] [ A^{-1} = \begin{bmatrix} \frac{2}{37} & \frac{5}{37} \ \frac{7}{37} & -\frac{1}{37} \end{bmatrix} ]

Now, we can find (x) using (x = A^{-1}B): [ x = \begin{bmatrix} \frac{2}{37} & \frac{5}{37} \ \frac{7}{37} & -\frac{1}{37} \end{bmatrix} \begin{bmatrix} 1 \ 44 \end{bmatrix} ] [ x = \begin{bmatrix} \frac{2}{37} \times 1 + \frac{5}{37} \times 44 \ \frac{7}{37} \times 1 + (-\frac{1}{37}) \times 44 \end{bmatrix} ] [ x = \begin{bmatrix} \frac{2}{37} + \frac{220}{37} \ \frac{7}{37} - \frac{44}{37} \end{bmatrix} ] [ x = \begin{bmatrix} \frac{222}{37} \ -\frac{37}{37} \end{bmatrix} ] [ x = \begin{bmatrix} 6 \ -1 \end{bmatrix} ]

So, the solution to the system of equations is (a = 6) and (b = -1).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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