Math Problem Statement

tried that and it didn't like the answer....

Solution

Let's re-evaluate the problem step-by-step.

We have the matrix equation: [2676]X+[4738]=[9185]\begin{bmatrix} -2 & -6 \\ -7 & -6 \end{bmatrix} X + \begin{bmatrix} 4 & -7 \\ -3 & -8 \end{bmatrix} = \begin{bmatrix} -9 & 1 \\ -8 & -5 \end{bmatrix}

First, isolate XX: [2676]X=[9185][4738]\begin{bmatrix} -2 & -6 \\ -7 & -6 \end{bmatrix} X = \begin{bmatrix} -9 & 1 \\ -8 & -5 \end{bmatrix} - \begin{bmatrix} 4 & -7 \\ -3 & -8 \end{bmatrix}

Calculate the matrix on the right side: [9185][4738]=[941+78+35+8]=[13853]\begin{bmatrix} -9 & 1 \\ -8 & -5 \end{bmatrix} - \begin{bmatrix} 4 & -7 \\ -3 & -8 \end{bmatrix} = \begin{bmatrix} -9 - 4 & 1 + 7 \\ -8 + 3 & -5 + 8 \end{bmatrix} = \begin{bmatrix} -13 & 8 \\ -5 & 3 \end{bmatrix}

So the equation becomes: [2676]X=[13853]\begin{bmatrix} -2 & -6 \\ -7 & -6 \end{bmatrix} X = \begin{bmatrix} -13 & 8 \\ -5 & 3 \end{bmatrix}

Next, find the inverse of the matrix [2676]\begin{bmatrix} -2 & -6 \\ -7 & -6 \end{bmatrix}.

The inverse of a 2x2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} is 1adbc[dbca]\frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.

For [2676]\begin{bmatrix} -2 & -6 \\ -7 & -6 \end{bmatrix}: Determinant=(2×6)(6×7)=1242=30\text{Determinant} = (-2 \times -6) - (-6 \times -7) = 12 - 42 = -30

The inverse is: 130[6672]=[1515730115]\frac{1}{-30} \begin{bmatrix} -6 & 6 \\ 7 & -2 \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & -\frac{1}{5} \\ -\frac{7}{30} & \frac{1}{15} \end{bmatrix}

Now, multiply the inverse by the matrix [13853]\begin{bmatrix} -13 & 8 \\ -5 & 3 \end{bmatrix}: X=[1515730115][13853]X = \begin{bmatrix} \frac{1}{5} & -\frac{1}{5} \\ -\frac{7}{30} & \frac{1}{15} \end{bmatrix} \begin{bmatrix} -13 & 8 \\ -5 & 3 \end{bmatrix}

Calculate each element of the resulting matrix:

  1. First element: X11=(15×13)+(15×5)=135+55=135+1=1355=85=85=1.6X_{11} = \left(\frac{1}{5} \times -13\right) + \left(-\frac{1}{5} \times -5\right) = -\frac{13}{5} + \frac{5}{5} = -\frac{13}{5} + 1 = -\frac{13 - 5}{5} = -\frac{8}{5} = -\frac{8}{5} = -1.6

  2. Second element: X12=(15×8)+(15×3)=8535=835=55=1X_{12} = \left(\frac{1}{5} \times 8\right) + \left(-\frac{1}{5} \times 3\right) = \frac{8}{5} - \frac{3}{5} = \frac{8 - 3}{5} = \frac{5}{5} = 1

  3. Third element: X21=(730×13)+(115×5)=9130515=91301030=8130=2710=2.7X_{21} = \left(-\frac{7}{30} \times -13\right) + \left(\frac{1}{15} \times -5\right) = \frac{91}{30} - \frac{5}{15} = \frac{91}{30} - \frac{10}{30} = \frac{81}{30} = \frac{27}{10} = 2.7

  4. Fourth element: X22=(730×8)+(115×3)=5630+315=5630+630=5030=53X_{22} = \left(-\frac{7}{30} \times 8\right) + \left(\frac{1}{15} \times 3\right) = -\frac{56}{30} + \frac{3}{15} = -\frac{56}{30} + \frac{6}{30} = -\frac{50}{30} = -\frac{5}{3}

Therefore, the correct solution for XX should be: X=[1.612.753]X = \begin{bmatrix} -1.6 & 1 \\ 2.7 & -\frac{5}{3} \end{bmatrix}

The answer you entered was incorrect, likely due to a calculation or simplification error. If you re-enter the values:

  • X11=1.6X_{11} = -1.6
  • X12=1X_{12} = 1
  • X21=2.7X_{21} = 2.7
  • X22=53X_{22} = -\frac{5}{3}

This should be the correct form. Double-check each step and make sure to use the exact fractions when entering the values.

Would you like more details or have any questions?

Here are 5 questions you might want to ask:

  1. How can I verify the correctness of a matrix inverse?
  2. Can you explain the determinant calculation in more detail?
  3. How does matrix multiplication work?
  4. What are some common mistakes when solving matrix equations?
  5. Can you show me how to solve another matrix equation?

Tip: Always double-check your calculations, especially when working with fractions and negative signs, to avoid common arithmetic errors.

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Math Problem Analysis

Mathematical Concepts

Matrix Algebra
Matrix Equations
Matrix Inverses

Formulas

Matrix multiplication
Inverse of a 2x2 matrix

Theorems

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Suitable Grade Level

Undergraduate level