Math Problem Statement

Diketahui A = [[1, 2], [-4, 3]], B = [[-3, 2], [-1, 5]], C = [[-3, -4], [1, 2]], dan A^T + 3B^T - 3C - 2X = 0. Jika A^T, B^T, dan X berturut-turut transpose matriks A, B, dan X, tentukan matriks X^T.

Solution

The problem provided involves matrix operations. Here's a breakdown of the question from the image:

  1. We are given the matrices:

    • A=(1243)A = \begin{pmatrix} 1 & 2 \\ -4 & 3 \end{pmatrix}
    • B=(3215)B = \begin{pmatrix} -3 & 2 \\ -1 & 5 \end{pmatrix}
    • C=(3412)C = \begin{pmatrix} -3 & -4 \\ 1 & 2 \end{pmatrix}
  2. We also have the equation: AT+3BT3C2X=0A^T + 3B^T - 3C - 2X = 0 Where ATA^T and BTB^T represent the transpose of matrices AA and BB, respectively, and XX is an unknown matrix that we need to find.

Step-by-Step Approach:

  1. Transpose of Matrices AA and BB:

    • The transpose of matrix AA (ATA^T) is: AT=(1423)A^T = \begin{pmatrix} 1 & -4 \\ 2 & 3 \end{pmatrix}
    • The transpose of matrix BB (BTB^T) is: BT=(3125)B^T = \begin{pmatrix} -3 & -1 \\ 2 & 5 \end{pmatrix}
  2. Matrix Multiplications:

    • Calculate 3BT3B^T: 3BT=3×(3125)=(93615)3B^T = 3 \times \begin{pmatrix} -3 & -1 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix} -9 & -3 \\ 6 & 15 \end{pmatrix}
    • Calculate 3C3C: 3C=3×(3412)=(91236)3C = 3 \times \begin{pmatrix} -3 & -4 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} -9 & -12 \\ 3 & 6 \end{pmatrix}
  3. Substitute into the Equation: The original equation is: AT+3BT3C2X=0A^T + 3B^T - 3C - 2X = 0 Substituting the values: (1423)+(93615)(91236)2X=0\begin{pmatrix} 1 & -4 \\ 2 & 3 \end{pmatrix} + \begin{pmatrix} -9 & -3 \\ 6 & 15 \end{pmatrix} - \begin{pmatrix} -9 & -12 \\ 3 & 6 \end{pmatrix} - 2X = 0

  4. Simplify the Left Side: Combine the matrices: (1423)+(93615)=(87818)\begin{pmatrix} 1 & -4 \\ 2 & 3 \end{pmatrix} + \begin{pmatrix} -9 & -3 \\ 6 & 15 \end{pmatrix} = \begin{pmatrix} -8 & -7 \\ 8 & 18 \end{pmatrix} Now subtract 3C3C: (87818)(91236)=(15512)\begin{pmatrix} -8 & -7 \\ 8 & 18 \end{pmatrix} - \begin{pmatrix} -9 & -12 \\ 3 & 6 \end{pmatrix} = \begin{pmatrix} 1 & 5 \\ 5 & 12 \end{pmatrix} So, the equation becomes: (15512)2X=0\begin{pmatrix} 1 & 5 \\ 5 & 12 \end{pmatrix} - 2X = 0

  5. Solve for XX: Rearrange the equation: 2X=(15512)2X = \begin{pmatrix} 1 & 5 \\ 5 & 12 \end{pmatrix} Divide both sides by 2: X=12×(15512)=(0.52.52.56)X = \frac{1}{2} \times \begin{pmatrix} 1 & 5 \\ 5 & 12 \end{pmatrix} = \begin{pmatrix} 0.5 & 2.5 \\ 2.5 & 6 \end{pmatrix}

Thus, the matrix XX is: X=(0.52.52.56)X = \begin{pmatrix} 0.5 & 2.5 \\ 2.5 & 6 \end{pmatrix}

Would you like more details or have any questions?

Here are 5 related questions to consider:

  1. What is the process for finding the transpose of a matrix?
  2. How would you perform matrix multiplication and addition?
  3. How do you handle matrix equations like the one in the problem?
  4. What properties of matrices are important when working with transposes?
  5. How does scalar multiplication work in matrix operations?

Tip: When solving matrix equations, always work step by step, simplifying matrices at each stage to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Matrix Transpose
Matrix Multiplication
Matrix Equation

Formulas

Transpose of a matrix: A^T
Matrix multiplication: kA
Solving matrix equations

Theorems

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Suitable Grade Level

Grades 10-12