Math Problem Statement

B=(312−6)B=(321−6)

)

(

C=(4−10253)C=(405−123)

)

Hasil dari B-1 x C adalah?

Select one:

a.

(

B−1C=(1,10,35−0,2−0,31,8−0,2)B−1C=(1,1−0,21,80,35−0,3−0,2)

)

b.

(

B−1C=(1,10,350,2−0,31,8−0,2)B−1C=(1,10,21,80,35−0,3−0,2)

)

c.

(

B−1C=(1,10,350,2−0,3−1,8−0,2)B−1C=(1,10,2−1,80,35−0,3−0,2)

)

d.

(

B−1C=(1,10,350,2−0,31,80,2)B−1C=(1,10,21,80,35−0,30,2)

)

Clear my choice

Solution

To solve for B1×CB^{-1} \times C, we need to follow these steps:

  1. Determine BB and CC:

    From the problem, we are given two matrices: B=(31263216)B = \begin{pmatrix} 312 & -6 \\ 321 & -6 \end{pmatrix} and C=(410253405)C = \begin{pmatrix} 4 & -10 \\ 253 & 405 \end{pmatrix}

  2. Find the inverse of matrix BB, B1B^{-1}: To find the inverse of a 2x2 matrix B=(abcd)B = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, the formula is: B1=1adbc(dbca)B^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

    Substituting values from BB: B=(31263216)B = \begin{pmatrix} 312 & -6 \\ 321 & -6 \end{pmatrix} where a=312a = 312, b=6b = -6, c=321c = 321, and d=6d = -6.

    The determinant det(B)\text{det}(B) is: det(B)=(312×6)(6×321)=1872+1926=54\text{det}(B) = (312 \times -6) - (-6 \times 321) = -1872 + 1926 = 54

    Now, using the inverse formula: B1=154(66321312)B^{-1} = \frac{1}{54} \begin{pmatrix} -6 & 6 \\ -321 & 312 \end{pmatrix} This simplifies to: B1=(1919107185227)B^{-1} = \begin{pmatrix} -\frac{1}{9} & \frac{1}{9} \\ -\frac{107}{18} & \frac{52}{27} \end{pmatrix}

  3. Multiply B1B^{-1} by CC: Now that we have B1B^{-1}, we can multiply it by CC: B1×C=(1919107185227)×(410253405)B^{-1} \times C = \begin{pmatrix} -\frac{1}{9} & \frac{1}{9} \\ -\frac{107}{18} & \frac{52}{27} \end{pmatrix} \times \begin{pmatrix} 4 & -10 \\ 253 & 405 \end{pmatrix}

    Perform the matrix multiplication:

    For the first row, first column: (19×4)+(19×253)=49+2539=2499=27.67\left(-\frac{1}{9} \times 4 \right) + \left( \frac{1}{9} \times 253 \right) = -\frac{4}{9} + \frac{253}{9} = \frac{249}{9} = 27.67

    For the first row, second column: (19×10)+(19×405)=109+4059=4159=46.11\left(-\frac{1}{9} \times -10 \right) + \left( \frac{1}{9} \times 405 \right) = \frac{10}{9} + \frac{405}{9} = \frac{415}{9} = 46.11

    For the second row, first column: (10718×4)+(5227×253)=42818+1315627=23.78+487.26=463.48\left(-\frac{107}{18} \times 4 \right) + \left( \frac{52}{27} \times 253 \right) = -\frac{428}{18} + \frac{13156}{27} = -23.78 + 487.26 = 463.48

    For the second row, second column: (10718×10)+(5227×405)=107018+2106027=59.44+780.00=839.44\left(-\frac{107}{18} \times -10 \right) + \left( \frac{52}{27} \times 405 \right) = \frac{1070}{18} + \frac{21060}{27} = 59.44 + 780.00 = 839.44

  4. Final Result: The resulting matrix B1×CB^{-1} \times C is approximately: (27.6746.11463.48839.44)\begin{pmatrix} 27.67 & 46.11 \\ 463.48 & 839.44 \end{pmatrix}

However, the multiple choices you have given don't exactly match this result. If you meant to provide an approximation, then you might look for values close to the ones we've computed.

Would you like more clarification or assistance on the steps?

Follow-up Questions:

  1. Would you like help with matrix multiplication in general?
  2. Do you need clarification on how to find the inverse of a matrix?
  3. How do you typically approach matrix problems like this?
  4. Are you interested in learning about different methods for solving similar problems?
  5. Do you need further assistance with matrix operations?

Tip: When calculating the inverse of a matrix, always ensure your determinant is non-zero. If it is zero, the matrix doesn't have an inverse!

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Math Problem Analysis

Mathematical Concepts

Matrix Inversion
Matrix Multiplication
Determinants

Formulas

Inverse of a 2x2 matrix: B^{-1} = (1/det(B)) * [[d, -b], [-c, a]]
Matrix multiplication formula

Theorems

Matrix inversion
Matrix multiplication

Suitable Grade Level

Grades 10-12