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Solve for X using the matrices below.2A + 4B = −2XA = 6 −5 7 03 −4 and B = 0 7 8 0−4 −1.X =

Question

Solve for X using the matrices below.

2A+4B=−2X 2A + 4B = -2X

A=(6−57 03−4) A = \begin{pmatrix} 6 & -5 & 7 \\\ 0 & 3 & -4 \end{pmatrix}

B=(078 0−4−1) B = \begin{pmatrix} 0 & 7 & 8 \\\ 0 & -4 & -1 \end{pmatrix}

X= X =

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Solution

1. Break Down the Problem

We need to solve the equation 2A+4B=−2X 2A + 4B = -2X for X X . Given matrices are: A=(6−5703−4)andB=(0780−4−1) A = \begin{pmatrix} 6 & -5 & 7 \\ 0 & 3 & -4 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 0 & 7 & 8 \\ 0 & -4 & -1 \end{pmatrix}

2. Relevant Concepts

  1. Matrix addition.
  2. Scalar multiplication of matrices.
  3. Rearrangement of the equation to solve for X X .

3. Analysis and Detail

First, calculate 2A 2A and 4B 4B :

  • Calculate 2A 2A : 2A=2⋅(6−5703−4)=(12−101406−8) 2A = 2 \cdot \begin{pmatrix} 6 & -5 & 7 \\ 0 & 3 & -4 \end{pmatrix} = \begin{pmatrix} 12 & -10 & 14 \\ 0 & 6 & -8 \end{pmatrix}

  • Calculate 4B 4B : 4B=4⋅(0780−4−1)=(028320−16−4) 4B = 4 \cdot \begin{pmatrix} 0 & 7 & 8 \\ 0 & -4 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 28 & 32 \\ 0 & -16 & -4 \end{pmatrix}

  • Combine 2A+4B 2A + 4B : 2A+4B=(12−101406−8)+(028320−16−4)=(1218460−10−12) 2A + 4B = \begin{pmatrix} 12 & -10 & 14 \\ 0 & 6 & -8 \end{pmatrix} + \begin{pmatrix} 0 & 28 & 32 \\ 0 & -16 & -4 \end{pmatrix} = \begin{pmatrix} 12 & 18 & 46 \\ 0 & -10 & -12 \end{pmatrix}

Now we have: 2A+4B=(1218460−10−12) 2A + 4B = \begin{pmatrix} 12 & 18 & 46 \\ 0 & -10 & -12 \end{pmatrix}

3. Solve for X X

From the equation 2A+4B=−2X 2A + 4B = -2X , we have: −2X=(1218460−10−12) -2X = \begin{pmatrix} 12 & 18 & 46 \\ 0 & -10 & -12 \end{pmatrix} This implies: X=−12(1218460−10−12)=(−6−9−23056) X = -\frac{1}{2} \begin{pmatrix} 12 & 18 & 46 \\ 0 & -10 & -12 \end{pmatrix} = \begin{pmatrix} -6 & -9 & -23 \\ 0 & 5 & 6 \end{pmatrix}

4. Verify and Summarize

To verify, we substitute X X back into the original equation 2A+4B=−2X 2A + 4B = -2X : −2X=−2(−6−9−23056)=(1218460−10−12) -2X = -2 \begin{pmatrix} -6 & -9 & -23 \\ 0 & 5 & 6 \end{pmatrix} = \begin{pmatrix} 12 & 18 & 46 \\ 0 & -10 & -12 \end{pmatrix} The calculations are consistent.

Final Answer

X=(−6−9−23056) X = \begin{pmatrix} -6 & -9 & -23 \\ 0 & 5 & 6 \end{pmatrix}

This problem has been solved

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