langoemmanuel304
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Find [tex]\( A \)[/tex] and [tex]\( B \)[/tex] from the following relations:

[tex]\[ 2A - B = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} \][/tex]

and

[tex]\[ 2B + A = \begin{bmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{bmatrix} \][/tex]

Sagot :

GinnyAnswer
Sure! Let's find the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] based on the given equations:

1. [tex]\( 2A - B = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} \)[/tex]
2. [tex]\( 2B + A = \begin{bmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{bmatrix} \)[/tex]

To solve these equations step-by-step:

### Step 1: Isolate [tex]\( A \)[/tex]

First, we can manipulate these equations to isolate one of the matrices. Let's solve for [tex]\( A \)[/tex] first.

Equation 1: [tex]\( 2A - B = \mathbf{X} \)[/tex], where [tex]\( \mathbf{X} = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} \)[/tex].

Multiply the whole equation by 2:
[tex]\[ 4A - 2B = 2\mathbf{X} \][/tex]

This results in:
[tex]\[ 4A - 2B = \begin{bmatrix} 12 & -12 & 0 \\ -8 & 4 & 2 \end{bmatrix} \][/tex]

Equation 2: [tex]\( 2B + A = \mathbf{Y} \)[/tex], where [tex]\( \mathbf{Y} = \begin{bmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{bmatrix} \)[/tex].

Add the two equations together:
[tex]\[ 4A - 2B + 2B + A = \begin{bmatrix} 12 & -12 & 0 \\ -8 & 4 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{bmatrix} \][/tex]
[tex]\[ 5A = \begin{bmatrix} 15 & -10 & 5 \\ -10 & 5 & -5 \end{bmatrix} \][/tex]

Now, divide both sides by 5:
[tex]\[ A = \begin{bmatrix} 3 & -2 & 1 \\ -2 & 1 & -1 \end{bmatrix} \][/tex]

### Step 2: Solve for [tex]\( B \)[/tex]

Now that we have [tex]\( A \)[/tex], substitute it back into one of the original equations to solve for [tex]\( B \)[/tex].

We'll use the equation [tex]\( 2A - B = \mathbf{X} \)[/tex]:
[tex]\[ 2A - B = \mathbf{X} \][/tex]
[tex]\[ 2 \begin{bmatrix} 3 & -2 & 1 \\ -2 & 1 & -1 \end{bmatrix} - B = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} \][/tex]
[tex]\[ \begin{bmatrix} 6 & -4 & 2 \\ -4 & 2 & -2 \end{bmatrix} - B = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} \][/tex]

Rearrange to isolate [tex]\( B \)[/tex]:
[tex]\[ B = \begin{bmatrix} 6 & -4 & 2 \\ -4 & 2 & -2 \end{bmatrix} - \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} \][/tex]
[tex]\[ B = \begin{bmatrix} 0 & 2 & 2 \\ 0 & 0 & -3 \end{bmatrix} \][/tex]

### Final Matrices

Therefore, the solutions for [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are:
[tex]\[ A = \begin{bmatrix} 3 & -2 & 1 \\ -2 & 1 & -1 \end{bmatrix} \][/tex]
[tex]\[ B = \begin{bmatrix} 0 & 2 & 2 \\ 0 & 0 & -3 \end{bmatrix} \][/tex]
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