langoemmanuel304
Answered

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

[tex]\[
\begin{array}{r}
2A - B = \left[\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]
\text{ and }
2B + A = \left[\begin{array}{ccc}
3 & 2 & 5 \\
-2 & 1 & -7
\end{array}\right]
\end{array}
\][/tex]

Sagot :

GinnyAnswer
To find matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex] from the given relations:
[tex]\[ 2A - B = \begin{pmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{pmatrix} \][/tex]
[tex]\[ 2B + A = \begin{pmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{pmatrix} \][/tex]

We need to solve these two equations step-by-step.

### Step 1: Express [tex]\(B\)[/tex] in terms of [tex]\(A\)[/tex]
From the first equation:
[tex]\[ 2A - B = \begin{pmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{pmatrix} \][/tex]

Rearrange this to solve for [tex]\(B\)[/tex]:
[tex]\[ B = 2A - \begin{pmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{pmatrix} \][/tex]

### Step 2: Substitute [tex]\(B\)[/tex] into the second equation
Now substitute [tex]\(B\)[/tex] from above into the second equation [tex]\(2B + A = \begin{pmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{pmatrix} \)[/tex].

[tex]\[ 2\left(2A - \begin{pmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{pmatrix}\right) + A = \begin{pmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{pmatrix} \][/tex]
[tex]\[ 4A - 2\begin{pmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{pmatrix} + A = \begin{pmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{pmatrix} \][/tex]
[tex]\[ 4A - \begin{pmatrix} 12 & -12 & 0 \\ -8 & 4 & 2 \end{pmatrix} + A = \begin{pmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{pmatrix} \][/tex]
[tex]\[ 5A - \begin{pmatrix} 12 & -12 & 0 \\ -8 & 4 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{pmatrix} \][/tex]
[tex]\[ 5A = \begin{pmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{pmatrix} + \begin{pmatrix} 12 & -12 & 0 \\ -8 & 4 & 2 \end{pmatrix} \][/tex]
[tex]\[ 5A = \begin{pmatrix} 3+12 & 2-12 & 5+0 \\ -2-8 & 1+4 & -7+2 \end{pmatrix} \][/tex]
[tex]\[ 5A = \begin{pmatrix} 15 & -10 & 5 \\ -10 & 5 & -5 \end{pmatrix} \][/tex]

### Step 3: Solve for [tex]\(A\)[/tex]
[tex]\[ A = \frac{1}{5} \begin{pmatrix} 15 & -10 & 5 \\ -10 & 5 & -5 \end{pmatrix} \][/tex]
[tex]\[ A = \begin{pmatrix} 3 & -2 & 1 \\ -2 & 1 & -1 \end{pmatrix} \][/tex]

### Step 4: Solve for [tex]\(B\)[/tex] using [tex]\(A\)[/tex]
Now that we have [tex]\(A\)[/tex], substitute it back into the expression for [tex]\(B\)[/tex]:
[tex]\[ B = 2A - \begin{pmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{pmatrix} \][/tex]
[tex]\[ B = 2 \begin{pmatrix} 3 & -2 & 1 \\ -2 & 1 & -1 \end{pmatrix} - \begin{pmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 6 & -4 & 2 \\ -4 & 2 & -2 \end{pmatrix} - \begin{pmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 6-6 & -4+6 & 2-0 \\ -4+4 & 2-2 & -2-1 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 0 & 2 & 2 \\ 0 & 0 & -3 \end{pmatrix} \][/tex]

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