langoemmanuel304
Answered

Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.