langoemmanuel304
Answered

Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.