Question

If A=begin{pmatrix}8 & 0 4 & -2 3&6 end{pmatrix} text{ and } B=begin{pmatrix}2 & -2 4 & 2 -5&1 end{pmatrix} ,then find the matrix X such that 2A+3X=5B

Matrices
ANSWERED
asked 2021-02-02
If \(A=\begin{pmatrix}8 & 0 \\ 4 & -2 \\ 3&6 \end{pmatrix} \text{ and } B=\begin{pmatrix}2 & -2 \\ 4 & 2 \\ -5&1 \end{pmatrix}\) ,then find the matrix X such that \(2A+3X=5B\)

Answers (1)

2021-02-03
Step 1
We have to find the matrix X such that \(2A+3X=5B\) where matrices are given as: \(A=\begin{pmatrix}8 & 0 \\ 4 & -2 \\ 3&6 \end{pmatrix} \text{ and } B=\begin{pmatrix}2 & -2 \\ 4 & 2 \\ -5&1 \end{pmatrix}\)
Here we have to use operation of matrices, in operation of matrices we focus on corresponding elements of the matrix.
Example:
\(2\begin{pmatrix}a & b \\c & d \end{pmatrix}=\begin{pmatrix}2a & 2b \\ 2c & 2d \end{pmatrix} , \frac{1}{2}\begin{pmatrix}a & b \\c & d \end{pmatrix}=\begin{pmatrix}\frac{a}{2} & \frac{b}{2} \\\frac{c}{2} & \frac{d}{2} \end{pmatrix}\)
for addition of two matrices,
\(\begin{pmatrix}a & b \\c & d \end{pmatrix}+\begin{pmatrix}x & y \\ z & w \end{pmatrix} = \begin{pmatrix}a+x & b+y \\ c+z & d+w \end{pmatrix}\)
Step 2
Putting given value of matrices into the expression, we get
\(2A+3X=5B\)
\(2\begin{pmatrix}8 & 0 \\ 4 & -2 \\ 3&6 \end{pmatrix}+3X=5\begin{pmatrix}2 & -2 \\ 4 & 2 \\ -5&1 \end{pmatrix}\)
\(\begin{pmatrix}2\times8 & 2\times0 \\ 2\times4 & 2(-2) \\ 2\times3&2\times6 \end{pmatrix}+3X=\begin{pmatrix}5\times2 & 5(-2) \\ 5\times4 & 5\times2 \\ 5(-5)&5\times1 \end{pmatrix}\)
\(\begin{pmatrix}16 & 0 \\ 8 & -4 \\ 6&12 \end{pmatrix}+3X=\begin{pmatrix}10 & -10 \\ 20 & 10 \\ -25&5 \end{pmatrix}\)
\(3X=\begin{pmatrix}10 & -10 \\ 20 & 10 \\ -25&5 \end{pmatrix}-\begin{pmatrix}16 & 0 \\ 8 & -4 \\ 6&12 \end{pmatrix}\)
\(=\begin{pmatrix}10-16 & -10-0 \\ 20-8 & 10+4 \\ -25-6&5-12 \end{pmatrix}\)
\(3X=\begin{pmatrix}-6 & -10 \\ 12 & 14 \\ -31&-7 \end{pmatrix}\)
\(X=\frac{1}{3} \begin{pmatrix}-6 & -10 \\ 12 & 14 \\ -31&-7 \end{pmatrix}\)
\(X= \begin{pmatrix}\frac{-6}{3} & \frac{-10}{3} \\ \frac{12}{3} & \frac{14}{3} \\ \frac{-31}{3}&\frac{-7}{3} \end{pmatrix}\)
\(=\begin{pmatrix}-2 & \frac{-10}{3} \\ 4 & \frac{14}{3} \\ \frac{-31}{3}&\frac{-7}{3} \end{pmatrix}\)
Hence, \(X=\begin{pmatrix}-2 & \frac{-10}{3} \\ 4 & \frac{14}{3} \\ \frac{-31}{3}&\frac{-7}{3} \end{pmatrix}\)
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