Question

Solve for X in the equation, given3X + 2A = BA=begin{bmatrix}-4 & 0 1 & -5-3&2 end{bmatrix} text{ and } B=begin{bmatrix}1 & 2 -2 & 1 4&4 end{bmatrix}

Matrices
ANSWERED
asked 2020-10-25

Solve for X in the equation, given
\(3X + 2A = B\)
\(A=\begin{bmatrix}-4 & 0 \\1 & -5\\-3&2 \end{bmatrix} \text{ and } B=\begin{bmatrix}1 & 2 \\ -2 & 1 \\ 4&4 \end{bmatrix}\)

Expert Answers (1)

2020-10-26

Step 1
Given that:
The matrix are :
\(A=\begin{bmatrix}-4 & 0 \\1 & -5\\-3&2 \end{bmatrix} \text{ and } B=\begin{bmatrix}1 & 2 \\ -2 & 1 \\ 4&4 \end{bmatrix}\)
Step 2
Use :
To add or subtract the matrices, they must be the same size as one another. In addition or subtraction of matrices, just add or subtract the corresponding entries in each matrix. To multiply a matrix by some scalar ( that is number ) c, simply multiply each entry in the matrix by c.
Step 3
a) \(3 X + 2 A = B.\)
To solve the above equation for X:
Let, \(3X + 2 A = B\)
Plug the values of matrix A and B in the above equation,
To get, \(3X+2A=B\)
\(3X+2\begin{bmatrix}-4 & 0 \\1 & -5\\-3&2 \end{bmatrix}=\begin{bmatrix}1 & 2 \\ -2 & 1 \\ 4&4 \end{bmatrix}\)
Since, \(2\begin{bmatrix}-4 & 0 \\1 & -5\\-3&2 \end{bmatrix}=\begin{bmatrix}-4(2) & 0(2) \\ 1(2) & -5(2) \\ -3(2)&2(2) \end{bmatrix}\)
\(=\begin{bmatrix}-8 & 0 \\2 & -10\\-6&4 \end{bmatrix}\)
Step 4
Plug \(2\begin{bmatrix}-4 & 0 \\1 & -5\\-3&2 \end{bmatrix} =\begin{bmatrix}-8 & 0 \\2 & -10\\-6&4 \end{bmatrix}\) from above equation.
To get ,
\(3X+\begin{bmatrix}-8 & 0 \\2 & -10\\-6&4 \end{bmatrix}=\begin{bmatrix}1 & 2 \\-2 & 1\\4&4 \end{bmatrix}\)
Subtract on both side by the matrix \(\begin{bmatrix}-8 & 0 \\2 & -10\\-6&4 \end{bmatrix}\)
\(3x=\begin{bmatrix}1 & 2 \\-2 & 1\\4&4 \end{bmatrix}-\begin{bmatrix}-8 & 0 \\2 & -10\\-6&4 \end{bmatrix}\)
\(3X=\begin{bmatrix}1-(-8) & 2-(0) \\-2-(2) & 1-(-10)\\4-(-6)&4-4 \end{bmatrix}\)
\(3X=\begin{bmatrix}9 & 2 \\-4 & 11\\10&0 \end{bmatrix}\)
Divide on both side by 3,
To get,
\(X=\frac{1}{3}\begin{bmatrix}9 & 2 \\-4 & 11\\10&0 \end{bmatrix}\)
Therefore
a)\(3X+2A=B\)
\(X=\frac{1}{3}\begin{bmatrix}9 & 2 \\-4 & 11\\10&0 \end{bmatrix}\)

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