# Solve for x and y begin{bmatrix}x & 2y 4 & 6 end{bmatrix}=begin{bmatrix}2 & -2 2x & -6y end{bmatrix} 3begin{bmatrix}x & y y & x end{bmatrix}=begin{bma

generals336 2020-11-08 Answered
Solve for x and y $\left[\begin{array}{cc}x& 2y\\ 4& 6\end{array}\right]=\left[\begin{array}{cc}2& -2\\ 2x& -6y\end{array}\right]$
$3\left[\begin{array}{cc}x& y\\ y& x\end{array}\right]=\left[\begin{array}{cc}6& -9\\ -9& 6\end{array}\right]$
$2\left[\begin{array}{cc}x& y\\ x+y& x-y\end{array}\right]=\left[\begin{array}{cc}2& -4\\ -2& 6\end{array}\right]$
$\left[\begin{array}{cc}x& y\\ -y& x\end{array}\right]-\left[\begin{array}{cc}y& x\\ x& -y\end{array}\right]=\left[\begin{array}{cc}4& -4\\ -6& 6\end{array}\right]$
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## Expert Answer

Demi-Leigh Barrera
Answered 2020-11-09 Author has 97 answers
Step 1
Answer(43):
Given that,
$\left[\begin{array}{cc}x& 2y\\ 4& 6\end{array}\right]=\left[\begin{array}{cc}2& -2\\ 2x& -6y\end{array}\right]$
Two matrices are said to be equal if each corresponding entry is equal. Therefore, equating the corresponding entries of the left and right sides matrice $x=2$
And
$2y=-2$
$y=\frac{-2}{2}$
$y=-1$
Hence, the solution of the given equation is x=2, y=-1
Step 2
Answer(44):
Given that, $3\left[\begin{array}{cc}x& y\\ y& x\end{array}\right]=\left[\begin{array}{cc}6& -9\\ -9& 6\end{array}\right]$
Since $k\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}a\cdot k& b\cdot k\\ c\cdot k& d\cdot k\end{array}\right]$
So , $\left[\begin{array}{cc}3x& 3y\\ 3y& 3x\end{array}\right]=\left[\begin{array}{cc}6& -9\\ -9& 6\end{array}\right]$
Two matrices are said to be equal if each corresponding entry is equal.
Therefore, equating the corresponding entries of the left and right sides matrices.
$3x=6$
$x=\frac{6}{3}$
$x=2$
$3y=-9$
$y=-\frac{9}{3}$
$y=-3$
Hence, the solution of the given equation is x=2, y=-3
Step 3
Answer(45):
Given that, $2\left[\begin{array}{cc}x& y\\ x+y& x-y\end{array}\right]=\left[\begin{array}{cc}2& -4\\ -2& 6\end{array}\right]$
Since $k\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}a\cdot k& b\cdot k\\ c\cdot k& d\cdot k\end{array}\right]$
So , $\left[\begin{array}{cc}2x& 2y\\ 2\left(x+y\right)& 2\left(x-y\right)\end{array}\right]=\left[\begin{array}{cc}2& -4\\ -2& 6\end{array}\right]$
Two matrices are said to be equal if each corresponding entry is equal.
Therefore, equating the corresponding entries of the left and right sides matrices
$2x=2$
$x=\frac{2}{2}$
$x=1$
$2y=-4$
$y=-\frac{4}{2}$
$y=-2$
Hence, the solution of the given equation is x=1, y=-2
Step 4
Answer(46):
Given that, $\left[\begin{array}{cc}x& y\\ -y& x\end{array}\right]-\left[\begin{array}{cc}y& x\\ x& -y\end{array}\right]=\left[\begin{array}{cc}4& -4\\ -6& 6\end{array}\right]$
To find the solution, Apply the matrices property $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]-\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]=\left[\begin{array}{cc}a-p& b-q\\ c-r& d-s\end{array}\right]$
$\left[\begin{array}{cc}x-y& y-x\\ -y-x& x=y\end{array}\right]=\left[\begin{array}{cc}4& -4\\ -6& 6\end{array}\right]$
Two matrices are said to be equal if each corresponding entry is equal.
Therefore, equating the corresponding entries of the left and right sides matrices.
(1) x-y=4
(2) x+y=6
Now, add the equation (1) and (2).
$2x=10$
$x=5$
And, subtract the equation (1) and (2).
$-2y=4-6$
$-2y=-2$
$y=1$
Hence, the solution of the given equation is x=5, y=1
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Jeffrey Jordon
Answered 2022-01-23 Author has 2047 answers

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