Question

Find the matrices $X$ and $Y$ , if $2X-Y=\left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]$ and $X+2Y=\left[\begin{array}{ccc}3& 2& 5\\ -2& 1& -7\end{array}\right]$

Answer 1

Hint: This problem can be treated as a linear equation in two variables in form of $X$ and $Y$ and solved just like how we solve a linear equation


Complete step-by-step answer:
The given matrices are
$2X-Y=\left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]\cdots \left(1\right)$
$X+2Y=\left[\begin{array}{ccc}3& 2& 5\\ -2& 1& -7\end{array}\right]\cdots \left(2\right)$
For performing the addition or subtraction of two matrices, the order of the matrices should be the same . The order of both the matrices is $2\times 3$ .
Let’s suppose the matrices be $X=\left[\begin{array}{ccc}p& q& r\\ s& t& u\end{array}\right]$ and $Y=\left[\begin{array}{ccc}a& b& c\\ d& e& f\end{array}\right]$
After analyzing the two equations, it is clear that matrix $Y$ can be eliminated for the calculation of matrix $X$ by multiplying equation (1) by 2 and adding equation (1) and (2).
Multiplying equation (1) by (2),
$2\left(2X-Y\right)=2\left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]$
If a matrix is multiplied by a scalar quantity, then the scalar is to be multiplied by each and every term of the matrix as shown below,
$4X-2Y=\left[\begin{array}{ccc}12& -12& 0\\ -8& 4& 2\end{array}\right]\cdots \left(3\right)$
Adding equation (2 ) and (3)
$$\left(X+2Y\right)+\left(4X-2Y\right)=\left[\begin{array}{ccc}3& 2& 5\\ -2& 1& -7\end{array}\right]+\left[\begin{array}{ccc}12& -12& 0\\ -8& 4& 2\end{array}\right]$$
For adding the two matrices , add their corresponding terms.
$\begin{array}{c}5X=\left[\begin{array}{ccc}3+12& 2-12& 5-0\\ -2-8& 1+4& -7+2\end{array}\right]\\ 5X=\left[\begin{array}{ccc}15& -10& 5\\ -10& 5& -5\end{array}\right]\cdots \left(4\right)\end{array}$
Now it is clear that in equation (4), we can take out 5 as a common factor.
$5X=5\left[\begin{array}{ccc}3& -2& 1\\ -2& 1& -1\end{array}\right]\cdots \left(5\right)$
Cancelling 5 from both sides of LHS and RHS in equation (5), matrix $X$ is obtained as
$X=\left[\begin{array}{ccc}3& -2& 1\\ -2& 1& -1\end{array}\right]\cdots \left(6\right)$
Substitute the value of matrix $X$ in equation (1),
$$2\left[\begin{array}{ccc}3& -2& 1\\ -2& 1& -1\end{array}\right]-Y=\left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]\cdots \left(7\right)$$
Multiply all the terms of the matrix $X$ by 2 as given in the equation (7)
$$\left[\begin{array}{ccc}6& -4& 2\\ -4& 2& -2\end{array}\right]-Y=\left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]$$
Now rearrange the terms and calculate the value of matrix $Y$ as,
$Y=\left[\begin{array}{ccc}6& -4& 2\\ -4& 2& -2\end{array}\right]-\left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]$
Now subtract the 2 matrices by subtracting their corresponding terms,
$Y=\left[\begin{array}{ccc}0& 2& 2\\ 0& 0& -3\end{array}\right]$
Hence, the value of matrix $X=\left[\begin{array}{ccc}3& -2& 1\\ -2& 1& -1\end{array}\right]$ and $Y=\left[\begin{array}{ccc}0& 2& 2\\ 0& 0& -3\end{array}\right]$..


Note: The important concepts to be remembered are
1)Two matrices can be added or subtracted only when they have the same order.
2)Multiplication of a matrix by a scalar, leads to multiplication of its terms.