Question

For what values of x and y are the following matrices equal?

{2x + 1}&{3y} \\\ 0&{{y^2} - 5y} \end{array}} \right],{\text{B}} = \left[ {\begin{array}{*{20}{c}} {x + 3}&{{y^2} + 2} \\\ 0&{ - 6} \end{array}} \right]$$. Also, find the value corresponding to $x-y$?

Answer 1

Here, we will proceed by checking the orders of the given matrices and then specifying the two necessary conditions for two matrices to be equal. Finally, we will obtain the required values of x and y by equating the corresponding elements of the two given matrices.

Complete step-by-step answer:
As we know that for any two matrices A and B to be equal, the below mentioned two conditions should always hold true.
Firstly, the order of both the matrices should be same i.e., if the order of matrix A is $m \times n$ then the order to the matrix B should always be equal to $m \times n$ in order to have equal matrices A and B.
Secondly, each corresponding element should be the same in both the matrices which need to be equal i.e., all the corresponding elements of matrices A and B should be equal.
Given two matrices A and B which are as following:

{2x + 1}&{3y} \\\ 0&{{y^2} - 5y} \end{array}} \right],{\text{B}} = \left[ {\begin{array}{*{20}{c}} {x + 3}&{{y^2} + 2} \\\ 0&{ - 6} \end{array}} \right]$$ Clearly, the order of matrix A is $2 \times 2$ and the order of matrix B is also $ 2\times 2 $. So, the first condition for given matrices A and B to be equal matrices is satisfied. For the second condition, the corresponding elements of the given matrices A and B should be equal. i.e., $ 2x + 1 = x + 3 \\\ \Rightarrow 2x - x = 3 - 1 \\\ \Rightarrow x = 2 $ $ 3y = {y^2} + 2 \\\ \Rightarrow {y^2} - 3y + 2 = 0 $ By factoring the above quadratic equation into linear factors, we get $ \Rightarrow {y^2} - y - 2y + 2 = 0$ By taking y common from first two terms and -2 common from last two terms on the LHS of the above equation, we get $ \Rightarrow y\left( {y - 1} \right) - 2\left( {y - 1} \right) = 0$ By taking (y-1) common from two terms on the LHS of the above equation, we get $ \Rightarrow \left( {y - 1} \right)\left( {y - 2} \right) = 0$ Either $ y - 1 = 0 \\\ \Rightarrow y = 1 $ or $ y - 2 = 0 \\\ \Rightarrow y = 2 $ and $${y^2} - 5y = - 6$$ $$ \Rightarrow {y^2} - 5y + 6 = 0$$ By factoring the above quadratic equation into linear factors, we get $ \Rightarrow {y^2} - 2y - 3y + 6 = 0 $ By taking y common from first two terms and -3 common from last two terms on the LHS of the above equation, we get $ \Rightarrow y\left( {y - 2} \right) - 3\left( {y - 2} \right) = 0$ By taking (y-2) common from two terms on the LHS of the above equation, we get $ \Rightarrow \left( {y - 2} \right)\left( {y - 3} \right) = 0$ Either $ y - 2 = 0 \\\ \Rightarrow y = 2 $ or $ y - 3 = 0 \\\ \Rightarrow y = 3 $ In one case, we are getting the values of y as 1 or 2 but in the second case, we are getting the values of y as 2 or 3. Both of these cases should be satisfied so the correct value of y will be the one common to both the cases. So, the required value of y is 2 i.e., $y = 2$ Therefore, the required values of x and y are 2 and 2 respectively. Also,$ x-y = 2-2 = 0$ **Hence, the value of $x-y$ is zero (0).** **Note:** The order of any matrix A can be represented as $m\times n$ where m denotes the number of rows in the matrix A and n denotes the number of columns in the matrix A. In this particular problem, by equating the corresponding elements of matrices A and B we have taken the intersection of the values of y.