Question

If $2\left[ \begin{matrix} 3 & 4 \\\ 5 & x \\\ \end{matrix} \right]+\left[ \begin{matrix} 1 & y \\\ 0 & 1 \\\ \end{matrix} \right]=\left[ \begin{matrix} 7 & 0 \\\ 10 & 5 \\\ \end{matrix} \right]$ on matrices. Find the value of x & y.

Answer 1

- Hint:Any two matrices can be added by adding the corresponding elements of both matrices. And if two matrices of the same order will be equal if their corresponding elements are equal. Use these properties of the matrix to solve the given expression.

Complete step-by-step solution -

We have $2\left[ \begin{matrix} 3 & 4 \\\ 5 & x \\\ \end{matrix} \right]+\left[ \begin{matrix} 1 & y \\\ 0 & 1 \\\ \end{matrix} \right]=\left[ \begin{matrix} 7 & 0 \\\ 10 & 5 \\\ \end{matrix} \right]..............(i)$
Here, we need to determine values of x and y.
So, we can observe that all the matrices involved in equation (i) are of same order i.e. $\left( 2\times 2 \right)$ . So, we can add both the matrices in the L.H.S. of the equation and hence, equate the result to the matrix given in R.H.S. of the equation.
As we know, multiplying of any constant/variable values with any matrix, will multiply to all the elements inside of that matrix. So, we can multiply ‘2’ in the first matrix i.e. with all of it’s elements.
Hence, we can get equation (i) as,

& \left[ \begin{matrix} 2\times 3 & 2\times 8 \\\ 2\times 5 & 2\times x \\\ \end{matrix} \right]+\left[ \begin{matrix} 1 & y \\\ 0 & 1 \\\ \end{matrix} \right]=\left[ \begin{matrix} 7 & 0 \\\ 10 & 5 \\\ \end{matrix} \right] \\\ & \Rightarrow \left[ \begin{matrix} 6 & 8 \\\ 10 & 2x \\\ \end{matrix} \right]+\left[ \begin{matrix} 1 & y \\\ 0 & 1 \\\ \end{matrix} \right]=\left[ \begin{matrix} 7 & 0 \\\ 10 & 5 \\\ \end{matrix} \right]\ldots \ldots \left( ii \right) \\\ \end{aligned}$$ Now, we know that addition of two matrices with the same order can be done by adding the each element of one matrix to the corresponding (same position) element of another matrix. So, it suggests that the adding process of two matrices can be given as, $\left[ \begin{matrix} m & n \\\ p & q \\\ \end{matrix} \right]+\left[ \begin{matrix} a & c \\\ b & d \\\ \end{matrix} \right]=\left[ \begin{matrix} m+a & n+c \\\ p+b & q+d \\\ \end{matrix} \right]\ldots \ldots \left( iii \right)$ Now, we can add both the matrices of the left hand side of the equation (ii), with the same approach as described in equation (iii). So, we can get equation (ii) as, $\begin{aligned} & \left[ \begin{matrix} 6 & 8 \\\ 10 & 2x \\\ \end{matrix} \right]+\left[ \begin{matrix} 1 & y \\\ 0 & 1 \\\ \end{matrix} \right]=\left[ \begin{matrix} 7 & 0 \\\ 10 & 5 \\\ \end{matrix} \right] \\\ & \Rightarrow \left[ \begin{matrix} 6+1 & 8+y \\\ 10+0 & 2x+1 \\\ \end{matrix} \right]=\left[ \begin{matrix} 7 & 0 \\\ 10 & 5 \\\ \end{matrix} \right] \\\ \end{aligned}$ $\left[ \begin{matrix} 7 & 8+y \\\ 10 & 2x+1 \\\ \end{matrix} \right]=\left[ \begin{matrix} 7 & 0 \\\ 10 & 5 \\\ \end{matrix} \right]\ldots \ldots \left( iv \right)$ As we know two or more matrices can be equal if all of its corresponding elements belonging to them are equal. It means if we have two equal matrices as, $\left[ \begin{matrix} m & n \\\ p & q \\\ \end{matrix} \right]=\left[ \begin{matrix} a & c \\\ b & d \\\ \end{matrix} \right]$ Then, we get, $m=a,n=c,p=b,q=d.$ Hence, we can equate the corresponding elements of equation (iv) and hence, we get $7=7,8+y=0,10=10,2x+1=5$ So, we get equation of ‘x’ and ‘y’ as, $8+y=0$ and $2x+1=5$ $y=-8$ and $2x=4$ $y=-8$ and $x=2$ _Hence, values of x and y are 2,-8 respectively._ Note: Another for the question would be that we can transfer one of the matrices in LHS to the RHS as well. Do multiply all terms in the first matrix by 2. Be careful with this step. One may go wrong if he/she solve problem in a way as $2\left( 3x-20 \right)+\left( 1-0 \right)=\left( 35-0 \right)$ , which is done with the help of expansion of determinant, so, be clear that matrix do not follow the expansion rules of determinant. Matrix cannot be solved to get one single value of it. So, be clear with the matrix and Determinant property.