Question

Solve the following matrix equation for $\left[ \begin{matrix} x & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\\ -2 & 0 \\\ \end{matrix} \right]=0$.

Answer 1

According to the multiplication rule of matrices, it will be clear how two matrices are multiplied. If a matrix A of order $m\times n$ and a matrix B of order $p\times q$ can be multiplied if the value of n and value of p are equal and the resultant matrix C is an order of $m\times q$. While multiplying two matrices, to have an element of ${{i}^{th}}$ row and ${{j}^{th}}$ column, we should multiply ${{i}^{th}}$ row of first matrix with ${{j}^{th}}$ column of second matrix. In this way, two matrices are multiplied. We know that if two matrices $\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} \\\ {{a}_{21}} & {{a}_{22}} \\\ \end{matrix} \right]$ and $\left[ \begin{matrix} {{b}_{11}} & {{b}_{12}} \\\ {{b}_{21}} & {{b}_{22}} \\\ \end{matrix} \right]$ are said to be equal, if each and every element in the matrix are equal. By using these concepts, we can find the value of x in matrix equation $\left[ \begin{matrix} x & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\\ -2 & 0 \\\ \end{matrix} \right]=0$.

Complete step-by-step solution
Before solving the problem, we should know how matrices are multiplied. If a matrix A of order $m\times n$ and a matrix B of order $p\times q$ can be multiplied if the value of n and value of p is equal and the resultant matrix C is an order of $m\times q$.
From the question, it is clear that we should find matrix $\left[ \begin{matrix} x & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\\ -2 & 0 \\\ \end{matrix} \right]$. It is clear that the matrix $\left[ \begin{matrix} x & 1 \\\ \end{matrix} \right]$ is an order of $1\times 2$. We can also say that the order of matrix $\left[ \begin{matrix} 1 & 0 \\\ -2 & 0 \\\ \end{matrix} \right]$ is $2\times 2$. Then the product of $\left[ \begin{matrix} x & 1 \\\ \end{matrix} \right]$ and $\left[ \begin{matrix} 1 & 0 \\\ -2 & 0 \\\ \end{matrix} \right]$. Let us compare $m\times n$ with $1\times 2$ and $p\times q$ with $2\times 2$. Then we get

& m=1....(1) \\\ & n=2......(2) \\\ & p=2.....(3) \\\ & q=2.......(4) \\\ \end{aligned}$$ From equation (1) and equation (4), it is clear that the value of m and value of q are equal. So, we can say that the order of product of matrix $$\left[ \begin{matrix} x & 1 \\\ \end{matrix} \right]$$ and matrix $$\left[ \begin{matrix} 1 & 0 \\\ -2 & 0 \\\ \end{matrix} \right]$$ is equal to $$1\times 2$$. While multiplying two matrices, to have an element of $${{i}^{th}}$$ row and $${{j}^{th}}$$ column, we should multiply $${{i}^{th}}$$ row of first matrix with $${{j}^{th}}$$ column of second matrix. Now by using this concept, we will find the product of matrix $$\left[ \begin{matrix} x & 1 \\\ \end{matrix} \right]$$ and matrix $$\left[ \begin{matrix} 1 & 0 \\\ -2 & 0 \\\ \end{matrix} \right]$$ $$\begin{aligned} & \Rightarrow \left[ \begin{matrix} x & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\\ -2 & 0 \\\ \end{matrix} \right]=\left[ \begin{matrix} x.1-1.2 & x.0+1.0 \\\ \end{matrix} \right] \\\ & \Rightarrow \left[ \begin{matrix} x & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\\ -2 & 0 \\\ \end{matrix} \right]=\left[ \begin{matrix} x-2 & 0 \\\ \end{matrix} \right].......(5) \\\ \end{aligned}$$ So, from equation (5) we can say that the product of matrix $$\left[ \begin{matrix} x & 1 \\\ \end{matrix} \right]$$ and matrix $$\left[ \begin{matrix} 1 & 0 \\\ -2 & 0 \\\ \end{matrix} \right]$$ is equal to $$\left[ \begin{matrix} x-2 & 0 \\\ \end{matrix} \right]$$. We know that if two matrices $$\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} \\\ {{a}_{21}} & {{a}_{22}} \\\ \end{matrix} \right]$$ and $$\left[ \begin{matrix} {{b}_{11}} & {{b}_{12}} \\\ {{b}_{21}} & {{b}_{22}} \\\ \end{matrix} \right]$$ are said to be equal, if each and every element in the matrix are equal. So, we can say that $$\begin{aligned} & {{a}_{11}}={{b}_{11}} \\\ & {{a}_{12}}={{b}_{12}} \\\ & {{a}_{21}}={{b}_{21}} \\\ & {{a}_{22}}={{b}_{22}} \\\ \end{aligned}$$ By using this concept, we should equate the matrix $$\left[ \begin{matrix} x-2 & 0 \\\ \end{matrix} \right]$$ and null matrix. In a null matrix, all the elements are equal to zero. So, it is clear that $$\begin{aligned} & \Rightarrow x-2=0 \\\ & \Rightarrow x=2....(6) \\\ \end{aligned}$$ So, from equation (6), the value of x is equal to 2. **Note:** Students may have a misconception that we can equate the elements of two matrices if both the matrices have a different order. But this misconception is wrong. So, students should have a clear view of this concept. Some students may not know that all the elements of the null matrix are equal to zero. If we did not know this point, then we cannot solve this problem.