Question

If we have a matrix as $A=\left(\begin{matrix} 0 & 2q & r \\\ p & q & -r \\\ p & -q & r \\\ \end{matrix} \right)$ . If $A{{A}^{T}}={{I}_{3}}$, then $\left| p \right|$ is equal to

Answer 1

If the rows and columns of a matrix A are interchanged, then that matrix is defined as the transpose of matrix A. The transpose of a matrix A is defined as ${{A}^{T}}$. 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 is equal and the resultant matrix C is an order of $n\times p$. 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. If two matrices 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 $\left| p \right|$.

Complete step-by-step solution:
Let us consider

0 & 2q & r \\\ p & q & -r \\\ p & -q & r \\\ \end{matrix} \right).....(1)$$ Now we should find the transpose of matrix A. If the rows and columns of a matrix A are interchanged, then that matrix is defined as the transpose of matrix A. The transpose of a matrix A is defined as $${{A}^{T}}$$. Now by using this concept, we should find the transpose of matrix A. From the definition of transpose of matrix A, we can say that $${{A}^{T}}=\left( \begin{matrix} 0 & p & p \\\ 2q & q & -q \\\ r & -r & r \\\ \end{matrix} \right)$$. Let us consider $${{A}^{T}}=\left( \begin{matrix} 0 & p & p \\\ 2q & q & -q \\\ r & -r & r \\\ \end{matrix} \right).....(2)$$ From the question, we were given that $$A{{A}^{T}}={{I}_{3}}$$. Let us consider $$A{{A}^{T}}={{I}_{3}}.....(3)$$ Now let us substitute equation (1) and equation (2) in equation (3), then we get $$\begin{aligned} & \Rightarrow A{{A}^{T}}=\left( \begin{matrix} 0 & 2q & r \\\ p & q & -r \\\ p & -q & r \\\ \end{matrix} \right)\left( \begin{matrix} 0 & p & p \\\ 2q & q & -q \\\ r & -r & r \\\ \end{matrix} \right) \\\ & \Rightarrow {{I}_{3}}=\left( \begin{matrix} 0 & 2q & r \\\ p & q & -r \\\ p & -q & r \\\ \end{matrix} \right)\left( \begin{matrix} 0 & p & p \\\ 2q & q & -q \\\ r & -r & r \\\ \end{matrix} \right) \\\ \end{aligned}$$ Now we should know the multiplication rule of matrices. According to multiplication rule of matrices, it will be clear how two matrices are multiplied. If a matrix A of order $$m\times n$$and a matrix B of order $$p\times q$$ can be multiplied if the value of n and value of p are equal and the resultant matrix C is an order of $$n\times p$$. While multiplying two matrices, to have an element of $${{i}^{th}}$$ row and $${{j}^{th}}$$ column, we should multiply $${{i}^{th}}$$ row of first matrix with $${{j}^{th}}$$ column of second matrix. In this way, two matrices are multiplied. As both the matrices are of order $$3\times 3$$, then the obtained matrix is also of a matrix of order $$3\times 3$$. $$\Rightarrow {{I}_{3}}=\left( \begin{matrix} (0)(0)+(2q)(2q)+(r)(r) & (0)(p)+(2q)(q)+(r)(-r) & (p)(0)+(-q)(2q)+(r)(r) \\\ (p)(0)+(q)(2q)+(-r)(r) & (p)(p)+(q)(q)+(-r)(-r) & (p)(p)+(-q)(q)+(r)(-r) \\\ (0)(p)+(p)(-q)+(r)(r) & (p)(p)+(-q)(q)+(r)(-r) & (p)(p)+(-q)(-q)+(r)(r) \\\ \end{matrix} \right)$$ $$\Rightarrow {{I}_{3}}=\left( \begin{matrix} 4{{q}^{2}}+{{r}^{2}} & 2{{q}^{2}}-{{r}^{2}} & -2{{q}^{2}}+{{r}^{2}} \\\ 2{{q}^{2}}-{{r}^{2}} & {{p}^{2}}+{{q}^{2}}+{{r}^{2}} & {{p}^{2}}-{{q}^{2}}-{{r}^{2}} \\\ -pq+{{r}^{2}} & {{p}^{2}}-{{q}^{2}}-{{r}^{2}} & {{p}^{2}}+{{q}^{2}}+{{r}^{2}} \\\ \end{matrix} \right)$$ $$\Rightarrow \left( \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right)=\left( \begin{matrix} 4{{q}^{2}}+{{r}^{2}} & 2{{q}^{2}}-{{r}^{2}} & -2{{q}^{2}}+{{r}^{2}} \\\ 2{{q}^{2}}-{{r}^{2}} & {{p}^{2}}+{{q}^{2}}+{{r}^{2}} & {{p}^{2}}-{{q}^{2}}-{{r}^{2}} \\\ -pq+{{r}^{2}} & {{p}^{2}}-{{q}^{2}}-{{r}^{2}} & {{p}^{2}}+{{q}^{2}}+{{r}^{2}} \\\ \end{matrix} \right)....(4)$$ We know that if two matrices are said to be equal, if each and every element in the matrix are equal. Now by using this concept, let us consider $$\begin{aligned} & 4{{q}^{2}}+{{r}^{2}}=1.....(5) \\\ & 2{{q}^{2}}-{{r}^{2}}=0......(6) \\\ & {{p}^{2}}+{{q}^{2}}+{{r}^{2}}=1.....(7) \\\ \end{aligned}$$ Now let us subtract equation (5) and equation (6), then we get $$\begin{aligned} & \Rightarrow \left( 4{{q}^{2}}+{{r}^{2}} \right)-\left( 2{{q}^{2}}-{{r}^{2}} \right)=1-0 \\\ & \Rightarrow 2{{q}^{2}}+2{{r}^{2}}=1 \\\ & \Rightarrow 2\left( {{q}^{2}}+{{r}^{2}} \right)=1 \\\ \end{aligned}$$ Now by using cross multiplication, we get $$\Rightarrow {{q}^{2}}+{{r}^{2}}=\dfrac{1}{2}....(8)$$ Now let us substitute equation (8) in equation (7), then we get $$\begin{aligned} & \Rightarrow {{p}^{2}}+\dfrac{1}{2}=1 \\\ & \Rightarrow {{p}^{2}}=\dfrac{1}{2} \\\ & \Rightarrow p=\pm \sqrt{\dfrac{1}{2}}....(9) \\\ \end{aligned}$$ From equation (9), it is clear that the value of p is equal to $$\pm \sqrt{\dfrac{1}{2}}$$. From the question, it is clear we should find the value of $$\left| p \right|$$. We know that the definition of modulus function. So, the modulus function of number x is defined as follows: $$\left| x \right|=\left\\{ \begin{aligned} & x,x>0 \\\ & -x,x<0 \\\ \end{aligned} \right.$$. So, now by using the concept of modulus function, we should find the value of $$\left| p \right|$$. $$\begin{aligned} & \Rightarrow \left| p \right|=\left| \dfrac{\pm 1}{\sqrt{2}} \right| \\\ & \Rightarrow \left| p \right|=\dfrac{1}{\sqrt{2}}....(10) \\\ \end{aligned}$$ **Hence, it is clear that option A is correct.** **Note:** Students may have a misconception that While multiplying two matrices, to have an element of $${{i}^{th}}$$ row and $${{j}^{th}}$$ column, we should multiply $${{i}^{th}}$$ column of first matrix with $${{j}^{th}}$$ row of second matrix. If this misconception is followed, then we cannot the correct answer. So, this misconception should be avoided