Question

Let $A=\left( \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right)$ and $B=\left( \begin{matrix} p \\\ q \\\ \end{matrix} \right)\ne \left( \begin{matrix} 0 \\\ 0 \\\ \end{matrix} \right)$ such that $AB=B$ and $a+d=2$ , then find the value of $\left( ad-bc \right)$.

Answer 1

For answering this question we will use the given information in the question stated as $A=\left( \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right)$,$B=\left( \begin{matrix} p \\\ q \\\ \end{matrix} \right)\ne \left( \begin{matrix} 0 \\\ 0 \\\ \end{matrix} \right)$,$AB=B$ and $a+d=2$. And then multiply ${{A}^{-1}}$ on both sides of $AB=B$ we need to find the determinant of $A$ which will be given as $\left( ad-bc \right)$ .

Complete step-by-step solution:
Now from the question we have $A=\left( \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right)$,$B=\left( \begin{matrix} p \\\ q \\\ \end{matrix} \right)\ne \left( \begin{matrix} 0 \\\ 0 \\\ \end{matrix} \right)$,$AB=B$ and $a+d=2$.
By multiplying ${{A}^{-1}}$ on both sides of the equation we will have
$\begin{aligned} & \text{ }{{A}^{-1}}AB={{A}^{-1}}B \\\ & \Rightarrow B={{A}^{-1}}B \\\ \end{aligned}$
Now, let us solve the given equation $AB=B$
$\begin{aligned} & \left( \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right)\left( \begin{matrix} p \\\ q \\\ \end{matrix} \right)=\left( \begin{matrix} p \\\ q \\\ \end{matrix} \right) \\\ & \Rightarrow \left( \begin{matrix} ap+bq \\\ cp+dq \\\ \end{matrix} \right)=\left( \begin{matrix} p \\\ q \\\ \end{matrix} \right) \\\ \end{aligned}$
From the equations we will have
$ap+bq=p$ …………. (1)
$cp+dq=q$ …………….. (2)
By solving the equation $AB=B$ we get the above two more equations.
And by simplifying the previous equation $\text{B =}{{A}^{-1}}B \text{ using } {{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( \begin{matrix} d & -b \\\ -c & a \\\ \end{matrix} \right)$
Now, let us find the value of $B$, by further solving the equation
$\begin{aligned} & \text{B =}{{A}^{-1}}B \\\ & \Rightarrow \left( \begin{matrix} p \\\ q \\\ \end{matrix} \right)=\dfrac{1}{\left| A \right|}\left( \begin{matrix} d & -b \\\ -c & a \\\ \end{matrix} \right)\left( \begin{matrix} p \\\ q \\\ \end{matrix} \right) \\\ & \Rightarrow \left( \begin{matrix} p \\\ q \\\ \end{matrix} \right)=\dfrac{1}{\left| A \right|}\left( \begin{matrix} dp-bq \\\ -cp+aq \\\ \end{matrix} \right) \\\ \end{aligned}$
Now from this we will have the following equations.
$\left| A \right|p=dp-bq$ ………… (3)
$\left| A \right|q=-cp+aq$ …………… (4)
By further solving we get the above equations.
Now, by adding the equations (1) and (3) we will have
$\begin{aligned} & ap+bq+dp-bq=p+\left| A \right|p \\\ & \Rightarrow ap+dp=p\left( 1+\left| A \right| \right) \\\ & \Rightarrow a+d=1+\left| A \right| \\\ \end{aligned}$
From the question we have $a+d=2$ by using this equation we will have
$\begin{aligned} & \Rightarrow 1=\left| A \right| \\\ & \Rightarrow ad-bc=1 \\\ \end{aligned}$
Hence we can conclude that the value of $ad-bc$ is $1$.

Note: While answering questions of this type we should be sure with the calculations and formulae of the matrices. The inverse of matrix $A=\left( \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right)$ is given as ${{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( \begin{matrix} d & -b \\\ -c & a \\\ \end{matrix} \right)$ because $A{{A}^{-1}}=I$ which can be simply written as $\left( \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right){{A}^{-1}}=\left( \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right)$ .