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 $ad-bc$.

Answer 1

We start solving this problem by first multiplying the matrices A and B. Then we equate the result to matrix B as we are given that AB=B. Then we equate the corresponding elements in the both matrices and then we get two equations with variables p and q. Solving them we get an equation with a, b, c and d. Then by substituting the value of $a+d$ given and solving it we cam find the value of $ad-bc$.

Complete step by step answer:
We are given that $A=\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} p \\\ q \\\ \end{matrix} \right]\ne \left[ \begin{matrix} 0 \\\ 0 \\\ \end{matrix} \right]$.
We are also given that AB=B and $a+d=2$.
As we are given that AB=B, let us multiply the matrices A and B and then equate the obtained result to B.
So, let us now consider the product AB.
$\begin{aligned} & \Rightarrow AB=\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]\left[ \begin{matrix} p \\\ q \\\ \end{matrix} \right] \\\ & \Rightarrow AB=\left[ \begin{matrix} ap+bq \\\ cp+dq \\\ \end{matrix} \right] \\\ \end{aligned}$
Now let us equate it to matrix B. Then we get,
$\Rightarrow \left[ \begin{matrix} ap+bq \\\ cp+dq \\\ \end{matrix} \right]=\left[ \begin{matrix} p \\\ q \\\ \end{matrix} \right]$
So, now let us equate the first element in the both matrices.

& \Rightarrow ap+bq=p \\\ & \Rightarrow bq=\left( 1-a \right)p \\\ & \Rightarrow q=\dfrac{\left( 1-a \right)p}{b}.........\left( 1 \right) \\\ \end{aligned}$$ Now let us equate the second term in the above matrices. Then we get, $$\begin{aligned} & \Rightarrow cp+dq=q \\\ & \Rightarrow \left( 1-d \right)q=cp \\\ & \Rightarrow q=\dfrac{cp}{1-d}.........\left( 2 \right) \\\ \end{aligned}$$ Now from equations (1) and (2) we get, $$\begin{aligned} & \Rightarrow \dfrac{\left( 1-a \right)p}{b}=\dfrac{cp}{1-d} \\\ & \Rightarrow \left( 1-a \right)\left( 1-d \right)p=bcp \\\ & \Rightarrow \left( 1-d-a+ad \right)p=bcp \\\ & \Rightarrow \left( 1-\left( d+a \right)+ad-bc \right)p=0 \\\ \end{aligned}$$ As, we are given that $p\ne 0$, we get, $$\Rightarrow 1-\left( a+d \right)+ad-bc=0$$ We are given that $a+d=2$, so let us substitute it in the above equation. Then we get, $$\begin{aligned} & \Rightarrow 1-2+ad-bc=0 \\\ & \Rightarrow -1+ad-bc=0 \\\ & \Rightarrow ad-bc=1 \\\ \end{aligned}$$ So, we get the value of $\left( ad-bc \right)$ as 1. **So, the correct answer is “1”.** **Note:** We can also solve this question by writing p in terms of q in equations (1) and (2) and then solving it as below. Writing p in terms of q we get the equations (1) and (2) as, $$\Rightarrow p=\dfrac{bq}{1-a}.........\left( 3 \right)$$ $$\Rightarrow p=\dfrac{\left( 1-d \right)q}{c}..........\left( 4 \right)$$ Equating them we get, $$\begin{aligned} & \Rightarrow \dfrac{bq}{1-a}=\dfrac{\left( 1-d \right)q}{c} \\\ & \Rightarrow cbq=\left( 1-a \right)\left( 1-d \right)q \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow cbq=\left( 1-a-d+ad \right)q \\\ & \Rightarrow \left( 1-a-d+ad-bc \right)q=0 \\\ \end{aligned}$$ As $q\ne 0$, we get, $$\Rightarrow 1-\left( a+d \right)+ad-bc=0$$ We are given that $a+d=2$, so let us substitute it in the above equation. Then we get, $$\begin{aligned} & \Rightarrow 1-2+ad-bc=0 \\\ & \Rightarrow -1+ad-bc=0 \\\ & \Rightarrow ad-bc=1 \\\ \end{aligned}$$ So, we get the value of $\left( ad-bc \right)$ as 1.