Answer
414.9k+ views
Hint: 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.
$\begin{align}
& \Rightarrow AB=\left[ \begin{matrix}
a & b \\
c & d \\
\end{matrix} \right]\left[ \begin{matrix}
p \\
q \\
\end{matrix} \right] \\
& \Rightarrow AB=\left[ \begin{matrix}
ap+bq \\
cp+dq \\
\end{matrix} \right] \\
\end{align}$
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.
\[\begin{align}
& \Rightarrow ap+bq=p \\
& \Rightarrow bq=\left( 1-a \right)p \\
& \Rightarrow q=\dfrac{\left( 1-a \right)p}{b}.........\left( 1 \right) \\
\end{align}\]
Now let us equate the second term in the above matrices. Then we get,
\[\begin{align}
& \Rightarrow cp+dq=q \\
& \Rightarrow \left( 1-d \right)q=cp \\
& \Rightarrow q=\dfrac{cp}{1-d}.........\left( 2 \right) \\
\end{align}\]
Now from equations (1) and (2) we get,
\[\begin{align}
& \Rightarrow \dfrac{\left( 1-a \right)p}{b}=\dfrac{cp}{1-d} \\
& \Rightarrow \left( 1-a \right)\left( 1-d \right)p=bcp \\
& \Rightarrow \left( 1-d-a+ad \right)p=bcp \\
& \Rightarrow \left( 1-\left( d+a \right)+ad-bc \right)p=0 \\
\end{align}\]
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,
\[\begin{align}
& \Rightarrow 1-2+ad-bc=0 \\
& \Rightarrow -1+ad-bc=0 \\
& \Rightarrow ad-bc=1 \\
\end{align}\]
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,
\[\begin{align}
& \Rightarrow \dfrac{bq}{1-a}=\dfrac{\left( 1-d \right)q}{c} \\
& \Rightarrow cbq=\left( 1-a \right)\left( 1-d \right)q \\
\end{align}\]
\[\begin{align}
& \Rightarrow cbq=\left( 1-a-d+ad \right)q \\
& \Rightarrow \left( 1-a-d+ad-bc \right)q=0 \\
\end{align}\]
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,
\[\begin{align}
& \Rightarrow 1-2+ad-bc=0 \\
& \Rightarrow -1+ad-bc=0 \\
& \Rightarrow ad-bc=1 \\
\end{align}\]
So, we get the value of $\left( ad-bc \right)$ as 1.
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.
$\begin{align}
& \Rightarrow AB=\left[ \begin{matrix}
a & b \\
c & d \\
\end{matrix} \right]\left[ \begin{matrix}
p \\
q \\
\end{matrix} \right] \\
& \Rightarrow AB=\left[ \begin{matrix}
ap+bq \\
cp+dq \\
\end{matrix} \right] \\
\end{align}$
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.
\[\begin{align}
& \Rightarrow ap+bq=p \\
& \Rightarrow bq=\left( 1-a \right)p \\
& \Rightarrow q=\dfrac{\left( 1-a \right)p}{b}.........\left( 1 \right) \\
\end{align}\]
Now let us equate the second term in the above matrices. Then we get,
\[\begin{align}
& \Rightarrow cp+dq=q \\
& \Rightarrow \left( 1-d \right)q=cp \\
& \Rightarrow q=\dfrac{cp}{1-d}.........\left( 2 \right) \\
\end{align}\]
Now from equations (1) and (2) we get,
\[\begin{align}
& \Rightarrow \dfrac{\left( 1-a \right)p}{b}=\dfrac{cp}{1-d} \\
& \Rightarrow \left( 1-a \right)\left( 1-d \right)p=bcp \\
& \Rightarrow \left( 1-d-a+ad \right)p=bcp \\
& \Rightarrow \left( 1-\left( d+a \right)+ad-bc \right)p=0 \\
\end{align}\]
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,
\[\begin{align}
& \Rightarrow 1-2+ad-bc=0 \\
& \Rightarrow -1+ad-bc=0 \\
& \Rightarrow ad-bc=1 \\
\end{align}\]
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,
\[\begin{align}
& \Rightarrow \dfrac{bq}{1-a}=\dfrac{\left( 1-d \right)q}{c} \\
& \Rightarrow cbq=\left( 1-a \right)\left( 1-d \right)q \\
\end{align}\]
\[\begin{align}
& \Rightarrow cbq=\left( 1-a-d+ad \right)q \\
& \Rightarrow \left( 1-a-d+ad-bc \right)q=0 \\
\end{align}\]
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,
\[\begin{align}
& \Rightarrow 1-2+ad-bc=0 \\
& \Rightarrow -1+ad-bc=0 \\
& \Rightarrow ad-bc=1 \\
\end{align}\]
So, we get the value of $\left( ad-bc \right)$ as 1.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)