Question

Let $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ and $B=A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ are two matrices such that AB = BA and $c\ne 0$, then value of $\frac{a-d}{3b-c}$ is :

• A. 0
• B. 2
• C. -2
• D. -1

Answer 1

The correct option is D

-1

$AB=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}a+2c& b+2d\\ 3a+4c& 3b+4d\end{array}\right]BA=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}a+3b& 2a+4b\\ c+3d& 2c+4d\end{array}\right]$
Given that AB = BA,
How do we use the fact that two matrices are equal. In such a case we can equate the values of corresponding elements of both matrices.
First lets equate the element present in row 1 and column 1 of both matrices.
a + 2c = a + 3b------------------(1)
$\Rightarrow$ 2c = 3b $\Rightarrow b\ne 0$(since it is given that c is non zero)
Next we will equate the element present in row 1 and column 2 of both matrices
b + 2d = 2a + 4b
$\Rightarrow$ 2a - 2d = -3b ------------------(2)
Now we will equate the element present in row 2 and column 1 of both matrices
3a+4c=c+3d
$\Rightarrow$ 3a+3c =3d.-----------------------(3)
Solve the three equations (1),(2) and (3) to obtain values of a ,c and d in terms of b. In the required expression substitute the values thus obtained as below.
$\frac{a-d}{3b-c}=\frac{-\frac{3}{2}b}{3b-\frac{3}{2}b}=-1$