Question

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

• A. 0.0
• B. 1
• C. $-1$
• D. 2

Answer 1

The correct option is C $-1$

$AB=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}a+2c& b+2d\\ 3a+4c& 3b+4d\end{array}\right]BA=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}a+3b& 2a+4b\\ c+3d& 2c+4d\end{array}\right]$
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.
$\frac{a-d}{3b-c}=\frac{-\frac{3}{2}b}{3b-\frac{3}{2}b}=-1$