For what value of x, the matrix 
is singular?
Hence 
=(5-x)×4-(x+1)×2 (Expanding along R1)
|A|=18-6x
For A to be a singular matrix, |A| has to be 0.
Therefore, 18-6x=0 or x=3.
1
For what value of x, the matrix is singular?
Hence
=(5-x)×4-(x+1)×2 (Expanding along R1)
|A|=18-6x
For A to be a singular matrix, |A| has to be 0.
Therefore, 18-6x=0 or x=3.