Given system of equations can be written as
AX=B
where A=⎡⎢⎣1111−1121−1⎤⎥⎦
X=⎡⎢⎣xyz⎤⎥⎦ ;B=⎡⎢⎣621⎤⎥⎦
Here, |A|=6≠0
Hence, A−1 exists.
Now, adjA=CT=⎡⎢⎣0332−3120−2⎤⎥⎦T
⇒adjA=⎡⎢⎣0223−3031−2⎤⎥⎦
A−1=adjA|A|
⇒A−1=16⎡⎢⎣0223−3031−2⎤⎥⎦
Solution to the system is given by
X=A−1B
⇒⎡⎢⎣xyz⎤⎥⎦=16⎡⎢⎣0223−3031−2⎤⎥⎦⎡⎢⎣621⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=16⎡⎢⎣61218⎤⎥⎦
⇒x=1,y=2,z=3
Comparing with given values , we get
k=1;m=2;n=3
n−(k×m)=3−2=1