प्रारम्भिक स्तम्भ संक्रिया `C_(2) rarr C_(2)-2C_(1)` का प्रयोग आव्यूह समीकरण `[(1,3),(2,4)]=[(1,1),(0,1)][(3,1),(2,4)]` में करने पर हमें प्राप्त होता है :

On using elementary operation C_(2) to C_(2) - 2C_(1) in the following matrix equation : [ (1,-3),(2,4)] = [ (1,1),(0,1)] [ (3,1),(2,4)] we have :

Use the elementary column operations C_(2) to C_(2)-2C_(1) in the matrix equation [{:(1,-3),(2,4):}]=[{:(1,-1),(0,1):}][{:(3,1),(2,4):}]

On usign elementry column operation C_(2)rArrC_(2)-2C_(1) in the following matrix equation [{:(1,-3),(2,4):}]=[{:(1,01),(0,1):}][{:(3,1),(2,4):}] , we have

On usign elementry column operation C_(2)rArrC_(2)-2C_(1) in the following matrix equation [{:(1,-3),(2,4):}]=[{:(1,01),(0,1):}][{:(3,1),(2,4):}] , we have

On using elementary column operations C_(2)toC_(2)-2C_(1) in the following matrix equation [{:(1,-3),(2,4):}]=[{:(1,-1),(0,1):}]*[{:(3,1),(2,4):}] we have :

[(1,3,-2),(-3,0,-5),(2,5,0)] =A [(1,0,0),(0,1,0),(0,0,1)] then, C_(2) rarr C_(2)-3C_(1) and C_(3) rarr C_(3) +2C_(1) gives... [(1,0,0),(3,-9,11),(-2,1,4)] =A [(-1,3,2),(0,-1,0),(0,0,-1)] [(1,0,0),(-3,-1,-4),(2,-9,11)] =A [(1,-3,2),(0,1,0),(0,0,1)] [(1,0,0),(-3,1,4),(-2,9,-11)] =A [(-1,3,2),(0,-1,0),(0,0,-1)] [(1,0,0),(-3,9,-11),(2,-1,4)] =A [(1,-3,2),(0,1,0),(0,0,1)]

If A = {:[(1,2),(0,1)], B = [(1,-3),(2,4)] and C = [(1,-1),(-1,2)] then prove that : BB' = 10C

If A=[(4,1,0),(1,-2,2)],B=[(2,0,-1),(3,1,4)],C=[(1),(2),(-1)] and (3B-2A)C+2X=O then X=

If A = {:[(1,2),(0,1)], B = [(1,-3),(2,4)] and C = [(1,-1),(-1,2)] then prove that : A(BC) = (AB)C

If A = {:[(1,2),(0,1)], B = [(1,-3),(2,4)] and C = [(1,-1),(-1,2)] then prove that : A(B+C) = AB+AC