Solve the equation of x,y, z and t, if 2[xzyt]+3[1−102]=3[3546]
Given, 2[xzyt]+3[1−102]=3[3546]⇒[2x2z2y2t]+[3−306]=[9151218]⇒[2x+32z−32y+02t+6]=[9151218]
By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get
2x+3 =9, 2y+0=12, 2z-3 =15 and 2t+6=18
⇒x=9−32,y=122,z=15+32 and t=18−62
⇒x=3,y=6,z=9 and t=6.