3[xz​yt​]=[x−1​62t​]+[4z+t​x+y3​][3x3z​3y3t​]=[x+4−1+z+t​6+x+y2t+3​] On comparing two matrices are equal if their corresponding elements are equal 3x=x+4⇒x=23t=2t+3⇒t=33z=−1+z+t⇒3z=−1+z+3⇒z=13y=6+x+y⇒ 2y=6+2⇒y=4∴(x,y,z,t)=(2,4,1,3)