Solve x y z and t in the matrix equation below [3x t+
Given,
[3xy−xt+z2t−z]=[31723]\begin{bmatrix} 3x& y-x \\t+\dfrac{z}{2}& t-z\end{bmatrix} =\begin{bmatrix} 3& 1\\\dfrac{7}{2}& 3\end{bmatrix}[3xt+2zy−xt−z]=[32713]
On comparing we get-
3x=3⇒x=13x=3\Rightarrow x=13x=3⇒x=1
Also, y−x=1⇒y=1+1⇒y=2y-x=1\Rightarrow y=1+1\Rightarrow y=2y−x=1⇒y=1+1⇒y=2
Also,
t+z2=72 −(1)t+\dfrac{z}{2}=\dfrac{7}{2}~~~~~~~~~~~~~-(1)t+2z=27 −(1)
t−z=3 −(2)t-z=3~~~~~~~~-(2)t−z=3 −(2)
Solving eqn.(1) and (2) and we get-
z=13,t=103z=\dfrac{1}{3} , t=\dfrac{10}{3}z=31,t=310
Hence Value of x=1,y=2,z=13x=1,y=2,z=\dfrac{1}{3}x=1,y=2,z=31 and t=103t=\dfrac{10}{3}t=310
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