HOMEWORK: Need assistance inputing Mesh Analysis complex equations into TI89
Compare each of your loop equations to what I show below.
$$\text{Bottom left loop: }\text{ }-(-6j) + (i_1-i_3)(-\frac{2}{3}j) +  (i_1-i_2)=0 $$
$$\text{Upper left loop: }\text{ }5i_1+(i_3-i_4)(4j)+(i_3-i_1)(-\frac{2}{3}j)=0 $$
$$\text{Upper right loop: }\text{ }(i_4-i_2)(-\frac{1}{2}j)+(i_4-i_3)(4j)=0 $$
$$\text{Bottom right loop: }\text{ }9+(i_2-i_1)+(i_2-i_4)(-\frac{1}{2}j)=0 $$
Now, collecting terms and writing in matrix form,
$$Ax=b$$
$$\begin{bmatrix}(1-\frac{2}{3}j)&amp;-1&amp;\frac{2}{3}j&amp;0\\(5+\frac{2}{3}j)&amp;0&amp;3\frac{1}{3}j&amp;-4j\\0&amp;\frac{1}{2}j&amp;-4j&amp;3\frac{1}{2}j\\-1&amp;(1-\frac{1}{2}j)&amp;0&amp;\frac{1}{2}j\end{bmatrix}\begin{bmatrix}i_1\\i_2\\i_3\\i_4\end{bmatrix}=\begin{bmatrix}-6j\\0\\0\\-9\\\end{bmatrix}$$
Then solve however you want (e.g. in your TI-89).  I used this nice online complex capable simultaneous equation solver.  The results are,
$$\begin{bmatrix}i_1\\i_2\\i_3\\i_4\end{bmatrix}=\begin{bmatrix}-1.8-1.2j\\-7-0.033j\\-9.05+6.608j\\-9.34+7.556j\end{bmatrix}\text{amps}$$
And, thanks to that handy 1Ω resistor we can easily find Vx,
$$V_x=(i_1-i_2)(1Ω)=5.33\angle-12.65⁰\text{ V}$$
The above is a phasor which we can convert to,
$$V_x(t)=5.33 cos(2000t-12.65⁰)\text{ V}$$
p.s.  I recommend you take the time to enter your work in MathJax for 2 reasons.  It is much easier for people to read (vs. a camera snap) and it gives you a chance to slow down and check your work. This neat tool lets you draw symbols and it tells you the corresponding Latex code.