Thank you! Your method gives the correct result. However, when I change some parameters to get a new matrix, the solution for x becomes a complex transcendental equation, and it is not possible to get a solution for x. How do I find the solution X to the system of linear equations in this case? The parameters also satisfy -1 < a < 1, -0.5 < b < 0.5, c > 0, f ≠ 0, and 0 < x < 1.
Here is the new matrix:
mat={{(-2 + x) Cos[1/2 \[Pi] (-2 + x)] -
f (-2 + x) Sin[1/2 \[Pi] (-2 + x)], -f (-2 + x) Cos[
1/2 \[Pi] (-2 + x)] + (2 - x) Sin[1/2 \[Pi] (-2 + x)],
x Cos[(\[Pi] x)/2] -
f (-2 + x) Sin[(\[Pi] x)/2], -f (-2 + x) Cos[(\[Pi] x)/2] -
x Sin[(\[Pi] x)/2], 0, 0, 0,
0}, {(-2 + x) Cos[1/2 \[Pi] (-2 + x)], (2 - x) Sin[
1/2 \[Pi] (-2 + x)],
x Cos[(\[Pi] x)/2], -x Sin[(\[Pi] x)/2], (2 - x) Cos[
3/2 \[Pi] (-2 + x)], (2 - x) Sin[3/2 \[Pi] (-2 + x)], -x Cos[(
3 \[Pi] x)/2], -x Sin[(3 \[Pi] x)/2]}, {1/
2 (-2 + a c - b c) (-2 + x) Cos[1/2 \[Pi] (-2 + x)],
1/2 (2 - a c + b c) (-2 + x) Sin[1/2 \[Pi] (-2 + x)],
1/2 (-2 + a c - b c) (-1 - (-2 + c - b c)/(2 - a c + b c) +
x) Cos[(\[Pi] x)/2],
1/2 (2 - a c + b c) (-1 - (-2 + c - b c)/(2 - a c + b c) +
x) Sin[(\[Pi] x)/2], (-2 + x) Cos[
3/2 \[Pi] (-2 + x)], (-2 + x) Sin[
3/2 \[Pi] (-2 + x)], (-1 + 1/2 (2 - c - a c) + x) Cos[(3 \[Pi] x)/
2], (-1 + 1/2 (2 - c - a c) + x) Sin[(3 \[Pi] x)/2]}, {Sin[
1/2 \[Pi] (-2 + x)], Cos[1/2 \[Pi] (-2 + x)], Sin[(\[Pi] x)/2],
Cos[(\[Pi] x)/2], Sin[3/2 \[Pi] (-2 + x)], -Cos[3/2 \[Pi] (-2 + x)],
Sin[(3 \[Pi] x)/2], -Cos[(3 \[Pi] x)/2]}, {0, 1, 0, 1, 0, -1,
0, -1}, {-2 + x, 0, x, 0, 2 - x, 0, -x, 0}, {0,
1/2 (2 - a c + b c) (-2 + x), 0,
1/2 (2 - a c + b c) (-1 + (-2 + c - b c)/(2 - a c + b c) + x), 0,
2 - x, 0, 1 + 1/2 (2 - c - a c) - x}, {1/2 (2 - a c + b c) (-2 + x),
0, 1/2 (2 - a c + b c) (-1 - (-2 + c - b c)/(2 - a c + b c) + x),
0, 2 - x, 0, 1 + 1/2 (-2 + c + a c) - x, 0}}
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