Find the matrix X for which:
.
Let A =
B =
C = ![]()
Then The given equations becomes as
AXB = C
= X = A – 1CB – 1
|A| = 35 – 14 = 21
|B| = – 1 + 2 = 1
A – 1 = ![]()
B – 1 = ![]()
= X = A – 1CB – 1 = ![]()
= ![]()
= ![]()
= ![]()
Hence, X = ![]()
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