Supposing the right hand side is another [tex]n\times n[/tex] matrix [tex]M[/tex], we have
[tex]C^{-1}(A+X)B^{-1}=M[/tex]
[tex](CC^{-1})(A+X)B^{-1}=CM[/tex]
[tex](A+X)B^{-1}=CM[/tex]
[tex](A+X)(B^{-1}B)=CMB[/tex]
[tex]A+X=CMB[/tex]
[tex]X=CMB-A[/tex]
Invertible matrices are matrices whose inverse exist. The solution of the given matrix equation exists and is: [tex]X = CB - A[/tex]
An invertible matrix is a matrix whose inverse matrix exists.
If A is a matrix and A⁻¹ is its inverse, then
[tex]A \tiems A^{-1} = I[/tex]
where I is identity matrix of same shape as A was.
The given matrix equation is solved as:
[tex]C^{-1} (A+X) B^{-1} = I_n\\\\\text{Multiplying C matrix on left of both sides, and B on right of both sides}\\\\CC^{-1} (A+X) B^{-1} B = CI_nB\\I_n \times (A+X) \times I_n = CI_nB\\A + X = CB \text{\:\:\:(Property of identity matrix multipication)}\\\\\text{Adding -A matrix on both the sides}\\\\A + X + (-A) = CB + (-A)\\X = CB - A[/tex]
Thus, the solution of the given matrix equation exists and is: [tex]X = CB - A[/tex]
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