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can you explain me the easiest way for solving this matrix equation? I have this matrix:

a={{-4,1,3,0},{-2,7,1,-1},{1,-1,9,-3},{-1,0,5,-10}

and the Equation is:

2a-x+IdentityMatrix[4]==0

Where x is the Unknown Matrix

or the better question apart from this upper exercise would be, how can i solve matrix equations that have this form A+X=...(0, or any other matrix)

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    $\begingroup$ So, try replacing x with 2 a + IdentityMatrix[4]... $\endgroup$ Jun 12, 2016 at 15:18
  • $\begingroup$ I know, thats the "common way", but is there like a command like at the polynomials for example: Solve[x+1==0,x] but just for matrix solving? $\endgroup$ Jun 12, 2016 at 15:21
  • $\begingroup$ Array[x, {4,4}] /. Solve[Flatten[ Thread /@ Thread[2 a - Array[x, {4, 4}] + IdentityMatrix[4] == 0]], Flatten@Array[x, {4, 4}]] or Solve[Flatten[ Table[2 a[[i, j]] - x[i, j] + Boole[i == j] == 0, {i, 4}, {j, 4}]], Flatten[Array[x, {4, 4}]]] $\endgroup$ Jun 12, 2016 at 16:14

1 Answer 1

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The following is a general technique, using Solve

a = {{-4, 1, 3, 0}, {-2, 7, 1, -1}, {1, -1, 9, -3}, {-1, 0, 5, -10}};
X = Array[x, {4, 4}]
sol = Solve[2 a - X + IdentityMatrix[4] == Array[0 &, {4, 4}], Flatten[X]]
X /. Flatten[sol] // MatrixForm
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