2007-02-03 07:43:55 · 2 answers · asked by Rosy 3 in Science & Mathematics ➔ Mathematics
Just so it fits on here, I'm going to assume A is a 2 by 2 matrix (though the proof can be generalized to n x n matrices).
Let A =
[a11 a12]
[a21 a22]
Then
(c + d)A =
[(c + d)a11 (c + d)a12]
[(c + d)a21 (c + d)a22]
Expanding within the matrix, we have
[c(a11) + d(a11) c(a12) + d(a12)]
[c(a21) + d(a21) c(a22) + d(a22)]
Now that we have an addition within the matrix, we can decompose this into the sum of two matrices.
[c(a11) c(a12)] + [d(a11) d(a12)]
[c(a21) c(a22)] [d(a21) d(a22)]
Factoring a scalar out of each matrix, we have,
c [a11 a12] + d[a11 a12]
. .[a21 a22] . . . .[a21 a22]
which translates into
cA + dA
2007-02-03 08:00:02 · answer #1 · answered by Puggy 7 · 1⤊ 0⤋
its vice-versa.. if u look at (cA+dA) and u write down the matrix of A.. u basically end up taking the matrix common as u would do for any algebraic equation.. and u will get back to ur question (c+d)A.. it slike (c+d)5= 5c+5d.. u hav to hav faith in it
2007-02-03 07:49:36 · answer #2 · answered by Anonymous · 0⤊ 0⤋