Stitz-Zeager_College_Algebra_e-book

# Within the scope of this text we will prove a few

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Unformatted text preview: below 3 −1 1 x 8 2 −1 X = y B = 4 A= 1 2 3 −4 z 10 We now consider the matrix equation AX = B . AX = B 3 −1 1 x 1 2 −1 y 2 3 −4 z 3x − y + z x + 2y − z 2x + 3 y − 4z 8 = 4 10 8 = 4 10 We see that ﬁnding a solution (x, y, z ) to the original system corresponds to ﬁnding a solution X for the matrix equation AX = B . If we think about solving the real number equation ax = b, we would simply ‘divide’ both sides by a. Is it possible to ‘divide’ both sides of the matrix equation AX = B by the matrix A? This is the central topic of Section 8.4. 8.3 Matrix Arithmetic 8.3.1 489 Exercises 1. Using the matrices A= 12 34 0 −3 −5 2 B= 10 − 11 0 2 3 59 5 C= 7 −13 1 −4 E= 0 0 D= 3 6 8 0 2 3 4 −9 0 −5 compute the following matrices or state that they are undeﬁned. (a) (b) (c) (d) (e) 7B − 4A AB BA E+D ED (f) (g) (h) (i) (j) (k) E 2 + 5E − 36I3 CD + 2I2 A A − 4I2 A2 − B 2 (A + B )(A − B ) A2 − 5A − 2I2 (l) EDC (m) CDE (n) ABCEDI2 2. Let A= abc def E1 = 01 10 E2 = 50 01 E3 = 1 −2 0 1 Compute E1 A, E2...
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