By Otto and Bretscher and Kyle Burke

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**Additional resources for Instructor's Solutions Manual for Linear Algebra with Applications**

**Example text**

Vm ], B = [w1 . . wm ], and x = . . . xm x1 Then (A + B)x = [v1 + w1 . . vm + wm ] . . = x1 (v1 + w1 ) + · · · + xm (vm + wm ) and xm x1 x1 Ax + Bx = [v1 . . vm ] . . + [w1 . . wm ] . . = x1 v1 + · · · + xm vm + x1 w1 + · · · + xm wm . xm xm The two results agree, as claimed. 15 we draw a typical vector in this plane. 55 We are looking for constants a and b such that a 2 + b 5 = 8 . 9 6 3 7 a + 4b = 7 The resulting system 2a + 5b = 8 has the unique solution a = −1, b = 2, so that 8 is indeed a linear 9 3a + 6b = 9 1 4 combination of the vector 2 and 5 .

43 a T (x) = 3 · x2 = 2x1 + 3x2 + 4x3 = [2 3 4] x2 x3 4 x3 The transformation is indeed linear, with matrix [2 3 4]. v1 b If v = v2 , then T is linear with matrix [v1 v2 v3 ], as in part (a). v3 a a x1 x1 x1 c Let [a b c] be the matrix of T . Then T x2 = [a b c] x2 = ax1 + bx2 + cx3 = b · x2 , so that v = b c c x3 x3 x3 does the job. 44 T x2 = v2 × x2 = v3 x1 − v1 x3 = v3 −v2 v1 x2 − v2 x1 x3 v3 x3 0 −v3 v2 v3 0 −v1 .

V3 a a x1 x1 x1 c Let [a b c] be the matrix of T . Then T x2 = [a b c] x2 = ax1 + bx2 + cx3 = b · x2 , so that v = b c c x3 x3 x3 does the job. 44 T x2 = v2 × x2 = v3 x1 − v1 x3 = v3 −v2 v1 x2 − v2 x1 x3 v3 x3 0 −v3 v2 v3 0 −v1 . 3. Part a holds, since L(T (v + w)) = L(T (v) + T (w)) = L(T (v)) + L(T (w)), and part b also works, because L(T (kv)) = L(kT (v)) = kL(T (v)). 47 Write w as a linear combination of v1 and v2 : w = c1 v1 + c2 v2 .