This classic treatment of linear algebra presents the fundamentals in the clearest possible way, examining basic ideas by means of computational examples and geometrical interpretation. It proceeds from familiar concepts to the unfamiliar, from the concrete to the abstract. Readers consistently praise this outstanding text for its expository style and clarity of presentation.

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* Clear, accessible, step-by-step explanations make the material crystal clear.

* Established the intricate thread of relationships between systems of equations, matrices, determinants, vectors, linear transformations and eigenvalues.

each of whose terms is the product of n entries from the given matrix. Each of these entries is either a constant or is of the form λ – a . The only term ij with a λ in each factor of the product is (λ – a )(λ – a ) … (λ – a ) 11 22 nn Therefore, this term must produce the highest power of λ in the characteristic polynomial. This power is clearly n and the coefficient of λ n is 1. 200 Exercise Set 7.1 17. The characteristic equation of A is λ2 – ( a + d)λ + ad – bc = 0 This is a

2 − 1 10 − 5 0 0 0 2 0 5 2 Thus, D(3 – 4 sin x + sin 2 x + 5 cos 2 x) = –4 cos x – 10 sin 2 x + 2 cos 2 x. SUPPLEMENTARY EXERCISES 8 3. By the properties of an inner product, we have T(v + w) = 〈v + w, v 〉v 0 0 = (〈v, v 〉 + 〈w, v 〉)v 0 0 0 = 〈v, v 〉v + 〈w, v 〉v 0 0 0 0 = T(v) + T(w) and T( kv) = 〈 kv, v 〉v = k〈v, v 〉v = kT(v) 0 0 0 0 Thus T is a linear operator on V. 5. (a) The matrix for T with respect to the

( x 2 + y 2 + z 2) i i i 1 + c x + c y + c z + c = 0 of the sphere yields four equations, which together with the 2 3 4 5 above sphere equation form a homogeneous linear system for c , … , c with a nontrivial 1 5 solution. Thus the determinant of this system is zero, which is Eq. (12). Exercise Set 11.1 345 10. Upon substitution of the coordinates of the three points ( x , y ), ( x , y ) and ( x , y ), we 1 1 2 2 3 3 obtain the equations: c y + c x 2 + c x + c = 0. 1 2 3

(22) and (23) we have M = 2 M – M = –4. 1 2 3 M = 2 M – M = 68. 5 4 3 Using (14) to solve for the a ’s, b ’s, c ’s, and d ’s we have i i i i a = ( M – M )/6 h = 3 1 2 1 a = ( M – M )/6 h = 3 2 3 2 a = ( M – M )/6 h = 3 3 4 3 a = ( M – M )/6 h = 3 4 5 4 b = M /2 = –21 1 1 b = M /2 = 171 2 2 b = M /2 = 161 3 3 b = M /2 = 251 4 4 360 Exercise Set 11.5 c = ( y – y )/ h – ( M + 2 M ) h/6 = 55 1 2 1 2 1 c = ( y – y )/ h – ( M + 2 M ) h/6 = 10 2 3 2

1 2 − 1 2 − 1 Add –1 times Row 2 to 0 1 1 5 − 3 Row 3 and multiply 0 0 0 0 0 Row 2 by –1. 44 Exercise Set 1.6 or 1 0 3 1 − 2 7 0 1 1 5 − 3 Add twice Row 2 to Row 3. 0 0 0 0 0 Thus if we let x = t, we have for Part (a) x = –12 – 3 t and x = –5 – t, while for 3 1 2 Part (b) x = 7 – 3 t and x = 3 – t. 1 2 17. The augmented matrix for this system of equations is 1 2 − 5 b 1 4 5 − 8