Exercises on Cramer’s rule, inverse matrix, and volume (5.3 #8. Introduction to Linear Algebra: Strang) Suppose ⎡ ⎤ 1 1 4 A = ⎣ 1 2 2 ⎦. 1 2 5 Problem 20.1: Find its cofactor matrix C and multiply AC T to ﬁnd det( A). ⎡ ⎤ 6 −3 0 · · ⎦ and AC T = C=⎣ · . · · · If you change a1,3 = 4 to 100, why is det( A) unchanged? Solution: AC T : We ﬁll in the cofactor matrix C and then multiply to obtain ⎡ ⎤ 6 −3 0 1 −1 ⎦ C=⎣ 3 −6 2 1 and ⎡ ⎤⎡ ⎤ ⎡ ⎤ 6 1 1 4 3 0 0 3 −6 1 2 ⎦ = ⎣ 0 3 0 ⎦ = 3I. AC T = ⎣ 1 2 2 ⎦ ⎣ −3 1 2 5 0 0 3 1 0 −1 Since AC T = det( A) I, we have det( A) = 3. If 4 is changed to 100, det( A) is unchanged because the cofactor of that entry is 0, and thus its value does not contribute to the determinant. Problem 20.2: (5.3 #28.) Spherical coordinates ρ, φ, θ satisfy x = ρ sin φ cos θ, y = ρ sin φ sin θ and z = ρ cos φ. Find the three by three matrix of partial derivatives: ⎡ ⎤ ∂x/∂ρ ∂x/∂φ ∂x/∂θ ⎣ ∂y/∂ρ ∂y/∂φ ∂y/∂θ ⎦ . ∂z/∂ρ ∂z/∂φ ∂z/∂θ Simplify its determinant to J = ρ2 sin φ. In spherical coordinates, dV = ρ2 sin φ dρ dφ dθ is the volume of an inﬁnitesimal “coordinate box.” Solution: The rows are formed by the partials of x, y, and z with respect to ρ, φ, and θ: ⎡ ⎤ sin φ cos θ ρ cos φ cos θ −ρ sin φ sin θ ⎣ sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ ⎦ . −ρ sin φ 0 cos φ Expanding its determinant J along the bottom row, we get: � � ρ cos φ cos θ −ρ sin φ sin θ J = cos φ ρ cos φ sin θ ρ sin φ cos θ � � sin φ cos θ −ρ sin φ sin θ −(−ρ sin φ) +0 sin φ sin θ ρ sin φ cos θ = cos φ(ρ2 cos φ sin φ cos2 θ + ρ2 cos φ sin φ sin2 θ ) +ρ sin φ(ρ sin2 φ cos2 θ + ρ sin2 φ sin2 θ ) = = = J = cos φ(ρ2 cos φ sin φ(cos2 θ + sin2 θ )) + ρ sin φ(ρ sin2 φ(cos2 θ + sin2 θ )) cos φ(ρ2 cos φ sin φ) + ρ2 sin3 φ ρ2 sin φ(cos2 φ + sin2 φ) ρ2 sin φ.