(a) Consider a linear operator T on R³ over R defined by T(x, y, z) = (2x, 4x – y, 2x + 3y – z). Is T invertible? If yes, justify…

(a) T(x, y, z) = (2x, 4x − y, 2x + 3y − z). Its standard matrix is

A = [[2, 0, 0], [4, −1, 0], [2, 3, −1]].

Using the determinant of a lower-triangular matrix,

det A = 2 × (−1) × (−1) = 2 ≠ 0.

Hence T is invertible.

Let T(x, y, z) = (X, Y, Z). Then

X = 2x ⇒ x = X/2.

Y = 4x − y ⇒ y = 4x − Y = 2X − Y.

Z = 2x + 3y − z ⇒ z = 2x + 3y − Z.

Substituting x and y:

z = X + 3(2X − Y) − Z = 7X − 3Y − Z.

Therefore

T⁻¹(X, Y, Z) = (X/2, 2X − Y, 7X − 3Y − Z).

Final answer (a): T is invertible since det A = 2 ≠ 0, and T⁻¹(X, Y, Z) = (X/2, 2X − Y, 7X − 3Y − Z).

(b) Given

u = (x + y)/(1 − xy), v = tan⁻¹ x + tan⁻¹ y.

By the quotient rule,

∂u/∂x = [(1)(1 − xy) − (x + y)(−y)]/(1 − xy)² = (1 + y²)/(1 − xy)².

Similarly,

∂u/∂y = [(1)(1 − xy) − (x + y)(−x)]/(1 − xy)² = (1 + x²)/(1 − xy)².

Also,

∂v/∂x = 1/(1 + x²), ∂v/∂y = 1/(1 + y²).

The Jacobian is

∂(u, v)/∂(x, y) = determinant of [[∂u/∂x, ∂u/∂y], [∂v/∂x, ∂v/∂y]]

= (∂u/∂x)(∂v/∂y) − (∂u/∂y)(∂v/∂x)

= [(1 + y²)/(1 − xy)²]·[1/(1 + y²)] − [(1 + x²)/(1 − xy)²]·[1/(1 + x²)]

= 1/(1 − xy)² − 1/(1 − xy)²

= 0.

Since the Jacobian vanishes identically wherever the derivatives exist, u and v are functionally related.

Now let x = tan α and y = tan β. Then

v = α + β,

and

u = (tan α + tan β)/(1 − tan α tan β) = tan(α + β) = tan v.

Thus the functional relation is

u = tan v, provided 1 − xy ≠ 0.

Equivalently, on branches,

v = tan⁻¹ u + kπ, k ∈ Z.

Final answer (b): ∂(u, v)/∂(x, y) = 0. u and v are functionally related, and the relation is u = tan v.

(c) The ‘image’ of the line in the plane is taken here in the standard sense of its mirror image (reflection). The given line is

r(t) = (3 − 6t, 2t, 3 + 2t),

so its direction vector is

d = (−6, 2, 2).

The plane is

3x + 4y − 5z + 26 = 0,

so its normal vector is

n = (3, 4, −5).

For reflection in the plane, the component of d along n reverses sign. First find the intersection of the line with the plane. Substitute the line into the plane:

3(3 − 6t) + 4(2t) − 5(3 + 2t) + 26 = 0

= 9 − 18t + 8t − 15 − 10t + 26

= 20 − 20t.

Thus t = 1. The point of intersection is

A = (3 − 6, 2, 3 + 2) = (−3, 2, 5).

Now reflect the direction vector d across the plane. Use

d′ = d − [2(d·n)/(n·n)]n.

Compute

d·n = (−6)(3) + (2)(4) + (2)(−5) = −18 + 8 − 10 = −20,

n·n = 3² + 4² + (−5)² = 9 + 16 + 25 = 50.

Therefore,

d′ = (−6, 2, 2) − [2(−20)/50](3, 4, −5)

= (−6, 2, 2) + (4/5)(3, 4, −5)

= (−6 + 12/5, 2 + 16/5, 2 − 4)

= (−18/5, 26/5, −2).

This is proportional to (−9, 13, −5).

Hence the reflected line passes through A = (−3, 2, 5) and has direction (−9, 13, −5).

Its parametric form is

x = −3 − 9s, y = 2 + 13s, z = 5 − 5s, s ∈ R.

Its Cartesian form is

(x + 3)/(−9) = (y − 2)/13 = (z − 5)/(−5).

Equivalently,

(x + 3)/9 = (y − 2)/(−13) = (z − 5)/5.

Check: the point (3, 0, 3) of the given line reflects to (3/5, −16/5, 7), which lies on this line (s = −0.4).

If 'image' is read instead as the orthogonal projection on the plane, the direction is d − [(d·n)/(n·n)]n ∝ (−4, 3, 0), giving (x + 3)/(−4) = (y − 2)/3, z = 5.

Final answer (c): The image line is

x = −3 − 9s, y = 2 + 13s, z = 5 − 5s, s ∈ R,

or in Cartesian form,

(x + 3)/9 = (y − 2)/(−13) = (z − 5)/5.