Questions on systems of equations and invariance

Question 1

Write the simultaneous equations 2x + y = 7 and x + 3y = 11 as a single matrix equation Mv = b, identifying M, v and b.

Accepted Answer

M has rows (2, 1) and (1, 3), v is the column (x, y) and b the column (7, 11). B1 B1 B1 for M, v and b.

Question 2

Using an inverse matrix, solve 2x + y = 7 and x + 3y = 11.

Accepted Answer

det M = 6 − 1 = 5, so M⁻¹ = (1/5) × rows (3, −1), (−1, 2). Then v = M⁻¹b = (1/5)(21 − 11, −7 + 22) = (2, 3). Check: 2 × 2 + 3 = 7 and 2 + 9 = 11. M1 for det M = 5, A1 for the inverse, M1 for v = M⁻¹b, A1 for (2, 3).

Question 3

Write the system x + 2y − z = 4, 3x − y + 2z = 1 and 2x + y + z = 7 as a single matrix equation Mv = b, identifying M, v and b.

Accepted Answer

M has rows (1, 2, −1), (3, −1, 2) and (2, 1, 1), v is the column (x, y, z) and b is the column (4, 1, 7). B1 B1 B1 for M, v and b. The method used for two equations extends unchanged to three: one row of M and one entry of b per equation, in the same order as the variables.

Question 4

State the equation that defines an invariant point v of a matrix M, and explain why the origin always satisfies it.

Accepted Answer

An invariant point satisfies Mv = v. B1 for the equation. Every linear transformation sends the origin to itself, since M multiplied by the zero column is the zero column whatever the entries of M are, so the origin solves Mv = v for every matrix M. B1 for the reason. Some matrices have no other invariant point; a few have a whole line of them, or every point.

Question 5

Three planes have no common point, yet no two of them are parallel. Describe the configuration they must form, explain why their pairwise intersection lines are parallel to each other, and state what this means for the coefficient matrix.

Accepted Answer

A prism: the three pairwise intersection lines are parallel and the planes fence off a triangular tube. Two of the planes meet in a line that the third misses everywhere, so the third is parallel to that line and meets each of the other two in a line parallel to it. The system is inconsistent and its coefficient matrix is singular. B1 for the prism, B1 for the third plane parallel to the line of the other two, B1 for the three parallel intersection lines, B1 for the singular coefficient matrix.

Question 6

Show that the line y = x is invariant under the transformation with matrix rows (4, 1) and (2, 3), and find the factor by which points on it are stretched.

Accepted Answer

Take (1, 1) on the line. It maps to (5, 5), still on y = x, and any multiple (t, t) maps to (5t, 5t) by linearity. The line maps into itself with stretch factor 5. M1 for mapping a general point of the line, A1 for the image (5t, 5t), A1 for the conclusion, B1 for the stretch factor 5. This is an invariant line rather than a line of fixed points. Every point on it moves, but none of them leaves the line.

Question 7

Find the second invariant line through the origin of the transformation with matrix rows (4, 1) and (2, 3).

Accepted Answer

Try y = mx. The image of (1, m) is (4 + m, 2 + 3m), which lies on y = mx when 2 + 3m = m(4 + m), so m² + m − 2 = 0, giving m = 1 or m = −2. The new line is y = −2x, stretched by factor 2, since (1, −2) maps to (2, −4). M1 for trying y = mx, A1 for m² + m − 2 = 0, A1 for y = −2x, B1 for the stretch factor 2.

Question 8

Explain the difference between an invariant line and a line of invariant points, and give a transformation with an example of each.

Accepted Answer

On an invariant line each point maps to a point of the same line, possibly elsewhere; a line of invariant points maps every point to itself. For a reflection, the mirror line is a line of invariant points, while any line perpendicular to the mirror is invariant without its points being fixed. B1 B1 for the two definitions, B1 B1 for the mirror line and a perpendicular line as the two examples.

Question 9

Find the invariant points of the matrix M with rows (2, −1) and (2, −1), and describe them geometrically.

Accepted Answer

Invariant points satisfy Mv = v, that is, (M − I)v = 0, where M − I has rows (1, −1) and (2, −2). M1 for forming M − I. Its determinant is 1 × (−2) − (−1) × 2 = 0, so it is singular and (M − I)v = 0 has non-zero solutions: from the first row, x − y = 0, so x = y. A1 for the singular check, M1 for solving to get x = y. Every point of y = x is invariant: (t, t) maps to (2t − t, 2t − t) = (t, t) for every t. A1 for identifying the whole line y = x as a line of invariant points, not only the origin.

Question 10

The matrix M with rows (1, 0, 1), (0, 1, 1) and (1, 1, 0) has inverse M⁻¹ = ½ × rows (1, −1, 1), (−1, 1, 1), (1, 1, −1). Use it to solve x + z = 9, y + z = 1 and x + y = 2.

Accepted Answer

v = M⁻¹b with b = (9, 1, 2): x = ½(9 − 1 + 2) = 5, y = ½(−9 + 1 + 2) = −3, z = ½(9 + 1 − 2) = 4. M1 for the multiplication, A1 for all three values. Check: 5 + 4 = 9, −3 + 4 = 1 and 5 − 3 = 2. B1 for the check against all three original equations.

Question 11

State, with a reason, whether the planes x + y + z = 3, 2x + 2y + 2z = 7 and x − y + z = 1 have any point in common.

Accepted Answer

The first two planes have normals (1, 1, 1) and (2, 2, 2), which are proportional, so the planes are parallel. B1 for spotting the proportional normals. Doubling the first equation's constant gives 6, not 7, so the two planes are parallel but not the same plane, and therefore never meet. B1 for the constants not matching. Since the first two planes alone share no point, no point can lie on all three, whatever the third plane is. B1 for the conclusion.

Question 12

The system x + y + z = 1, x + 2y + 3z = 4 and 2x + 3y + kz = 5 has a coefficient matrix M. Find the value of k for which M is singular. For that value of k, determine whether the system is consistent, and describe the geometry of the three planes.

Accepted Answer

Expanding along the top row, det M = (2k − 9) − (k − 6) + (3 − 4) = k − 4, so M is singular when k = 4. For any other k the inverse exists and there is a single point of intersection. With k = 4 the third equation reads 2x + 3y + 4z = 5, which is exactly the sum of the first two, so it carries no new information and the system is consistent. Solving the first two, subtracting gives y + 2z = 3, so y = 3 − 2z and then x = 1 − y − z = −2 + z. Writing z = λ, the solutions form the line r = (−2, 3, 0) + λ(1, −2, 1). Geometrically the three planes form a sheaf, meeting in that common line. M1 for expanding the determinant, A1 for det M = k − 4, A1 for k = 4, M1 for testing consistency, A1 for the third equation being the sum of the first two, M1 for solving the first two, A1 for the line of solutions, B1 for the sheaf. A singular matrix rules out a unique solution but says nothing about consistency; the sheaf here and an inconsistent prism both give det M = 0, so the equations still have to be tested.

Question 13

The planes x + y + z = 2 and 2x + y − z = 1 meet in a line. Find that line in the form r = a + λd, and find the value of μ for which the plane x + 2y + 4z = μ also contains it, so that the three planes form a sheaf.

Accepted Answer

Adding and subtracting the first two equations: eliminating z from the sum gives 3x + 2y = 3, and eliminating y from the difference gives −x + 2z = 1. M1 for a valid elimination. Writing x = 1 + 2λ: from −x + 2z = 1, z = 1 + λ; from 3x + 2y = 3, y = (3 − 3x)/2 = −3λ. A1 for the line r = (1, 0, 1) + λ(2, −3, 1). Substituting into the third plane's expression: x + 2y + 4z = (1 + 2λ) + 2(−3λ) + 4(1 + λ) = 5 + (2 − 6 + 4)λ = 5, independent of λ. M1 for the substitution, A1 for the λ-terms cancelling, A1 for μ = 5. Since every point of the line satisfies x + 2y + 4z = 5 whatever λ is, the whole line lies in that plane too, and the three planes meet along it: a sheaf. B1 for identifying the configuration as a sheaf. Had the λ-terms not cancelled, no single μ could work and no sheaf would be possible.

Question 14

M is the matrix with rows (5, −2) and (−2, 2). Find both invariant lines of M through the origin, find the stretch factor on each, and explain what is special about the one with stretch factor 1.

Accepted Answer

Try y = mx. The image of (1, m) is (5 − 2m, −2 + 2m), which lies on y = mx when −2 + 2m = m(5 − 2m), that is 2m² − 3m − 2 = 0. M1 for the substitution and the equation. Factorising, (2m + 1)(m − 2) = 0, so m = 2 or m = −½. A1 for both roots. On y = 2x, the point (1, 2) maps to (5 − 4, −2 + 4) = (1, 2): the point is unchanged, so the stretch factor is 1. M1 for testing a point of this line, A1 for the image (1, 2) and factor 1. On y = −x/2, the point (2, −1) maps to (10 + 2, −4 − 2) = (12, −6) = 6(2, −1), a stretch factor of 6. A1 for the image and factor 6. A stretch factor of 1 means every point of y = 2x maps to itself, so this invariant line is in fact a line of invariant points, not merely an invariant line; y = −x/2 is invariant but its points do move. B1 for identifying y = 2x as a line of invariant points.

Question 15

Three quantities x, y and z in a mixture satisfy x + y + z = 12, 2x − y + z = 5 and x + 2y − 3z = 1. Solve the system, and explain why the solution should still be checked against a real-world constraint such as non-negativity before it is reported as an answer to the mixture problem.

Accepted Answer

Write the system as one matrix equation Mv = b, with M the matrix with rows (1, 1, 1), (2, −1, 1) and (1, 2, −3), v the column (x, y, z) and b the column (12, 5, 1). M1 for the matrix equation, with M and b correct. Expanding along the first row, det M = 1(3 − 2) − 1(−6 − 1) + 1(4 + 1) = 13. M1 for the determinant. It is non-zero, so M⁻¹ exists and the three planes meet in a single point: there is exactly one solution to find. The cofactors, transposed, give M⁻¹ = (1/13) times the matrix with rows (1, 5, 2), (7, −4, 1) and (5, −1, −3). A1 for the inverse. Then v = M⁻¹b gives x = (12 + 25 + 2)/13 = 3, y = (84 − 20 + 1)/13 = 5 and z = (60 − 5 − 3)/13 = 4, so x = 3, y = 5, z = 4. A1 for the solution. Check: 3 + 5 + 4 = 12, 6 − 5 + 4 = 5 and 3 + 10 − 12 = 1, all three confirmed. B1 for the verification against all three original equations. The algebra alone cannot know that x, y and z stand for amounts in a mixture, so it could just as easily have returned a negative value; here all three are positive, which the context requires but the arithmetic does not guarantee. B1 for that closing observation.