"Matrices and Determinants" — Practice

Question 1

Evaluate the determinant of the matrix with rows (1, 3, 5), (2, 6, 10) and (7, 4, 9) without full expansion.

Accepted Answer

Look at the first two rows before computing anything. The second row is exactly twice the first, since 2, 6 and 10 are double 1, 3 and 5. A determinant with two proportional rows is zero, so the answer is 0 and no expansion is needed. Structurally, one row carries no information the other did not already supply, so the action collapses a dimension and the determinant records that collapse with a zero.

Question 2

Evaluate the determinant with rows (203, 204, 205), (206, 207, 208) and (209, 210, 211).

Accepted Answer

Direct expansion means six products of three-digit numbers. Instead use column operations, which leave the determinant unchanged. Replace the second column by itself minus the first, making every entry 1. Replace the third column by itself minus the original first column, making every entry 2. The second and third columns are now (1,1,1) and (2,2,2), which are proportional, so the determinant is zero. The whole evaluation takes two lines and no multiplication.

Question 3

If A is a three by three matrix with determinant 4, find the determinants of 2A, of the adjoint of A, and of A inverse.

Accepted Answer

For 2A, every one of the three rows is doubled and each contributes an independent factor of 2, so the determinant is 2 cubed times 4, which is 32. For the adjoint, the standard result gives the determinant of the adjoint as the determinant of A raised to the power n minus 1, which for n equal to 3 is 4 squared, that is 16. For the inverse, the multiplicative property applied to A times its inverse equalling the identity gives the product of the determinants as 1, so the determinant of the inverse is one quarter.

Question 4

Show that the points (2, 3), (4, 7) and (6, 11) are collinear, and find the area of the triangle with vertices (0, 0), (4, 0) and (0, 6).

Accepted Answer

For the first, evaluate the determinant with rows (2,3,1), (4,7,1) and (6,11,1). Expanding along the first row gives 2(7 minus 11) minus 3(4 minus 6) plus 1(44 minus 42), which is minus 8 plus 6 plus 2, that is 0. A zero determinant means zero area, so the points are collinear; indeed each satisfies y equal to 2x minus 1. For the second, the area is half the absolute value of the determinant with rows (0,0,1), (4,0,1) and (0,6,1). Expanding gives 0 minus 0 plus 1 times (24 minus 0), which is 24, so the area is 12 square units.

Question 5

Find the inverse of the two by two matrix with rows (3, 5) and (1, 2), and verify it.

Accepted Answer

The determinant is 3 times 2 minus 5 times 1, which is 6 minus 5, that is 1, so the matrix is non-singular and an inverse exists. For a two by two matrix the adjoint swaps the diagonal entries and negates the off-diagonal ones, giving rows (2, minus 5) and (minus 1, 3). Dividing by the determinant of 1 leaves the inverse equal to that adjoint. Verifying: multiplying the original by this gives first entry 3 times 2 plus 5 times minus 1, which is 1, and the remaining entries work out to give the identity matrix.

Question 6

Prove that the determinant of any skew-symmetric matrix of odd order is zero, and explain why the argument fails for even order.

Accepted Answer

Let A be skew-symmetric of order n, so the transpose of A equals minus A. Transposing never changes a determinant, so the determinant of A equals the determinant of its transpose, which equals the determinant of minus A. Scaling a matrix by minus one scales the determinant by minus one raised to the power n, since each of the n rows contributes a factor. So the determinant of A equals minus one to the n times the determinant of A. For odd n this says the determinant equals its own negative, forcing it to be zero. For even n the factor is plus one and the equation becomes the empty statement that the determinant equals itself, so nothing is proved and even-order skew-symmetric matrices generally have non-zero determinants.

Question 7

Give an example of two non-zero two by two matrices whose product is the zero matrix, and explain what this says about cancellation.

Accepted Answer

Take A with rows (1, 0) and (0, 0), and B with rows (0, 0) and (0, 1). Neither is the zero matrix. Their product has entry (1,1) equal to 1 times 0 plus 0 times 0, which is 0, and by similar computation every entry of the product is zero. So a product of non-zero matrices can be the zero matrix. The consequence for cancellation is direct: from AB equal to AC we cannot conclude B equals C, because AB minus AC equals A times (B minus C), and that can vanish with B minus C non-zero whenever A is singular. Cancellation is legitimate only when A has an inverse.

Question 8

Express the matrix with rows (2, 5) and (1, 4) as the sum of a symmetric and a skew-symmetric matrix.

Accepted Answer

Use the standard decomposition, taking half the sum of the matrix and its transpose for the symmetric part and half the difference for the skew part. The transpose has rows (2, 1) and (5, 4). Their sum has rows (4, 6) and (6, 8), and half of that is the symmetric matrix with rows (2, 3) and (3, 4). Their difference has rows (0, 4) and (minus 4, 0), and half of that is the skew-symmetric matrix with rows (0, 2) and (minus 2, 0). Adding the two recovers the original, and note that the skew part has zeros on its diagonal, as it must.

Question 9

For the system x + y + z = 6, x + 2y + 3z = 10, x + 2y + kz = m, find the conditions on k and m for a unique solution, no solution, and infinitely many solutions.

Accepted Answer

The coefficient determinant has rows (1,1,1), (1,2,3) and (1,2,k). Expanding gives 1 times (2k minus 6) minus 1 times (k minus 3) plus 1 times (2 minus 2), which simplifies to k minus 3. If k is not 3 the determinant is non-zero, the action is reversible and there is a unique solution whatever m may be. If k equals 3 the second and third equations become x plus 2y plus 3z equal to 10 and equal to m, with identical left sides. If m is not 10 they contradict each other and there is no solution. If m equals 10 the third equation merely repeats the second, leaving two independent equations in three unknowns and hence infinitely many solutions.

Question 10

For what values of k does the homogeneous system kx + y + z = 0, x + ky + z = 0, x + y + kz = 0 have a non-trivial solution?

Accepted Answer

A homogeneous system always has the trivial solution, so the question is when something else exists, and that happens exactly when the coefficient determinant is zero. The determinant with rows (k,1,1), (1,k,1) and (1,1,k) expands to k times (k squared minus 1) minus 1 times (k minus 1) plus 1 times (1 minus k). That is k cubed minus k minus k plus 1 plus 1 minus k, which simplifies to k cubed minus 3k plus 2. Factorising gives (k minus 1) squared times (k plus 2). So the determinant vanishes when k equals 1 or k equals minus 2, and those are the values admitting a non-trivial solution.

Question 11

Solve the system 2x + 3y = 8, x - y = 1 by the matrix inversion method.

Accepted Answer

Write the system as AX equal to B, where A has rows (2, 3) and (1, minus 1), X is the column of unknowns and B is the column (8, 1). The determinant of A is 2 times minus 1 minus 3 times 1, which is minus 5, and being non-zero it guarantees a unique solution. The adjoint has rows (minus 1, minus 3) and (minus 1, 2), so the inverse is that divided by minus 5. Multiplying the inverse by B gives x equal to (minus 8 minus 3) divided by minus 5, which is 11 over 5, and y equal to (minus 8 plus 2) divided by minus 5, which is 6 over 5. Substituting back into both original equations confirms the answer.

Question 12

A system of three linear equations in three unknowns has a coefficient determinant equal to zero, and the product of the adjoint with the constant column is also the zero column. A student concludes there are infinitely many solutions. Is the student right? Explain geometrically.

Accepted Answer

The student is not entitled to that conclusion. When the determinant vanishes and the adjoint product also vanishes, the standard test is inconclusive: the system may have infinitely many solutions or none at all, and only substituting back into the original equations settles it. Geometrically, each equation is a plane. A zero determinant says the three planes have lost independence, so they no longer meet at a single point. That leaves two possibilities: they may share a whole line or plane, giving infinitely many solutions, or they may be arranged with no common point at all, as in a triangular prism configuration. The determinant reports the lost independence but carries no information about whether the constant terms lined up consistently with it.

Practice — "Matrices and Determinants"

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