Matrices and Determinants Questions and Answers

Answer: A

The matrix is singular for \(x = 0\) and \(x = 3\).

The matrix is \(A - xI\) with \(A = \begin{bmatrix} 3 & 2 & 2 \\ 2 & 4 & 1 \\ -2 & -4 & -1 \end{bmatrix}\), so it is singular exactly when \(x\) is an eigenvalue of \(A\). Use the invariants of \(A\):

\[ \begin{aligned} \det A &= 3(0) - 2(0) + 2(0) = 0 \\ \text{trace} &= 3 + 4 - 1 = 6 \\ \text{sum of 2x2 minors} &= 8 + 1 + 0 = 9 \end{aligned} \]

The zero determinant gives one eigenvalue 0. The other two add to 6 and multiply to 9 (the sum of the principal minors equals \(\lambda_{1}\lambda_{2}\) when the third eigenvalue is 0), so both are 3. The characteristic equation is \(x(x - 3)^{2} = 0\), giving \(x = 0\) and \(x = 3\).

The values -3, 6 and -6 are not roots: 6 is the trace, the sum of all three eigenvalues, not one of them, and the negative values have the wrong sign.

Answer: A

The determinant is -81.

For square matrices of order n, \(|kA| = k^{n}|A|\) and \(|AB| = |A||B|\). Here n = 3:

\[ \begin{aligned} |3AB| &= 3^{3}\,|A|\,|B| \\ &= 27 \times (-1) \times 3 = -81 \end{aligned} \]

-9 comes from multiplying the determinant by 3 instead of \(3^{3}\), as if the scalar multiplied only one row. -27 leaves out |B|. 81 drops the minus sign of |A|.

Answer: B

The values are a = 1, b = 4.

Expanding, \((A + B)^{2} = A^{2} + AB + BA + B^{2}\), so the condition holds exactly when \(AB + BA = 0\). Matrix products do not commute, so AB + BA is not 2AB.

\[ \begin{aligned} AB &= \begin{bmatrix} a - b & 2 \\ 2a - b & 3 \end{bmatrix} \\ BA &= \begin{bmatrix} a + 2 & -a - 1 \\ b - 2 & 1 - b \end{bmatrix} \end{aligned} \]

\[ AB + BA = \begin{bmatrix} 2a - b + 2 & 1 - a \\ 2a - 2 & 4 - b \end{bmatrix} \]

The top-right entry gives a = 1 and the bottom-right gives b = 4; then 2a - b + 2 = 0 and 2a - 2 = 0 check. The other pairs fail at once: a = 4 or a = 0 or a = 2 leaves 1 - a non-zero.

Answer: B

The product is \(\begin{bmatrix} 64 & 46 & 70 \\ 36 & 39 & 40 \\ 116 & 107 & 128 \end{bmatrix}\).

\([\mathbf{P}]\) is 2 × 3, so its transpose is 3 × 2, and a 3 × 2 matrix times the 2 × 3 matrix \([\mathbf{Q}]\) gives a 3 × 3 matrix. The rows of \([\mathbf{P}]^{T}\) are (2, 4), (3, 1) and (7, 5); multiply each by the columns (8, 12), (11, 6) and (9, 13) of \([\mathbf{Q}]\):

\[ \begin{aligned} \text{row 1} &: 16 + 48,\ 22 + 24,\ 18 + 52 \\ &= 64,\ 46,\ 70 \\ \text{row 2} &: 24 + 12,\ 33 + 6,\ 27 + 13 \\ &= 36,\ 39,\ 40 \\ \text{row 3} &: 56 + 60,\ 77 + 30,\ 63 + 65 \\ &= 116,\ 107,\ 128 \end{aligned} \]

The first option is the transpose of the correct answer, \(([\mathbf{P}]^{T}[\mathbf{Q}])^{T} = [\mathbf{Q}]^{T}[\mathbf{P}]\). The 2 × 2 options cannot be right whatever their entries, because a 3 × 2 matrix times a 2 × 3 matrix is always 3 × 3.

Answer: A

The sum of the values of \(x\) is 5.

A singular matrix has zero determinant:

\[ \begin{aligned} x(x - 5) - (-3)(2) &= 0 \\ x^{2} - 5x + 6 &= 0 \end{aligned} \]

The roots are \(x = 2\) and \(x = 3\), and their sum is 5 (equally, the sum of the roots of \(x^{2} - 5x + 6\) is the negative of the coefficient of \(x\)).

6 is the product of the roots, not their sum. 0 and \(-1\) cannot arise: whatever the individual roots, their sum is fixed at 5 by the coefficient of \(x\) in the quadratic.

Answer: A

The value is \(-16\).

The determinant of a product is the product of the determinants, and transposing a matrix does not change its determinant:

\[ \begin{aligned} \det(AB) &= \det A \cdot \det B \\ -80 &= 5 \det B \\ \det B &= -16 = \det B^{T} \end{aligned} \]

\(-\frac{1}{16}\) is \(1/\det B\), the determinant of the inverse. \(-400\) comes from multiplying \(-80\) by 5 instead of dividing, and 28 has no basis in the rules above.

Answer: A

The eigen values are \(-2\) and \(-0.8\).

For any square matrix the product of the eigen values equals the determinant and their sum equals the trace. So the two values satisfy

\[ \begin{aligned} \lambda_{1} + \lambda_{2} &= -2.8 \\ \lambda_{1}\lambda_{2} &= 1.6 \\ \lambda^{2} + 2.8\lambda + 1.6 &= 0 \\ (\lambda + 2)(\lambda + 0.8) &= 0 \end{aligned} \]

Check each option against both conditions. \(-1.8\) and \(-1\) add to \(-2.8\) but multiply to 1.8. \(-3\) and 0.2 add to \(-2.8\) but multiply to \(-0.6\). \(-1.2\) and \(-1.6\) multiply to 1.92 and add to \(-2.8\), so they also fail the product test. Only \(-2\) and \(-0.8\) satisfy both.

Answer: B

The product of the eigenvalues is 2.

The product of the eigenvalues of a square matrix equals its determinant. Expanding along the first row:

\[ \begin{aligned} |P| &= 2[(-3)(-1) - (3)(2)] - 0 + 1[(4)(2) - 0] \\ &= 2(3 - 6) + 8 = 2 \end{aligned} \]

-6 is the first term alone, \(2(3 - 6)\), without the contribution of the corner element 1; 6 and -2 come from sign slips in the cofactors. The sum of the eigenvalues would be the trace, 2 - 3 - 1 = -2, which is the other quantity the distractor -2 matches.

Answer: D

The incorrect statement is All eigenvalues of P are real numbers.

P is the matrix of a rotation by 45 degrees about the y axis. Its columns are orthonormal, so P is orthogonal (\(P^{T}P = I\)), its inverse equals its transpose, and its determinant is

\[ |P| = 1 \times \left(\tfrac{1}{2} + \tfrac{1}{2}\right) = 1 \]

A rotation keeps only its axis fixed, so it has one real eigenvalue, 1. The others satisfy \(\lambda^{2} - \sqrt{2}\lambda + 1 = 0\), giving \(\lambda = (1 \pm i)/\sqrt{2} = e^{\pm i\pi/4}\), which are complex. So the first three statements are true and the fourth is the incorrect one.

Answer: B

The rank is 2.

Expand the determinant along the first row:

\[ \begin{aligned} |A| &= -4(-1 - 3) - 1(-1 + 7) - 1(3 + 7) \\ &= 16 - 6 - 10 = 0 \end{aligned} \]

So the rank is less than 3. The first two rows, (-4, 1, -1) and (-1, -1, -1), are not multiples of each other, so the rank is at least 2. Indeed the third row is a combination of the first two: -2 times the first row plus the second row gives (7, -3, 1). A 3 by 3 matrix cannot have rank 4, and rank 1 would need every row to be a multiple of one row.