Algebra of Matrices and Determinants Questions and Answers

Answer: C

\(P^{4052} = I\), the unit matrix of order 3.

Square \(P\). The first two rows are those of the identity, so they are unchanged; the third row of \(P^{2}\) is the third row of \(P\) times \(P\):

\[ \begin{aligned} (10, 100, -1)P &= (10 - 10,\ 100 - 100,\ 1) \\ &= (0, 0, 1) \end{aligned} \]

So \(P^{2} = I\), and since 4052 is even, \(P^{4052} = (P^{2})^{2026} = I\).

An odd power would give back \(P\), so the option \(P\) belongs to odd exponents. \(P^{T}\), \(-P^{T}\) and \(2P^{T}\) are not powers of \(P\) at all, since \(P\) is not symmetric.

Answer: A

The determinant is 7100.

Expand along the first row:

\[ \begin{aligned} \Delta &= 11(21 \cdot 31 - 1) - 1(31 - 1) + 1(1 - 21) \\ &= 11(650) - 30 - 20 \\ &= 7150 - 50 = 7100 \end{aligned} \]

A quick check: the product of the diagonal, \(11 \times 21 \times 31 = 7161\), is close, and the off-diagonal ones reduce it slightly. 6800, 6900 and 6700 are too small, and 7300 is larger than the diagonal product.

Answer: E

\(\alpha^{2} - \beta^{2} = 0\).

First, \(A^{2} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}^{2} = -I\). Expanding (the matrices \(I\) and \(A\) commute):

\[ \begin{aligned} (\alpha I + \beta A)^{2} &= \alpha^{2}I + 2\alpha\beta A + \beta^{2}A^{2} \\ &= (\alpha^{2} - \beta^{2})I + 2\alpha\beta A \end{aligned} \]

Setting this equal to \(A\) and comparing the coefficients of \(I\) and \(A\) gives \(\alpha^{2} - \beta^{2} = 0\) and \(2\alpha\beta = 1\).

The options 1, -1, 2 and -2 come from forgetting that \(A^{2} = -I\) or from comparing the wrong coefficients.

Answer: D

The true statement is \(A + A^{T}\) is symmetric and \(A - A^{T}\) is skew symmetric.

Take transposes, using \((A^{T})^{T} = A\):

\[ \begin{aligned} (A + A^{T})^{T} &= A^{T} + A = A + A^{T} \\ (A - A^{T})^{T} &= A^{T} - A = -(A - A^{T}) \end{aligned} \]

So the sum equals its transpose (symmetric) and the difference equals minus its transpose (skew symmetric). This is why every square matrix splits as \(\frac{1}{2}(A + A^{T}) + \frac{1}{2}(A - A^{T})\). The other options swap or repeat these labels, and \(-A - A^{T} = -(A + A^{T})\) is symmetric, not skew symmetric.

Answer: C

The determinant is 0.

Use \((p + q)^{2} - (p - q)^{2} = 4pq\). In each row \(p\) and \(q\) are reciprocals (\(10^{5}\) and \(10^{-5}\), and so on), so \(pq = 1\) and the first entry minus the second is 4.

Apply \(C_{1} \to C_{1} - C_{2}\):

\[ \begin{vmatrix} 4 & (10^{5} - 10^{-5})^{2} & 1 \\ 4 & (100^{6} - 100^{-6})^{2} & 1 \\ 4 & (6^{100} - 6^{-100})^{2} & 1 \end{vmatrix} \]

Now the first column is 4 times the third, and a determinant with two proportional columns is 0. The huge numbers never need to be evaluated.

Answer: E

\(AA^{T}\) equals \(\begin{pmatrix} \sec^{2}x & 0 \\ 0 & \sec^{2}x \end{pmatrix}\).

\(A^{T} = \begin{pmatrix} 1 & -\tan x \\ \tan x & 1 \end{pmatrix}\). Multiplying:

\[ \begin{aligned} (1,1) &: 1 + \tan^{2}x = \sec^{2}x \\ (1,2) &: -\tan x + \tan x = 0 \\ (2,1) &: -\tan x + \tan x = 0 \\ (2,2) &: \tan^{2}x + 1 = \sec^{2}x \end{aligned} \]

The off-diagonal entries cancel, so options with 1 or -1 off the diagonal are wrong. The result is \(\sec^{2}x\) times the identity.

Answer: C

\(a = 1\) and \(b = 19\).

Each row adds up to \(x + 19\). Add all columns into the first and take out that factor:

\[ \Delta = (x + 19)\begin{vmatrix} 1 & 8 & 9 \\ 1 & x + 6 & 9 \\ 1 & 8 & x + 7 \end{vmatrix} \]

Subtract the first row from the other two:

\[ \Delta = (x + 19)\begin{vmatrix} 1 & 8 & 9 \\ 0 & x - 2 & 0 \\ 0 & 0 & x - 2 \end{vmatrix} = (x + 19)(x - 2)^{2} \]

Comparing with \((x - 2)^{2}(ax + b)\) gives \(a = 1\), \(b = 19\). A quick check at \(x = 0\): the determinant is \(2(42 - 72) - 8(28 - 36) + 9(32 - 24) = 76 = 4 \times 19\).

Answer: E

\(\lambda\) is -1.

Expanding the determinant: \(f(x) = 2x^{3} - \sin 2\pi x\), which is indeed odd. Then \(g(x) = f(-x) = -f(x)\).

\[ \begin{aligned} f(1) &= 2 - \sin 2\pi = 2 \\ g(1) &= -f(1) = -2 \\ \lambda \cdot 2 \cdot (-2) &= 4 \\ \lambda &= -1 \end{aligned} \]

Forgetting that \(g(1) = -f(1)\) for an odd function gives \(4\lambda = 4\) and \(\lambda = 1\), the sign trap among the options.

Answer: B

The determinant of the inverse is \(\frac{1}{2}\).

For any invertible matrix, \(\det(A^{-1}) = \frac{1}{\det A}\).

\[ \begin{aligned} \det A &= (-4)(2) - (-5)(2) \\ &= -8 + 10 = 2 \\ \det(A^{-1}) &= \frac{1}{2} \end{aligned} \]

2 is the determinant of the matrix itself. A sign slip in \(-(-5)(2)\) gives \(-18\), and \(\pm\frac{1}{4}\) do not come from this matrix.

Answer: A

\(a + d\) is 0.

\(A\) is upper triangular, so every power of it is upper triangular with diagonal entries equal to the powers of the diagonal entries:

\[ A^{n} = \begin{bmatrix} 1 & * \\ 0 & i^{n} \end{bmatrix} \]

So \(a = 1\) and \(d = i^{42} = (i^{4})^{10} \cdot i^{2} = -1\), giving \(a + d = 1 - 1 = 0\).

The off-diagonal entry \(b\) does not affect \(a + d\), and since \(i^{42}\) is an even power of \(i\) it is real, which rules out the answers \(\pm i\).