Algebra of Matrices and Determinants Questions and Answers

Answer: A

The minimum value of \(|A|\) is 2.

Write \(s = \sin\theta\) and expand along the first row:

\[ \begin{aligned} |A| &= 1(1 + s^{2}) - s(-s + s) + 1(s^{2} + 1) \\ &= 1 + s^{2} + 0 + s^{2} + 1 \\ &= 2 + 2\sin^{2}\theta \end{aligned} \]

Since \(0 \le \sin^{2}\theta \le 1\), \(|A|\) ranges from 2 to 4, and its minimum, 2, occurs when \(\sin\theta = 0\). 4 is the maximum, and 0, 1 and -1 are below the smallest possible value.

Answer: C

The determinant is 0.

Expand the third column with the sine addition formula: \(\sin(\theta + 75^\circ) = \sin\theta\cos 75^\circ + \cos\theta\sin 75^\circ\).

\[ C_3 = \cos 75^\circ \, C_1 + \sin 75^\circ \, C_2 \]

The third column is a linear combination of the first two (with the same constants \(\cos 75^\circ\) and \(\sin 75^\circ\) in every row), so the operation \(C_3 \rightarrow C_3 - \cos 75^\circ C_1 - \sin 75^\circ C_2\) makes it a column of zeros and the determinant vanishes. No numerical evaluation is needed.

Answer: A

\(A^{10}\) equals \(A^{-1}\).

Multiply the given equation by \((A - I)\); the sum telescopes.

\[ \begin{aligned} (A - I)(I + A + \cdots + A^{10}) &= A^{11} - I \\ A^{11} - I &= O \Rightarrow A^{11} = I \end{aligned} \]

Since A is non-singular, multiply \(A^{11} = I\) by \(A^{-1}\): \(A^{10} = A^{-1}\). \(A^{10} = I\) would need \(A^{10}\) rather than \(A^{11}\) to equal I, and O is impossible because a power of a non-singular matrix is non-singular.