Matrices and Determinants Questions and Answers

Answer: C

Setting \( \lambda = 0 \) in the characteristic polynomial gives the determinant up to sign, which yields the relation:

\[ \lambda_1 \lambda_2 \cdots \lambda_n = \det A \qquad \lambda_1 + \lambda_2 + \cdots + \lambda_n = \operatorname{tr} A \]

Together the two relations check a computed spectrum in a couple of lines.

Answer: D

The characteristic determinant is itself triangular, so it factorises into the diagonal differences directly. It is why reduction to triangular form is so valuable, though the reduction must preserve similarity rather than merely rank.

Answer: A

Realness of the spectrum is what makes such matrices so tractable, and their eigenvectors for distinct eigenvalues are mutually orthogonal. Both facts underpin principal axis analysis and the theory of quadratic forms.

Answer: B

Substituting the matrix into its characteristic polynomial gives the zero matrix, which allows high powers to be reduced and the inverse to be expressed as a polynomial in the matrix. It is a standard route to computing an inverse without cofactors.

Answer: C

Similar matrices represent the same linear map in different bases, so they share their characteristic polynomial and hence their eigenvalues, trace and determinant. Diagonalisation is the search for a similar matrix that happens to be diagonal.

Answer: D

Distinct eigenvalues guarantee independence and so suffice, but they are not necessary, since a repeated eigenvalue may still furnish enough independent vectors. A matrix falling short of n independent eigenvectors is called defective and cannot be diagonalised.

Answer: A

For a two by two matrix the determinant is the product of the leading diagonal less that of the other:

\[ \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = (2)(5) - (3)(4) = 10 - 12 = -2 \]

A negative value is perfectly admissible and simply reflects the orientation the transformation produces.

Answer: B

Expanding along the first row:

\[ \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{vmatrix} = 1\begin{vmatrix} 5 & 6 \\ 8 & 10 \end{vmatrix} - 2\begin{vmatrix} 4 & 6 \\ 7 & 10 \end{vmatrix} + 3\begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} \]

\[ = 1(50-48) - 2(40-42) + 3(32-35) = 2 + 4 - 9 = -3 \]

Changing the last entry from 10 to 9 would make the third row the sum of twice the second less the first, and the determinant would vanish.

Answer: C

Each of the three rows contributes one factor of \( k \), so for a matrix of order \( n \):

\[ \det(kA) = k^{n}\det A \]

\[ \det(2A) = 2^{3}\times 5 = 8 \times 5 = 40 \]

Using a single factor of two and reporting 10 is the classic error with this property.

Answer: D

Since \( AA^{-1} = I \) and the identity has determinant one:

\[ \det A \cdot \det(A^{-1}) = 1 \quad\Rightarrow\quad \det(A^{-1}) = \frac{1}{\det A} = \frac{1}{4} = 0.25 \]

It also shows at once that a singular matrix, having zero determinant, can have no inverse.