Matrices and Determinants Questions and Answers

Answer: A

The trace is \( 4+3=7 \) and the determinant \( (4)(3)-(1)(2)=10 \), so:

\[ \lambda^{2} - (\operatorname{tr}A)\lambda + \det A = 0 \quad\Rightarrow\quad \lambda^{2} - 7\lambda + 10 = 0 \]

\[ (\lambda-5)(\lambda-2)=0 \quad\Rightarrow\quad \lambda = 5,\; 2 \]

They duly sum to 7 and multiply to 10, which is the check.

Answer: B

Building the characteristic equation from the trace and determinant directly:

\[ \lambda^{2} - 9\lambda + 20 = 0 \quad\Rightarrow\quad (\lambda-5)(\lambda-4)=0 \quad\Rightarrow\quad \lambda = 5,\; 4 \]

Checking that \( 5+4=9 \) and \( 5\times 4=20 \) confirms the roots without solving again.

Answer: C

A diagonal matrix is a special case of a triangular one, so its characteristic determinant factorises directly:

\[ \det(A-\lambda I) = (2-\lambda)(3-\lambda)(7-\lambda) = 0 \quad\Rightarrow\quad \lambda = 2,\;3,\;7 \]

Their sum \( 2+3+7=12 \) is the trace and their product \( 2\times3\times7=42 \) the determinant, as the general relations require.

Answer: D

The determinant is \( (2)(1)-(1)(1)=1 \), so the inverse is the adjoint itself:

\[ A^{-1} = \frac{1}{\det A}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \frac{1}{1}\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} \]

Checking:

\[ \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Answer: A

Correct answer: 2

The third row is a linear combination of the first two, so the three rows span at most a two-dimensional space and the rank is at most 2. It is exactly 2 provided the first two rows are themselves independent, and less only if they are not. The determinant is necessarily zero, since a matrix of order 3 with rank below 3 is singular.

Answer: B

Correct answer: a unique solution

The two ranks are equal, so the system is consistent, and their common value 3 equals the number of unknowns, so there are no free variables. The fourth equation is therefore not an extra constraint but a consequence of the other three - more equations than unknowns does not imply inconsistency, only that some are redundant.

Answer: C

Correct answer: it has no solution

When the augmented rank exceeds the coefficient rank the elimination produces a row of the form \( 0 = c \) with \( c \neq 0 \), which no assignment of the unknowns can satisfy. The system is inconsistent. This is the one case in which the solution set is genuinely empty; every other rank combination gives either one solution or infinitely many.

Answer: D

Correct answer: a one-parameter family of non-trivial solutions

The number of free variables is \( n - r = 3 - 2 = 1 \), so the solution set is a line through the origin - every scalar multiple of a single non-zero vector. It contains the trivial solution as the point where the parameter is zero, but it is not confined to it, since the rank is less than the number of unknowns.

Answer: A

Each unknown is the ratio of its own numerator determinant to the coefficient determinant:

\[ x = \frac{D_x}{D} = \frac{18}{6} = 3 \]

The rule requires \( D \neq 0 \), which holds here; where it vanishes the rank criteria must be used instead.

Answer: B

Taking determinants of \( A\,\mathrm{adj}A = (\det A) I \) for order \( n \):

\[ \det A \cdot \det(\mathrm{adj}A) = (\det A)^{n} \quad\Rightarrow\quad \det(\mathrm{adj}A) = (\det A)^{n-1} \]

\[ = 2^{3-1} = 2^{2} = 4 \]

For a 2 by 2 matrix the exponent is one and the adjoint's determinant simply equals the matrix's own.