Matrices and Determinants Questions and Answers

Answer: B

A non-zero coefficient determinant guarantees a unique solution, each unknown being a ratio of two determinants. When it vanishes the rule fails and the rank criteria must be used to decide between no solution and infinitely many.

Answer: C

Equivalently it is the number of non-zero rows when the matrix is reduced to echelon form, and the two definitions always agree. It measures how many of the rows are genuinely independent, which is what governs the solvability of the associated system.

Answer: D

Correct answer: none at all

Elementary row operations are invertible, so they cannot create or destroy linear dependence among the rows - the row space is unchanged and therefore so is its dimension. That invariance is exactly what makes Gaussian elimination a legitimate way of finding the rank: the echelon form has a rank that is easy to read off and is the same as the original’s.

Answer: A

If augmenting the matrix raises the rank, one equation contradicts the others and no solution exists. When the ranks agree the system is solvable, uniquely if that rank equals the number of unknowns and otherwise with infinitely many solutions.

Answer: B

Correct answer: when the common rank is less than the number of unknowns

Consistency requires the coefficient and augmented ranks to be equal. If that common rank \( r \) is less than the number of unknowns \( n \), then \( n-r \) of the unknowns can be assigned freely and the rest follow, giving an \( (n-r) \)-parameter family. The rank can never exceed \( n \), so the fourth option describes an impossibility.

Answer: C

Correct answer: the trivial solution with all unknowns zero

Setting every unknown to zero satisfies \( AX = 0 \) whatever \( A \) is, so a homogeneous system is never inconsistent. The only question is whether anything else satisfies it as well, and that is decided by the rank - which is why the interesting question about a homogeneous system is always whether non-trivial solutions exist.

Answer: D

Correct answer: the rank is less than the number of unknowns

With \( r < n \) there are \( n-r \) free variables, so solutions other than the zero vector exist. For a square system this is the same as saying the coefficient determinant vanishes. If instead \( r = n \), the only solution is the trivial one, and having more equations than unknowns does not by itself change anything.

Answer: A

The characteristic equation is a polynomial of degree equal to the order of the matrix, so an n by n matrix has n eigenvalues counted with multiplicity. A non-zero eigenvector exists only where that determinant vanishes.

Answer: B

Both quantities are the coefficient of the second-highest power in the characteristic polynomial, up to sign. It gives a quick arithmetic check on any set of computed eigenvalues, alongside the corresponding product relation.

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.