Matrices and Determinants Questions and Answers

Answer: C

A single interchange reverses the sign, so an even number of interchanges restores the original value. It is why a determinant with two identical rows must be zero, an interchange leaving it unchanged and yet obliged to change sign.

Answer: D

Interchanging the identical rows must change the sign while leaving the determinant unaltered, and only zero satisfies both. The same conclusion follows from the rows being linearly dependent, which makes the matrix singular.

Answer: A

This invariance is precisely what makes Gaussian elimination usable for evaluating determinants, since the value survives every such step. Only interchanges, which flip the sign, and scaling a row, which scales the value, alter it.

Answer: B

Expanding along the first column repeatedly leaves only the diagonal terms, since every other cofactor carries a zero factor. It is why reduction to triangular form is the standard method of evaluating a large determinant.

Answer: C

The multiplicative property holds for square matrices of the same order and is one of the most useful facts about determinants. It immediately shows that a product is singular if either factor is, and it gives the determinant of an inverse as the reciprocal.

Answer: D

Correct answer: the same as that of the matrix

\( |A^{T}| = |A| \). The determinant can be expanded along a row or along a column with the same result, and transposing simply exchanges the two, so the value cannot change. A useful consequence is that every row property of determinants has an identical column counterpart, which halves the number of rules to remember.

Answer: A

Correct answer: k raised to the power n, times the determinant of A

\[ |kA| = k^{n}|A| \] for a matrix of order \( n \), because multiplying the matrix by \( k \) multiplies every one of its \( n \) rows by \( k \), and each row scaling takes a factor of \( k \) out of the determinant. The common error is to take out a single factor, giving \( k|A| \), which is offered here.

Answer: B

The sign is positive when the sum of the row and column indices is even and negative when odd, giving the familiar chequerboard pattern. Expansion of the determinant is the sum of the products of a row's elements with their own cofactors.

Answer: C

Correct answer: as the transpose of the matrix of cofactors

Two steps in order: replace each entry by its cofactor - the minor with the sign \( (-1)^{i+j} \) - and then transpose. Omitting the signs gives the matrix of minors, and omitting the transpose gives the cofactor matrix itself; both are offered here. The adjoint earns its place through \( A^{-1} = \mathrm{adj}(A)/|A| \).

Answer: D

The relation holds whether or not the matrix is invertible, and dividing through by the determinant is exactly how the inverse formula arises. For a singular matrix it shows the product with the adjoint is the zero matrix.