Algebra of Matrices and Determinants Questions and Answers

Answer: A

The relation is \(P^{T} = P\).

Transposition distributes over a sum and leaves a scalar factor alone, and \((A^{T})^{T} = A\).

\[ P^{T} = \frac{1}{2}(A^{T} + A) = P \]

So P is the symmetric part of A, whatever A is; the entries of A play no role. \(P^{T} = -P\) describes the skew-symmetric part \(\frac{1}{2}(A - A^{T})\), and the multiples 2P, -2P and 3P cannot hold for a non-zero P.

Answer: D

The product that is not defined is \((AB)C\).

A product XY is defined when the number of columns of X equals the number of rows of Y. AB is \(3 \times 3\) times \(3 \times 2\), giving a \(3 \times 2\) matrix; multiplying it by the \(3 \times 1\) matrix C needs 2 = 3, which fails.

The others are defined: \(C^{T}AB\) is \((1 \times 3)(3 \times 3)(3 \times 2) = 1 \times 2\); \(A^{T}AB\) is \(3 \times 2\); \((AB)^{T}C\) is \((2 \times 3)(3 \times 1) = 2 \times 1\); and \(B^{T}C\) is \((2 \times 3)(3 \times 1) = 2 \times 1\).

Answer: A

\(|adj(adj A)|\) equals 6561.

For a square matrix of order n, \(|adj A| = |A|^{n-1}\), so applying it twice gives

\[ |adj(adj A)| = |A|^{(n-1)^{2}} = 9^{4} = 6561 \]

for \(n = 3\). Using \(|A|^{n-1}\) only once would give 81. The other options are near-misses of \(9^{4}\) that come from no rule.

Answer: C

The values are −2, −3.

The determinant is lower triangular (all entries above the diagonal are zero), so it equals the product of the diagonal entries:

\[ \begin{aligned} 1 \cdot (x + 2)(x + 3) &= 0 \\ &\Rightarrow x \\ &= -2 \text{ or } x \\ &= -3 \end{aligned} \]

The options 2, 3 and −2, 3 have the wrong signs, and no factor gives 1 or −1.

Answer: B

\(x = -14\).

\(I + A = \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix}\). The entry in row 2, column 2 of the product is row 2 of \(I + A\) times column 2 of the other matrix:

\[ x = 3(-3) + 5(-1) = -9 - 5 = -14 \]

Check the given entries: row 1 gives \(1(4) + 2(2) = 8\) and \(1(-3) + 2(-1) = -5\), and row 2, column 1 gives \(3(4) + 5(2) = 22\), all matching. 14 drops the sign, and ±12 forget the identity added to A.

Only one entry of the product is needed, so there is no need to multiply the whole matrices out.

Answer: A

\(a + b + c\) is 6.

Expand the determinant along the first row:

\[ \begin{aligned} \Delta &= a(bc - 1) - (c - 1) + (1 - b) \\ &= abc - a - b - c + 2 \end{aligned} \]

Setting \(\Delta = 2\) gives \(abc = a + b + c\). For positive integers the only solution, up to order, is 1, 2, 3, since \(1 \times 2 \times 3 = 1 + 2 + 3 = 6\). Taking all three equal to 1 gives 1 = 3, and larger values make the product outgrow the sum. So a + b + c = 6.

Answer: D

The value of \(\Delta\) is 0.

Expanding the determinant along the first row and collecting terms gives

\[ \Delta = -(x^{3} + y^{3} - 3xy + 1) \]

This is \(-(x^{3} + y^{3} + c^{3} - 3xyc)\) with \(c = 1\), and the identity \(p^{3} + q^{3} + r^{3} - 3pqr = (p + q + r)(\ldots)\) shows it vanishes whenever the three numbers add to zero. Here \(x + y + 1 = 0\) because \(x + y = -1\). Directly: \(x^{3} + y^{3} = (x + y)^{3} - 3xy(x + y) = -1 + 3xy\), so the bracket is \(-1 + 3xy - 3xy + 1 = 0\).

Answer: B

\(|adj(adj A)|\) equals 256.

A is upper triangular, so its determinant is the product of the diagonal: \(|A| = 1 \times 2 \times (-2) = -4\). For an n-by-n matrix, \(|adj A| = |A|^{n-1}\), and applying this twice:

\[ \begin{aligned} |adj(adj A)| &= |A|^{(n-1)^{2}} = (-4)^{4} \\ &= 256 \end{aligned} \]

With n = 3 the exponent is 4, an even power, so the sign disappears; -256 keeps a sign that should not be there. 16 is \(|adj A| = (-4)^{2}\), one step only.

Answer: A

A unique solution needs \(\lambda \neq 1\).

A linear system has a unique solution exactly when the determinant of its coefficient matrix is non-zero:

\[ \begin{aligned} \begin{vmatrix} 1 & -2 & 1 \\ 2 & -1 & 2 \\ 1 & 2 & \lambda \end{vmatrix} &= 1(-\lambda - 4) + 2(2\lambda - 2) + 1(5) \\ &= 3\lambda - 3 \end{aligned} \]

This is zero only for \(\lambda = 1\), so for a unique solution \(\lambda\) must not be 1. Any of the other listed values gives a non-zero determinant, so they are all allowed.

Answer: B

\(x = 4\), \(y = 2\) and \(t = -3\).

A symmetric matrix equals its transpose, so the entry in row \(i\), column \(j\) equals the entry in row \(j\), column \(i\):

\[ \begin{aligned} a_{13} &= a_{31}: \quad x = 4 \\ a_{21} &= a_{12}: \quad y = 2 \\ a_{32} &= a_{23}: \quad t = -3 \end{aligned} \]

4,2,3 drops the sign of \(-3\); 4,2,\(-7\) wrongly takes \(t\) from the diagonal entry \(-7\); 2,4,\(-7\) swaps \(x\) and \(y\) as well; 4,3,2 has no basis in the matrix.