Matrices and Determinants Questions and Answers

Answer: A

The system has no solution.

Subtract the first equation from the second:

\[ \begin{aligned} (3x - 3y + 4z) - (2x - y + z) &= 6 - 1 \\ x - 2y + 3z &= 5 \end{aligned} \]

But the third equation says \(x - 2y + 3z = 4\). The same expression cannot equal both 5 and 4, so the equations are inconsistent. In matrix terms, the coefficient matrix has rank 2 while the augmented matrix has rank 3.

A consistent system with a rank-deficient matrix would have infinitely many solutions, not exactly two or three, and a unique solution would need the coefficient matrix to be non-singular, which it is not here.

Answer: A

The system has infinitely many solutions.

It is homogeneous (every right-hand side is 0), so \(x = y = z = 0\) always works. It has other solutions as well exactly when the coefficient determinant is zero:

\[ \begin{aligned} \Delta &= 7(1 + 1) + 3(-3 - 1) + 1(-3 + 1) \\ &= 14 - 12 - 2 = 0 \end{aligned} \]

The rank is 2 (the first two rows are not multiples of each other), so the solutions form a line through the origin: for example \(x = 1, y = 2, z = -1\) and any multiple of it.

A unique solution would need \(\Delta \neq 0\). A homogeneous system can never have no solution, and a linear system cannot have exactly three.

Answer: D

The transpose of a symmetric matrix is \(A\) itself.

A matrix is symmetric precisely when \(a_{ij} = a_{ji}\) for all i and j, which is the statement \(A^{T} = A\). Transposing swaps rows and columns, and for a symmetric matrix that swap changes nothing.

\(A^{T} = -A\) defines a skew-symmetric matrix, not a symmetric one. \(A^{T} = A^{-1}\) defines an orthogonal matrix, and a symmetric matrix need not be invertible at all. The transpose of A is the null matrix only when A itself is the null matrix.

Answer: B

The system has infinitely many solutions.

It is homogeneous (all right-hand sides zero), so \(x = y = z = 0\) is always a solution, and there are other solutions exactly when the coefficient determinant vanishes:

\[ \begin{aligned} \det &= 2(1 \cdot 1 - 0) - 3(1 \cdot 1 - 0) + 1(1 - 0) \\ &= 2 - 3 + 1 = 0 \end{aligned} \]

Indeed the first equation is the sum of the others plus one copy of the second: \((x + y) + (x + y) + (y + z) = 2x + 3y + z\). With y = t, x = −t and z = −t for any t. A homogeneous system always has the zero solution, so 'no solution' is impossible, and a linear system cannot have a finite number greater than one.

Answer: C

The eigenvalues are 1 and 3.

The matrix is upper triangular, and the eigenvalues of a triangular matrix are its diagonal entries. Formally:

\[ \det\begin{bmatrix} 1 - \lambda & 2 \\ 0 & 3 - \lambda \end{bmatrix} = (1 - \lambda)(3 - \lambda) = 0 \]

Check: the eigenvalues sum to the trace, 1 + 3 = 4, and multiply to the determinant, 3. The pairs (0, 2), (2, 3) and (1, 2) fail one or both of these checks; the off-diagonal 2 does not affect the eigenvalues of a triangular matrix.

Answer: D

The FALSE statement is \(\lambda = d + b\) is an eigenvalue.

Multiply \(A\) by the vector of ones:

\[ A\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} a + b \\ c + d \end{bmatrix} = (a + b)\begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

because \(a + b = c + d\). So \([1\ \ 1]^{T}\) is an eigenvector with eigenvalue \(a + b\), and the first two statements are true.

The eigenvalues of a 2 x 2 matrix add up to its trace \(a + d\), so the other eigenvalue is \((a + d) - (a + b) = d - b\), and the third statement is true too. \(d + b\) could equal \(d - b\) only if \(b = 0\), which is excluded, and could equal \(a + b\) only in the special case \(d = a\). In general it is not an eigenvalue, so that statement is the false one.

Answer: C

The matrix is skew symmetric.

A matrix is skew symmetric when \(A^{T} = -A\), that is \(a_{ji} = -a_{ij}\) for every pair, which forces every diagonal entry to be zero. Here the diagonal is 0, 0, 0 and each mirrored pair has opposite signs: \(a_{12} = 2\) and \(a_{21} = -2\); \(a_{13} = -3\) and \(a_{31} = 3\); \(a_{23} = 4\) and \(a_{32} = -4\).

It is not symmetric, because that needs \(a_{ji} = a_{ij}\). It is not diagonal, since its off-diagonal entries are non-zero, and it is not triangular, because it has non-zero entries both above and below the diagonal.

Answer: C

The eigenvalues are 6.3 and 2.7.

They are the roots of the characteristic equation \(\det(A - \lambda I) = 0\):

\[ \begin{aligned} (5 - \lambda)(4 - \lambda) - 3 &= 0 \\ \lambda^{2} - 9\lambda + 17 &= 0 \\ \lambda &= \frac{9 \pm \sqrt{13}}{2} \approx 6.30,\ 2.70 \end{aligned} \]

A quick check: the eigenvalues must add up to the trace, 9, and multiply to the determinant, 17. 6.3 and 2.7 give 9 and 17.01. The negative pairs have the wrong sign of sum, and 2.3 and 6.7 add to 9 but multiply to about 15.4.

Answer: C

The rank is 3.

A 3 × 4 matrix has rank at most 3, so the question is whether the three rows are independent. Row-reduce: subtracting \(\frac{3}{5}\) of row 1 from row 2 and \(\frac{2}{5}\) of row 1 from row 3 gives

\[ \begin{bmatrix} 5 & 3 & -3 & -1 \\ 0 & 1/5 & -1/5 & -2/5 \\ 0 & -11/5 & 16/5 & 42/5 \end{bmatrix} \]

Adding 11 times row 2 to row 3 leaves \((0, 0, 1, 4)\), which is non-zero, so there are three pivots and the rank is 3.

Rank 4 is impossible for a matrix with only three rows, and ranks 1 or 2 would need a row to be a combination of the others, which the elimination rules out.

Answer: A

The determinant is unaltered if its columns and rows are interchanged, that is, on transposition: \(\det(A^{T}) = \det(A)\).

Here a “line” means a row or a column. Expanding along the first row of \(A\) gives the same terms as expanding along the first column of \(A^{T}\), which is why every property of rows also holds for columns.

The other operations change the determinant. If two parallel lines (two rows or two columns) are identical, the determinant becomes zero. Parallel lines of a matrix are separate rows or columns and cannot intersect, so that option describes nothing. Multiplying every element of one line by a factor \(k\) multiplies the determinant by \(k\).