Matrices and Determinants Questions and Answers

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 \(k\).

Answer: A

The matrix is singular for \(x = 0\) and \(x = 3\).

The matrix is \(A - xI\) with \(A = \begin{bmatrix} 3 & 2 & 2 \\ 2 & 4 & 1 \\ -2 & -4 & -1 \end{bmatrix}\), so it is singular exactly when \(x\) is an eigenvalue of \(A\). Use the invariants of \(A\):

\[ \begin{aligned} \det A &= 3(0) - 2(0) + 2(0) = 0 \\ \text{trace} &= 3 + 4 - 1 = 6 \\ \text{sum of 2x2 minors} &= 8 + 1 + 0 = 9 \end{aligned} \]

The zero determinant gives one eigenvalue 0. The other two add to 6 and multiply to 9 (the sum of the principal minors equals \(\lambda_{1}\lambda_{2}\) when the third eigenvalue is 0), so both are 3. The characteristic equation is \(x(x - 3)^{2} = 0\), giving \(x = 0\) and \(x = 3\).

The values -3, 6 and -6 are not roots: 6 is the trace, the sum of all three eigenvalues, not one of them, and the negative values have the wrong sign.

Answer: A

The determinant is -81.

For square matrices of order n, \(|kA| = k^{n}|A|\) and \(|AB| = |A||B|\). Here n = 3:

\[ \begin{aligned} |3AB| &= 3^{3}\,|A|\,|B| \\ &= 27 \times (-1) \times 3 = -81 \end{aligned} \]

-9 comes from multiplying the determinant by 3 instead of \(3^{3}\), as if the scalar multiplied only one row. -27 leaves out |B|. 81 drops the minus sign of |A|.

Answer: B

The values are a = 1, b = 4.

Expanding, \((A + B)^{2} = A^{2} + AB + BA + B^{2}\), so the condition holds exactly when \(AB + BA = 0\). Matrix products do not commute, so AB + BA is not 2AB.

\[ \begin{aligned} AB &= \begin{bmatrix} a - b & 2 \\ 2a - b & 3 \end{bmatrix} \\ BA &= \begin{bmatrix} a + 2 & -a - 1 \\ b - 2 & 1 - b \end{bmatrix} \end{aligned} \]

\[ AB + BA = \begin{bmatrix} 2a - b + 2 & 1 - a \\ 2a - 2 & 4 - b \end{bmatrix} \]

The top-right entry gives a = 1 and the bottom-right gives b = 4; then 2a - b + 2 = 0 and 2a - 2 = 0 check. The other pairs fail at once: a = 4 or a = 0 or a = 2 leaves 1 - a non-zero.