Algebra of Matrices and Determinants Questions and Answers

Answer: B

The matrix that can be obtained is \(\begin{bmatrix}1&1&1\\2&3&4\\1&2&1\end{bmatrix}\).

Every elementary row operation is reversible and multiplies the determinant by a non-zero number (swap: \(-1\); scaling a row by \(k \ne 0\): \(k\); adding a multiple of one row to another: 1). Starting from \(I\), whose determinant is 1, you can therefore reach exactly the invertible matrices, so the test is a non-zero determinant.

\[ \begin{aligned} \text{B: } &1(3 - 8) - 1(2 - 4) + 1(4 - 3) = -2 \\ \text{C: } &1(24 - 20) - 1(16 - 8) + 1(10 - 6) = 0 \\ \text{D: } &1(3 - 4) - 1(-3 - 0) + 1(-2 - 0) = 0 \end{aligned} \]

The matrix of all 1s has three equal rows, so its determinant is 0 as well. Only the second matrix is invertible, so it alone can be reached from the identity.

Answer: D

The value is 135.

For a 3×3 matrix, \(|kA| = k^{3}|A|\) and \(|\text{adj } B| = |B|^{2}\).

\[ \begin{aligned} |B| &= |3A| = 27 \times 5 = 135 \\ \frac{|\text{adj } B|}{|3A|} &= \frac{135^{2}}{135} = 135 \end{aligned} \]

Using \(|3A| = 3|A| = 15\) is the usual slip (the factor 3 multiplies all three rows), and 125 or 25 come from working with \(|A|\) alone.

Answer: E

\(\alpha + \beta = \) -2.

Multiply each row of the 3×2 matrix by the column (7, 8):

\[ \begin{aligned} \alpha &= (-1)(7) + 2(8) = 9 \\ \beta &= 3(7) + (-4)(8) = -11 \\ \text{row 3} &= (-5)(7) + 6(8) = 13 \end{aligned} \]

The third row reproduces the given 13, confirming the arithmetic. So \(\alpha + \beta = 9 - 11 = -2\). Sign slips in a single product produce values such as 18, 21 or -21.

Answer: A

\(k = \) 6.

A matrix is singular when its determinant is zero:

\[ \begin{aligned} (8 - k)(4 - k) - (2)(-2) &= 0 \\ k^{2} - 12k + 32 + 4 &= 0 \\ (k - 6)^{2} &= 0 \end{aligned} \]

So \(k = 6\) (a repeated root). Check: the matrix becomes \(\begin{bmatrix} 2 & 2 \\ -2 & -2 \end{bmatrix}\), whose rows are proportional. Subtracting 4 instead of adding it gives \(k^{2} - 12k + 28 = 0\), which has no integer root among 2 to 5.

Answer: D

\(m = \) -2.

A square system lacks a unique solution exactly when its coefficient determinant is zero:

\[ \begin{aligned} \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & -m \\ 3 & 5 & 3 \end{vmatrix} &= 1(9 + 5m) - 1(6 + 3m) + 1(10 - 9) \\ &= 4 + 2m \end{aligned} \]

Setting \(4 + 2m = 0\) gives \(m = -2\). (Then \(x = 1, y = z = 0\) is one solution among infinitely many, so the system is consistent but not uniquely solvable.) The value 2 comes from a sign slip in the \(-m\) entry.

Answer: A

\(x_{1} + x_{2} =\) 3.

\(X = A^{-1}B\) means \(AX = B\), that is \(x_{1} - x_{2} = 3\) and \(2x_{1} + x_{2} = 6\). Adding the equations:

\[ 3x_{1} = 9 \;\Rightarrow\; x_{1} = 3,\quad x_{2} = 0 \]

so \(x_{1} + x_{2} = 3\). (The same comes from \(A^{-1} = \frac{1}{3}\begin{bmatrix} 1 & 1 \\ -2 & 1 \end{bmatrix}\).) Values such as 6 or 9 come from multiplying by \(A\) instead of \(A^{-1}\).

Answer: A

\(\lambda =\) \(-2\).

A singular matrix has determinant zero:

\[ \begin{aligned} |A| &= 1(8) - (-1)(4\lambda) = 8 + 4\lambda \\ 8 + 4\lambda &= 0 \;\Rightarrow\; \lambda = -2 \end{aligned} \]

\(\lambda = 2\) comes from taking \(-(-1)(4\lambda)\) as \(-4\lambda\), the usual sign slip. The other values leave the determinant non-zero.

Check: with \(\lambda = -2\) the second row is \((-8, 8)\), which is \(-8\) times the first row \((1, -1)\), so the rows are dependent.

Answer: A

\(a =\) 5.

Expand along the first row:

\[ \begin{aligned} |A| &= a(4 - 0) + 1(0 + 1) - a(0 - 1) \\ &= 4a + 1 + a = 5a + 1 \\ 5a + 1 &= 26 \;\Rightarrow\; a = 5 \end{aligned} \]

Dropping the sign of the third cofactor gives \(3a + 1 = 26\), which has no integer answer among the options, and the other values do not satisfy \(5a + 1 = 26\).

Check: with \(a = 5\), the expansion gives \(5 \times 4 + 1 + 5 = 26\).

Answer: B

The order of A is \(2 \times 3\).

For BA to be defined, A must have as many rows as B has columns, which is 2. The product of a \(3 \times 2\) and a \(2 \times k\) matrix is \(3 \times k\), and it must be \(3 \times 3\), so k = 3.

A \(2 \times 2\) matrix would give a \(3 \times 2\) product, and every option with 3 rows cannot be multiplied after a matrix with 2 columns.

Answer: E

The determinant is 6.

Determinants multiply, and \(\det(Q^{-1}) = \frac{1}{\det Q}\).

\[ \begin{aligned} \det(QPQ^{-1}) &= \det Q \cdot \det P \cdot \frac{1}{\det Q} \\ &= \det P \end{aligned} \]

P is upper triangular, so its determinant is the product of the diagonal: \(1 \times 2 \times 3 = 6\). Q never needs to be inverted. 48 is det Q, and 24, 12 or 8 come from multiplying in parts of det Q that cancel.