Determinants: questions with answers

Collinearity means the determinant is zero, so $k - 11 = 0$ and $k = 11$. Setting it to $1$ instead -- as though the condition compared the determinant with one rather than with zero -- gives $k = 12$, which puts the three points a hair off the line.

From $A' = -A$ we get $|A| = (-1)^4|A|$. The order is even, so this reads $|A| = |A|$ and forces nothing. So the rule depends entirely on the order being odd or even -- it is not a blanket property of skew-symmetric matrices.

Scalars move freely through a matrix product, so the two of them collect together and the two matrices collect together: $\left(k \cdot \dfrac{1}{k}\right)(AA^{-1})$, which is $1 \times I$. Collecting the scalars as $k \cdot k$ instead would leave $k^2$, and the product would not be $I$ at all.

$A^{-1} = \dfrac{1}{|A|}\text{adj }A$. It comes straight from $A(\text{adj }A) = |A|I$ by dividing both sides by $|A|$, which is exactly the step that needs $|A| \neq 0$. Multiplying instead would give $A(|A|\text{adj }A) = |A|^2 I$, which is true but is not an inverse.

The bars are determinant notation, not absolute value: the chapter says $|A|$ is read as determinant of $A$ and not modulus of $A$. Here the leading diagonal gives 12 and the other 14, so $|A| = -2$ -- negative, and correctly so. The student's 2 is that value with its sign thrown away.

Expanding along any row or column gives the same value -- the chapter works three expansions in full and shows they agree, which is why it can say that expanding along any row or column gives the same value. The arithmetic on the way looks completely different, and that is the only thing that changes.

Transposing changes nothing, so $|A'| = |A|$; then $|B^{-1}| = \dfrac{1}{|B|}$ makes the two factors of $|B|$ cancel, and what is left is $|A| = 7$. Neither operation touches the sign, and the value of $|B|$ never enters the answer at all.

Expanding gives $|A| = x - 8$. Singular means $|A| = 0$, so $x = 8$. Solving $|A| = 1$ instead -- reading 'singular' as 'determinant one' -- gives $x = 9$, exactly one too large.

Area is a magnitude, so the formula takes the modulus: the area is 12 square units. A negative answer is not a smaller area, nor an area measured in some other direction; it is not an area at all.

An expansion pairs each element with its cofactor, and the middle position here is an odd one, so $A_{12} = -2$ while $M_{12} = 2$. Line 1 already writes minors where cofactors belong, and Line 2 substitutes them faithfully. The determinant is 11, not 23.

The subscripts are read in the order row-then-column, so $M_{23}$ deletes row $2$ and column $3$ and comes to 3. Line 1 already names the deletion the wrong way round, and Line 2 follows it faithfully. Reading the subscripts the other way round is the single commonest way to lose a minor.

It is the diagonal pair that is interchanged and the off-diagonal pair that changes sign. Line 1 interchanges the off-diagonal pair, which is the pair that should only change sign. The adjoint is $\begin{bmatrix} 2 & -4 \\ -3 & 1 \end{bmatrix}$.

There is no such thing as dividing by a matrix. Line 2 puts $A$ underneath, which is already the illegal step; Line 3 only carries it along. Every step here has to be a multiplication by $A^{-1}$, applied to each term on the same side.

It means a solution exists -- one or more. Infinitely many solutions is a consistent system, not an inconsistent one; the word says nothing about how many. Nor does it say anything about the constants: a system whose constants are all zero is consistent, because $x = y = 0$ always solves it.

$A$ stands to the left of $X$, so only a left multiplication cancels it. Line 2 multiplies on the right, and Line 3 simply reports what that produced. $BA^{-1}$ is not merely the wrong answer -- with $B$ of order $2 \times 1$ and $A^{-1}$ of order $2 \times 2$, that product does not exist.