Determinants and Cramer's Rule: Additional Practice (Set 2)

A unique solution requires a nonzero coefficient determinant.

$D = \begin{vmatrix} 2 & k \\ 3 & 6 \end{vmatrix} = 12 - 3k$

This is zero only at $k = 4$, so the system has a unique solution for every $k \ne 4$.

A unique solution requires a nonzero coefficient determinant.

$D = \begin{vmatrix} 1 & k \\ k & 9 \end{vmatrix} = 9 - k^2$

This is zero when $k = 3$ or $k = -3$, so the system has a unique solution for $k \ne 3$ and $k \ne -3$.

The determinant $\begin{vmatrix} 2 & 4 \\ 3 & 6 \end{vmatrix} = 0$, so inspect the equations. Dividing the first by $2$ gives $x + 2y = 3$, and dividing the second by $3$ gives $x + 2y = \tfrac{10}{3}$.

$x + 2y = 3 \quad\text{versus}\quad x + 2y = \tfrac{10}{3}$

The lines are parallel and distinct, so the system has no solution.

The determinant is again zero, so compare the equations. Multiplying the first by $-2$ gives $-2x + 4y = -6$, which conflicts with the second.

$-2x + 4y = -6 \quad\text{versus}\quad -2x + 4y = 1$

The lines are parallel and distinct, so there is no solution.

Expand along the first row, carefully tracking the signs of the negative entries.

$2(0 \cdot 1 - (-2)(1)) - (-1)(1 \cdot 1 - (-2)(4)) + 3(1 \cdot 1 - 0 \cdot 4) = 2(2) + 1(9) + 3(1) = 16$

The cofactor signs alternate along the first row, starting from plus.

$+, \ -, \ +$

The middle term is subtracted, which is the sign most often dropped by mistake.

Cramer's rule would need eleven $10 \times 10$ determinants, and a cofactor expansion of each fans out into an enormous number of terms.

$\text{cofactor expansion cost grows about like } n!$

Row reduction grows far more slowly, so it is the efficient choice for large systems.

Compute $D = \begin{vmatrix} 3 & -4 \\ 2 & 1 \end{vmatrix} = 3 + 8 = 11$ and $D_y = \begin{vmatrix} 3 & 18 \\ 2 & 1 \end{vmatrix} = 3 - 36 = -33$.

$y = \frac{D_y}{D} = \frac{-33}{11} = -3$

The second row is exactly twice the first, which forces the determinant to zero. Expanding confirms it.

$1(4 \cdot 5 - 6 \cdot 0) - 2(2 \cdot 5 - 6 \cdot 1) + 3(2 \cdot 0 - 4 \cdot 1) = 20 - 8 - 12 = 0$

A zero determinant only rules out the unique-solution case; it cannot by itself decide between no solution and infinitely many.

$D = 0 \;\Rightarrow\; \text{no unique solution guaranteed}$

So the only safe conclusion is that a unique solution is not guaranteed.

Only the biconditional holds. When $D = 0$ the solution is not unique, but it can be no solution or infinitely many, so neither one-directional claim about $D = 0$ is always true.

$\text{unique solution} \iff D \ne 0$

Expand along the first row.

$4(1 \cdot 3 - 2 \cdot 5) - 2(3 \cdot 3 - 2 \cdot 2) + 1(3 \cdot 5 - 1 \cdot 2) = 4(-7) - 2(5) + 13 = -25$