The determinant of the matrix  $A = \begin{pmatrix} -2 & 3 & 1 \\ p & 2 & 1 \\ 1 & 4 & 2\end{pmatrix}$ is -5. Find the value of p.

Analysis

To find the value of $p$, we need to calculate the determinant of the $3 \times 3$ matrix $A$ and set it equal to $-5$. The determinant of a matrix $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$ is calculated using the expansion formula: $|A| = a(ei - fh) - b(di - fg) + c(dh - eg)$

Concept

The determinant of a matrix is a scalar value that provides information about the matrix's properties. For a $3 \times 3$ matrix, we expand along the first row. Remember to alternate the signs of the elements: $(+), (-), (+)$.

Step-by-Step Solution

Given matrix $A = \begin{pmatrix} -2 & 3 & 1 \\ p & 2 & 1 \\ 1 & 4 & 2 \end{pmatrix}$ and $|A| = -5$.

1. Set up the expansion along the first row: $|A| = -2 \begin{vmatrix} 2 & 1 \\ 4 & 2 \end{vmatrix} - 3 \begin{vmatrix} p & 1 \\ 1 & 2 \end{vmatrix} + 1 \begin{vmatrix} p & 2 \\ 1 & 4 \end{vmatrix} = -5$

2. Calculate the $2 \times 2$ determinants:

• First part: $-2[(2 \times 2) - (1 \times 4)] = -2[4 - 4] = -2(0) = 0$
• Second part: $-3[(p \times 2) - (1 \times 1)] = -3[2p - 1] = -6p + 3$
• Third part: $+1[(p \times 4) - (2 \times 1)] = 1[4p - 2] = 4p - 2$

3. Combine the results: $0 + (-6p + 3) + (4p - 2) = -5$ $-6p + 4p + 3 - 2 = -5$ $-2p + 1 = -5$

4. Solve for $p$: $-2p = -5 - 1$ $-2p = -6$ $p = \frac{-6}{-2}$ $p = 3$

Conclusion

Comparing our result with the provided options, we find that $p = 3$ matches option C.

Exam Tip

When calculating determinants, always be extremely careful with the negative signs in the expansion formula (the middle term is always subtracted). A common mistake is forgetting to distribute the negative sign across the entire bracketed expression!