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.

The problem gives a \(3 \times 3\) matrix

\(A = \begin{pmatrix} -2 & 3 & 1 \\ p & 2 & 1 \\ 1 & 4 & 2\end{pmatrix}\)

and tells us that the determinant of this matrix is \(-5\). We are to find the value of \(p\).

Step 1: Formula for the determinant of a 3x3 matrix
For any matrix \[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \] the determinant is calculated as: \[ \det = aei + bfg + cdh - ceg - bdi - afh \]

Step 2: Enter the values from matrix A

• \(a = -2\), \(b = 3\), \(c = 1\)
• \(d = p\), \(e = 2\), \(f = 1\)
• \(g = 1\), \(h = 4\), \(i = 2\)

Plug them into the determinant formula: \[ \det(A) = (-2)(2)(2) + (3)(1)(1) + (1)(p)(4) - (1)(2)(1) - (3)(p)(2) - (-2)(1)(4) \]

Step 3: Simplify each term

• \((-2)(2)(2) = -8\)
• \((3)(1)(1) = 3\)
• \((1)(p)(4) = 4p\)
• \((1)(2)(1) = 2\)
• \((3)(p)(2) = 6p\)
• \((-2)(1)(4) = -8\)

\[ \det(A) = [-8 + 3 + 4p] - [2 + 6p - 8] \]

But recall, the formula is all the "positive" terms minus all the "negative" terms: \[ \det(A) = (-8) + 3 + 4p - 2 - 6p + 8 \]

Step 4: Combine like terms

• Combine the numbers:
  • \(-8 + 3 = -5\)
  • \(-5 - 2 = -7\)
  • \(-7 + 8 = 1\)
• Combine the \(p\) terms:
  • \(4p - 6p = -2p\)

So the determinant simplifies to: \[ \det(A) = 1 - 2p \]

Step 5: Set equal to -5 and solve for \(p\)

• Subtract 1 from both sides: \[ -2p = -5 - 1 = -6 \]
• Divide both sides by -2: \[ p = \frac{-6}{-2} = 3 \]

Conclusion: The value of \(p\) that makes the determinant equal to \(-5\) is \(3\).
This is because, with \(p=3\), plugging back into the formula gives \[ \det(A) = 1 - 2 \times 3 = 1 - 6 = -5 \] exactly as required.