If the determinant of the matrix A = \begin{pmatrix} k & 1 & 2 \\ 3 & -1 & 0 \\ 2 & 4 & 1 … — JAMB UTME

If the determinant of the matrix $A = \begin{pmatrix} k & 1 & 2 \\ 3 & -1 & 0 \\ 2 & 4 & 1 \end{pmatrix}$ is equal to $17$ , what is the value of $k$ ?

Answer: 8

Answer

The value of

k

8

Expanding the matrix determinant along the first row gives

\det(A) = k(-1) - 1(3) + 2(14) = -k + 25

. Setting

-k + 25 = 17

leads to

-k = -8

, so

k = 8

Step-by-Step Solution

Perform cofactor expansion along the first row of matrix

A

\det(A) = k((-1)(1) - (0)(4)) - 1((3)(1) - (0)(2)) + 2((3)(4) - (-1)(2))

Expanding along the first row uses the formula \det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}.

Evaluate the products and simplify the algebraic expression for the determinant.

\det(A) = -k - 3 + 28 = -k + 25

Simplifying each sub-determinant term yields a linear expression in

k

Equate the expression to

17

and solve for

k

-k + 25 = 17 \implies k = 8

Subtracting 25 from both sides gives

-k = -8

, which simplifies to

k = 8

Key Concept

Determinant of a 3x3 Matrix

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