For the matrix M = \begin{pmatrix} k & 3 & 1 \\ 2 & k & 0 \\ 1 & 5 & 2 \end{pmatrix}, the … — JAMB UTME

For the matrix $M = \begin{pmatrix} k & 3 & 1 \\ 2 & k & 0 \\ 1 & 5 & 2 \end{pmatrix}$ , the determinant of $M$ is equal to $4$ . What is the positive value of $k$ ?

Answer: 2

Answer

The positive value of

k

2

Expanding the determinant along the second row yields

\det(M) = -2(6-5) + k(2k-1) = 2k^2 - k - 2

. Setting this equal to

4

yields

2k^2 - k - 6 = 0

, which factors into

(2k+3)(k-2)=0

. The positive value is

2

Step-by-Step Solution

Evaluate the determinant of

M

in terms of

k

using row 2 cofactor expansion

\det(M) = 2k^2 - k - 2

Expanding along the second row gives

-2(6 - 5) + k(2k - 1) - 0 = -2 + 2k^2 - k

Set the determinant equal to the given value 4 and rearrange into standard quadratic form

2k^2 - k - 6 = 0

Subtracting 4 from both sides gives

2k^2 - k - 6 = 0

Solve the quadratic equation by factorization

k = -1.5 \text{ or } k = 2

Factoring

(2k + 3)(k - 2) = 0

gives roots

k = -1.5

and

k = 2

Select the positive root

k = 2

The question specifically asks for the positive value of

k

Key Concept

Evaluating a 3x3 matrix determinant using cofactor expansion and solving the resulting quadratic equation for an unknown parameter.

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic