If the matrix P = \begin{pmatrix} k & 3 & 1 \\ 2 & 1 & k \\ 4 & 2 & 0 \end{pmatrix} is sin… — JAMB UTME

If the matrix $P = \begin{pmatrix} k & 3 & 1 \\ 2 & 1 & k \\ 4 & 2 & 0 \end{pmatrix}$ is singular, find the non-zero value of $k$ .

Answer: 6

Answer

The non-zero value of

k

is 6.

For matrix P to be singular, its determinant must be 0. Expanding along row 3 yields 4(3k - 1) - 2(k^2 - 2) = 12k - 4 - 2k^2 + 4 = -2k^2 + 12k = 0. Factoring gives -2k(k - 6) = 0, which yields k = 0 or k = 6. The non-zero value is 6.

Step-by-Step Solution

Set the determinant of matrix P to 0

\det(P) = 0

A matrix is singular if and only if its determinant equals zero.

Evaluate the 3x3 determinant by expanding along the third row

4 \cdot (3k - 1) - 2 \cdot (k^2 - 2) + 0 = 0

Expanding along the third row takes advantage of the zero entry to simplify computation.

Expand and combine like terms

-2k^2 + 12k = 0

12k - 4 - 2k^2 + 4 reduces to -2k^2 + 12k.

Factor out common factors and solve for k

-2k(k - 6) = 0 \implies k = 0 \text{ or } k = 6

Applying the zero-product property.

Select the required root

k = 6

The question specifies the non-zero value of k.

Key Concept

Determinant of a 3x3 matrix and singular matrix condition

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic