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Zorluk: ZorMatrices and Determinants

If the matrix P=(k3121k420)P = \begin{pmatrix} k & 3 & 1 \\ 2 & 1 & k \\ 4 & 2 & 0 \end{pmatrix} is singular, find the non-zero value of kk.

Cevap: 6

Cevap

The non-zero value of kk 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.

Adım Adım Çözüm

1
Set the determinant of matrix P to 0
\det(P) = 0
A matrix is singular if and only if its determinant equals zero.
2
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.
3
Expand and combine like terms
-2k^2 + 12k = 0
12k - 4 - 2k^2 + 4 reduces to -2k^2 + 12k.
4
Factor out common factors and solve for k
-2k(k - 6) = 0 \implies k = 0 \text{ or } k = 6
Applying the zero-product property.
5
Select the required root
k = 6
The question specifies the non-zero value of k.

Anahtar Kavram

Determinant of a 3x3 matrix and singular matrix condition
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