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Zorluk: OrtaMatrices and Determinants

Given the matrix A=(12034120k)A = \begin{pmatrix} 1 & 2 & 0 \\ 3 & 4 & 1 \\ 2 & 0 & k \end{pmatrix}, if det(A)=10\det(A) = 10, what is the value of kk?

Cevap: -3

Cevap

The value of kk is 3-3.
Expanding the determinant along the first row gives 1(4k)2(3k2)+0=2k+41(4k) - 2(3k - 2) + 0 = -2k + 4. Setting 2k+4=10-2k + 4 = 10 leads directly to 2k=6-2k = 6, giving k=3k = -3.

Adım Adım Çözüm

1
Expand the determinant of matrix AA along the first row.
\det(A) = 1(4k - 0) - 2(3k - 2) + 0 = -2k + 4
Using cofactor expansion along the top row to find the expression for the determinant.
2
Set the calculated determinant equal to the given value and solve for kk.
-2k + 4 = 10 \implies -2k = 6 \implies k = -3
Equating the determinant algebraic expression to 10.

Anahtar Kavram

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