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Zorluk: OrtaMatrices and Determinants

Two 2×22 \times 2 matrices are given as A=(2−143)A = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix} and B=(102k)B = \begin{pmatrix} 1 & 0 \\ 2 & k \end{pmatrix}. If the determinant of the product matrix ABAB is equal to 3030, what is the value of kk?

Cevap: 3

Cevap

The value of k is 3.
The determinant of a product of square matrices equals the product of their individual determinants: det⁡(AB)=det⁡(A)⋅det⁡(B)\det(AB) = \det(A) \cdot \det(B). Evaluating det⁡(A)\det(A) yields (2)(3)−(−1)(4)=10(2)(3) - (-1)(4) = 10, and det⁡(B)\det(B) yields (1)(k)−(0)(2)=k(1)(k) - (0)(2) = k. Substituting these into det⁡(AB)=30\det(AB) = 30 gives 10k=3010k = 30, leading directly to k=3k = 3.

Adım Adım Çözüm

1
Evaluate the determinant of matrix A
\det(A) = 10
Using the 2×22 \times 2 determinant formula det⁡(abcd)=ad−bc\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc, we calculate det⁡(A)=(2)(3)−(−1)(4)=6+4=10\det(A) = (2)(3) - (-1)(4) = 6 + 4 = 10.
2
Evaluate the determinant of matrix B
\det(B) = k
Calculating the determinant of matrix BB yields det⁡(B)=(1)(k)−(0)(2)=k\det(B) = (1)(k) - (0)(2) = k.
3
Use the product property of determinants to solve for k
k = 3
Since det⁡(AB)=det⁡(A)⋅det⁡(B)\det(AB) = \det(A) \cdot \det(B), we have 10k=3010k = 30. Dividing both sides by 10 yields k=3k = 3.

Anahtar Kavram

Determinant of Matrix Product
Tahmini Süre:1m 30s
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