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Difficulty: Very hardMatrices and Determinants

Given the 3×33 \times 3 matrix M=(k213k−12141)M = \begin{pmatrix} k & 2 & 1 \\ 3 & k-1 & 2 \\ 1 & 4 & 1 \end{pmatrix}, where k>5k > 5. If det⁡(M2)=100\det(M^2) = 100 and det⁡(M)<0\det(M) < 0, what is the value of kk?

Answer: 7

Answer

7
Using the matrix determinant power identity det⁡(M2)=(det⁡(M))2=100\det(M^2) = (\det(M))^2 = 100 and the condition det⁡(M)<0\det(M) < 0, we find det⁡(M)=−10\det(M) = -10. Expanding det⁡(M)\det(M) along the first row yields det⁡(M)=k(k−9)−2(1)+1(13−k)=k2−10k+11\det(M) = k(k-9) - 2(1) + 1(13-k) = k^2 - 10k + 11. Setting this equal to −10-10 gives the quadratic equation k2−10k+21=0k^2 - 10k + 21 = 0, which factors as (k−3)(k−7)=0(k-3)(k-7) = 0. Given k>5k > 5, the unique solution is k=7k = 7.

Step-by-Step Solution

1
Apply determinant properties for matrix powers
det⁡(M)=−10\det(M) = -10
Because det⁡(M2)=(det⁡(M))2=100\det(M^2) = (\det(M))^2 = 100 and it is given that det⁡(M)<0\det(M) < 0, taking the negative square root gives det⁡(M)=−10\det(M) = -10.
2
Evaluate the determinant of matrix MM using first-row expansion
det⁡(M)=k2−10k+11\det(M) = k^2 - 10k + 11
Expanding along row 1 gives k((k−1)(1)−8)−2(3(1)−2(1))+1(3(4)−(k−1)(1))=k2−9k−2+13−k=k2−10k+11k((k-1)(1) - 8) - 2(3(1) - 2(1)) + 1(3(4) - (k-1)(1)) = k^2 - 9k - 2 + 13 - k = k^2 - 10k + 11.
3
Set up and simplify the quadratic equation for kk
k2−10k+21=0k^2 - 10k + 21 = 0
Equating k2−10k+11k^2 - 10k + 11 to −10-10 yields k2−10k+21=0k^2 - 10k + 21 = 0.
4
Solve the quadratic equation and enforce the inequality condition k>5k > 5
k=7k = 7
Factoring (k−3)(k−7)=0(k-3)(k-7) = 0 gives k=3k = 3 or k=7k = 7. Applying the restriction k>5k > 5 selects k=7k = 7.

Key Concept

Determinant Properties of Matrix Powers and 3x3 Matrix Expansion
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Topics

Matrices and Determinants

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