Question

Difficulty: HardMatrices and Determinants

If the matrix A=(1022k1314)A = \begin{pmatrix} 1 & 0 & 2 \\ 2 & k & 1 \\ 3 & 1 & 4 \end{pmatrix} is singular, what is the value of kk?

  1. 32\frac{3}{2}Answer
  2. B
    −32-\frac{3}{2}
  3. C
    72\frac{7}{2}
  4. D
    −12-\frac{1}{2}

Answer

32\frac{3}{2}
A matrix is singular if its determinant equals zero. Expanding the determinant of matrix AA along its first row gives 1(4k−1)−0+2(2−3k)=−2k+31(4k - 1) - 0 + 2(2 - 3k) = -2k + 3. Setting −2k+3=0-2k + 3 = 0 yields k=32k = \frac{3}{2}.

Step-by-Step Solution

1
Apply the condition for a singular matrix
A matrix is singular when its determinant equals zero, so det⁡(A)=0\det(A) = 0.
By definition, square matrices with zero determinant are singular.
2
Expand det⁡(A)\det(A) along the first row
det⁡(A)=1(4k−1)−0(8−3)+2(2−3k)=4k−1+4−6k=−2k+3\det(A) = 1(4k - 1) - 0(8 - 3) + 2(2 - 3k) = 4k - 1 + 4 - 6k = -2k + 3.
Laplace expansion along the first row simplifies computation due to the zero entry.
3
Solve for kk
−2k+3=0  ⟹  2k=3  ⟹  k=32-2k + 3 = 0 \implies 2k = 3 \implies k = \frac{3}{2}.
Isolating the variable gives the required value of kk.

Key Concept

Singular Matrix Condition and 3x3 Determinant Expansion
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