(D) Given the determinant equation: $\left| \begin{array}{ccc} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -1 \end{array} \right| = 0$ Expanding along the firs

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(D) Given the determinant equation: $\left| \begin{array}{ccc} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -1 \end{array} \right| = 0$ Expanding along the first row: $1(k(-1) - 3(-2)) - k(3(-1) - 2(-2)) + 3(3(3) - 2(k)) = 0$ $1(-k + 6) - k(-3 + 4) + 3(9 - 2k) = 0$ $-k + 6 - k(1) + 27 - 6k = 0$ $-k + 6 - k + 27 - 6k = 0$ $-8k + 33 = 0$ $8k = 33$ $k = \frac{33}{8}$ Since $\frac{33}{8}$ is not among the given options,the correct choice is $(d)$.

Class 12 Mathematics 3-4. Matrices and Determinants Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $\left| \begin{array}{ccc} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -1 \end{array} \right| = 0$,then the value of $k$ is

IIT 1979 All IIT

A
$-1$
B
$0$
C
$1$
D
None of these

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If the system of equations
$(k+1)^3 x + (k+2)^3 y = (k+3)^3$
$(k+1) x + (k+2) y = k+3$
$x + y = 1$
is consistent,then the value of $k$ is