The value of 1990 1991 19921991 1992 19931992 1993 1994 is

Let A=1990 1991 19921991 1992 19931992 1993 1994 Applying C2→C2−C1,C3→C3−C2 =199011199111199211=0(∴ Two columns are identical )

The value of 1990 1991 19921991 1992 19931992 1993 1994 is

Let A=1990 1991 19921991 1992 19931992 1993 1994 Applying C2→C2−C1,C3→C3−C2 =199011199111199211=0(∴ Two columns are identical )

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$The value of |\begin{array}{lll} 1990 1991 1992 \\ 1991 1992 1993 \\ 1992 1993 1994 \end{array}| is$

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1993

1991

1994

answer is D.

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Detailed Solution

$Let A = |\begin{array}{lll} 1990 1991 1992 \\ 1991 1992 1993 \\ 1992 1993 1994 \end{array}| Applying C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{2} \begin{array}{l} = |\begin{array}{lll} 1990 & 1 & 1 \\ 1991 & 1 & 1 \\ 1992 & 1 & 1 \end{array}| \\ = 0 (∴ Two columns are identical) \end{array}$