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15/01/2022 3:36 pm
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Let A = \(\begin{pmatrix} [x+1] & [x+2] & [x+3] \\[0.3em] [x] & [x+3] & [x+3] \\[0.3em] [x] & [x+2] & [x+4] \end{pmatrix}\), where [t] denotes the greatest integer less than or equal to t. If det(A) = 192, then the set of values of x is the interval:
(1) [68, 69)
(2) [62, 63)
(3) [65, 66)
(4) [60, 61)
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15/01/2022 3:42 pm
Correct answer: (2) [62, 63)
Explanation:
\(\begin{vmatrix} [x+1] & [x+2] & [x+3] \\[0.3em] [x] & [x+3] & [x+3] \\[0.3em] [x] & [x+2] & [x+4] \end{vmatrix}\) = 192
R1 → R1 - R3 & R2 → R2 - R3
\(\begin{bmatrix}1 & 0 & -1 \\[0.3em] 0 & 1 & -1 \\[0.3em] [x] & [x]+2 & [x]+4 \end{bmatrix}\) = 62
2[x] + 6 + [x] = 192
⇒ [x] = 62