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The value of the determinant `|(x+2,x+3,x+5),(x+4,x+6,x+9),(x+8,x+11,x+15)|` isA. 2B. `-2`C. 3D. `x -1` |
Answer» Correct Answer - B Applying `R_(2) rarr R_(2) - R_(1) and R_(3) rarr R_(3) - R_(2)`, we get `Delta = |(x +2,x +3,x +5),(2,3,4),(4,5,6)|` `rArr Delta = 2|(x,x,x +1),(2,3,4),(1,1,1)| " " [("Applying" R_(1) rarr R_(1) - R_(2)),(and R_(3) rarr R_(3) - R_(2))]` `rArr Delta = 2 |(x,0,1),(2,1,1),(1,0,0)| " " [("Applying " C_(2) rarr C_(2) - C_(1)),(and C_(3) rarr C_(3) - C_(2))]` `rArr Delta = -2` [Expanding along `R_(3)`] |
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