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Question

For how many values of x in the closed interval [4,1] the matrix 3x1231x+2x+312 is singular.

A
1
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B
2
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C
4
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D
3
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Solution

The correct option is A 1
A matrix is singular only if its determinant is zero
∣ ∣3x1231x+2x+312∣ ∣=03(2+x+2)+(x+1)(6(x+2)(x+3))+2(3+x+3)=03x+(x+1)(x25x)+2x=0x3+6x2=0x=0,6
0[4,1]
Hence, option A is correct.

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