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Find the values of $a$, $b$, and $c$ so that $\det(A) = ax^2 + bx + c$ $$\left[\begin{array}{rrr}x&-1&0\\1&x&-1\\1&1&3 \end{array}\right]‎$$

3 Answers 3

3

Determinant: $$D(A)= x(x•3 - (-1•1)) + (3+1)= 3x^2 +x +4$$

Hence we know that $$a=3, b=1, c=4$$

2

On finding determinant of matrix -

det A = $x[x(3) - 1(-1)] - (-1) [1(3) - 1(-1)] + 0[1(1) - 1(x)]$

det A = $x(3x + 1) + 1(3 + 1) + 0(1 - x)$

$= 3x^2 + x + 4$

On comparing,

a = 3, b = 1, c = 4.

2

Offbeat hint: let $P(x)=ax^2+bx+c=\det \left|\begin{matrix} x & -1 & 0 \\ 1 & x & -1 \\ 1 & 1 & 3\,\end{matrix}\right|\;$ then:

  • $$ c = P(0) = \det \left|\begin{matrix} 0 & -1 & 0 \\ 1 & 0 & -1 \\ 1 & 1 & 3\,\end{matrix}\right|\ = 4 $$

  • $$ a = \lim_{x \to \infty} \frac{1}{x^2}P(x) = \det \left|\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 3\,\end{matrix}\right|\ = 3 $$

  • $b$ is left as an exercise to the reader.

Note: the above is technically correct, though most likely not the expected answer. Yet since the OP gave no context, background, or even a hint of own effort, then it should be fair game, wouldn't it.

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    why the first value is C and second is A and third is B? istead of A, B C? – 2017-02-07
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    @Noname For the first value, just replace $x$ with $0$ everywhere. For the second one, divide the first two columns by $x$ then note that $\frac{1}{x} \to 0$ when $x \to \infty$. P.S. If you have to ask, you most likely do *not* want to use such an approach. I wrote that much at the end of the post itself. Next time you ask a question, you would improve your chances at good answers by providing more context - grade, course, your own attempts etc. That would allow potential answerers to gauge the type and level of answer you are looking for. – 2017-02-07
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    First, your answer compare to others was the only one which I understood completely. Second I didn't know how to do it at all, and it was my first question which I ask here. So sorry for any mistake. And lastly, I didn't know how to do this question at all... That's why I didn't put anything for answer from my side. But Thanks again for your help... – 2017-02-07
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    @Armando, right now, you've marked **my** answer as best answer. If you think instead this is the best answer, you should mark the answer as best answer. – 2017-02-07
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    @Programmer2134 Yours is the more typical approach, so it's not surprising that it was the accepted one. (Maybe you could have just elaborated a little more on the actual calculation of the determinant - first row expansion by minors, I guess.) – 2017-02-07