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Question:

I know how to solve this problem but I need a faster approach to this problem.

By using properties of determinants prove that the determinant $$\begin{vmatrix}a & \sin x & \cos x \\ -\sin x & -a & 1 \\ \cos x & 1 & a\end{vmatrix}$$ is independent of $x$.

My Approach:

\begin{align*} \text{LHS} &= \begin{vmatrix}a & \sin x & \cos x \\ -\sin x & -a & 1 \\ \cos x & 1 & a\end{vmatrix} \\&= -a^3 \begin{vmatrix}1 & \dfrac{\sin x}{a} & \dfrac{\cos x}{a} \\ \dfrac{\sin x}{a} & a & \dfrac{-1}{a} \\ \dfrac{\cos x}{a} & \dfrac{1}{a} & 1\end{vmatrix}\begin{array}{l}\text{Taking a common $a$ from $R_1$}\\\text{Taking a common $-a$ from $R_2$}\\\text{Taking a common $a$ from $R_3$}\end{array}\end{align*}

2nd part of my solution

Ѕᴀᴀᴅ
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  • "Your approach" seems unfinished - what you have still depends on $x$. Can you please post a complete question before anyone attempts to answer it? –  Feb 13 '18 at 05:28
  • Yes please wait. –  Feb 13 '18 at 05:28
  • @user8734617 I have edited it. –  Feb 13 '18 at 05:32
  • Try just taking the determinant without messing around with the terms or factoring, and then trying to cancel/simplify terms with $x$. You should only need to use one easy trig identity. – Carl Feb 13 '18 at 05:33
  • @Carl but the question demands using the properties. –  Feb 13 '18 at 05:34
  • @CalculusProgrammer You expand in your solution too, so if expanding is fine there's no reason not to expand immediately. Trying to do it without expanding seems silly. – Carl Feb 13 '18 at 05:36

1 Answers1

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If you use the "criss-cross" method for a $3\times3$ determinant $$\left|\matrix{a&b&c\cr d&e&f\cr g&h&i\cr}\right|=aei+bfg+cdh-ceg-afh-bdi$$ you can just write down the answer: $$\eqalign{\det(\hbox{your matrix}) &=-a^3+\cos x\sin x-\sin x\cos x+a\cos^2x+a\sin^2x-a\cr &=-a^3\ .\cr}$$ I really doubt you'll get anything faster than that.

If you haven't seen the "criss-cross" method before, it is easy to remember: copy out the first two columns again, add up these three products $$\def\c#1{\color{red}{#1}} \left|\matrix{\c a&b&c\cr d&\c e&f\cr g&h&\c i\cr}\right|\matrix{a&b\cr d&e\cr g&h\cr}\qquad \left|\matrix{a&\c b&c\cr d&e&\c f\cr g&h&i\cr}\right|\matrix{a&b\cr d&e\cr \c g&h\cr}\qquad \left|\matrix{a&b&\c c\cr d&e&f\cr g&h&i\cr}\right|\matrix{a&b\cr \c d&e\cr g&\c h\cr}$$ and subtract these three products $$\def\c#1{\color{blue}{#1}} \left|\matrix{a&b&\c c\cr d&\c e&f\cr \c g&h&i\cr}\right|\matrix{a&b\cr d&e\cr g&h\cr}\qquad \left|\matrix{a&b&c\cr d&e&\c f\cr g&\c h&i\cr}\right|\matrix{\c a&b\cr d&e\cr g&h\cr}\qquad \left|\matrix{a&b&c\cr d&e&f\cr g&h&\c i\cr}\right|\matrix{a&\c b\cr \c d&e\cr g&h\cr}$$ But NOTE that it only works for the $3\times3$ case.
David
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  • Your solution is great bro.You are really genius.May you get further success in this subject. –  Feb 13 '18 at 05:49