Name: The Eigenvectors and Eigenvalues
Uploaded: 2019-07-19
Duration: 17 min 15 s
Description: Studying linear algebra, we will inevitably stumble upon the concept of eigenvalues. These sound very exotic, but they are very important.

Question

If

a^{2} + b^{2} + c^{2} = 1

then show that the value of the determinant

|\begin{matrix} a^{2} + (b^{2} + c^{2}) \cos θ & b a (1 - \cos θ) & c a (1 - \cos θ) \\ a b (1 - \cos θ) & b^{2} + (c^{2} + a^{2}) \cos θ & c b (1 - \cos θ) \\ ac (1 - \cos θ) & b c (1 - \cos θ) & c^{2} + (a^{2} + b^{2}) \cos θ \end{matrix}|

simplifies to

\cos^{2} θ

Accepted Answer

Let $D = |\begin{matrix} a^{2} + (b^{2} + c^{2}) \cos θ & b a (1 - \cos θ) & c a (1 - \cos θ) \\ a b (1 - \cos θ) & b^{2} + (c^{2} + a^{2}) \cos θ & c b (1 - \cos θ) \\ ac (1 - \cos θ) & b c (1 - \cos θ) & c^{2} + (a^{2} + b^{2}) \cos θ \end{matrix}|$

Multiplying $C_{1} by a, C_{2} by b & C_{3} by c$ we get,

$D = \frac{1}{a b c} |\begin{matrix} a (a^{2} + (b^{2} + c^{2}) \cos θ) & b^{2} a (1 - \cos θ) & c^{2} a (1 - \cos θ) \\ a^{2} b (1 - \cos θ) & b (b^{2} + (c^{2} + a^{2}) \cos θ) & c^{2} b (1 - \cos θ) \\ a^{2} c (1 - \cos θ) & b^{2} c (1 - \cos θ) & c (c^{2} + (a^{2} + b^{2}) \cos θ) \end{matrix}|$

Taking common from each row $a from R_{1}, b from R_{2} & c from R_{3}$ we get,

$D = \frac{a b c}{a b c} |\begin{matrix} a^{2} + (b^{2} + c^{2}) \cos θ & b^{2} - b^{2} \cos θ & c^{2} - c^{2} \cos θ \\ a^{2} - a^{2} \cos θ & b^{2} + (c^{2} + a^{2}) \cos θ & c^{2} - c^{2} \cos θ \\ a^{2} - a^{2} \cos θ & b^{2} - b^{2} \cos θ & c^{2} + (a^{2} + b^{2}) \cos θ \end{matrix}|$

$C_{1} \to C_{1} + C_{2} + C_{3}$

$D = |\begin{matrix} a^{2} + b^{2} \cos θ + c^{2} \cos θ + b^{2} - b^{2} \cos θ + c^{2} - c^{2} \cos θ & b^{2} - b^{2} \cos θ & c^{2} - c^{2} \cos θ \\ a^{2} - a^{2} \cos θ + b^{2} + c^{2} \cos θ + a^{2} \cos θ + c^{2} - c^{2} \cos θ & b^{2} + (c^{2} + a^{2}) \cos θ & c^{2} - c^{2} \cos θ \\ a^{2} - a^{2} \cos θ + b^{2} - b^{2} \cos θ + c^{2} + a^{2} \cos θ + b^{2} \cos θ & b^{2} - b^{2} \cos θ & c^{2} + (a^{2} + b^{2}) \cos θ \end{matrix}|$

$∵ a^{2} + b^{2} + c^{2} = 1$

$D = |\begin{matrix} 1 & b^{2} - b^{2} \cos θ & c^{2} - c^{2} \cos θ \\ 1 & b^{2} + (c^{2} + a^{2}) \cos θ & c^{2} - c^{2} \cos θ \\ 1 & b^{2} - b^{2} \cos θ & c^{2} + (a^{2} + b^{2}) \cos θ \end{matrix}|$
$R_{1} \to R_{1} - R_{3}$ & $R_{2} \to R_{2} - R_{3}$ ,

$D = |\begin{matrix} 0 & 0 & - \cos θ \\ 0 & \cos θ & - \cos θ \\ 1 & b^{2} - b^{2} \cos θ & c^{2} + (a^{2} + b^{2}) \cos θ \end{matrix}| = - \cos θ (- \cos θ) = \cos^{2} θ$

Important Questions on Matrices and Determinants

Learn and Prepare for any exam you want !

If a2+b2+c2=1 then show that the value of the determinant a2+b2+c2cosθba(1-cosθ)ca(1-cosθ)ab(1-cosθ)b2+c2+a2cosθcb(1-cosθ)ac(1-cosθ)bc(1-cosθ)c2+a2+b2cosθ simplifies to cos2θ

Important Questions on Matrices and Determinants

Learn and Prepare for any exam you want !