Determinant Problem

Question

Question:

My problem:

I can't understand how does the hint work.I also want to know the best method to solve this problem.

Answer 1

There is a typo in the hint which should be $R_1-\sin x \cdot R_3$ and $R_2+a\cdot R_3$ (the first of these has $R_2$ in the text). This may explain your confusion. And the result is wrong even then - the hint has two significant errors.

The strategy being used is to use elementary row operations to make most of the second column zero, then to expand the determinant using the second column, because most of the components are zero.

Here it seems as easy just to expand the determinant using the first row.

Answer 2

Using the Laplace expansion formula for the first line (which is a property of determinants) to obtain:

$$\det \boldsymbol{A}=a(-a^2-1)-\sin x \left[-a\sin x -\cos x \right]+\cos x\left[-\sin x + a\cos x \right]$$ $$=-a^3-a+a\sin^2 x +\sin x \cos x- \sin x \cos x+a\cos^2 x$$ $$=-a^3-a+a(\sin^2 x +\cos^2 x)=-a^3-a+a\cdot 1=-a^3.$$

Edit for row operations:

If $R_1 = \begin{bmatrix} a & \sin x & \cos x\end{bmatrix}$ and $R_3 = \begin{bmatrix} \cos x & 1 & a\end{bmatrix}$ then $$-\sin x R_3= \begin{bmatrix} -\sin x\cos x & -\sin x & -a\sin x\end{bmatrix}$$

hence, if we add $-\sin x R_3$ to the first row $R_1$ we obtain:

$$R_1 -\sin xR_3=\begin{bmatrix} a-\sin x\cos x & 0 & \cos x-a\sin x\end{bmatrix}.$$

Then you obtain the new first row.

Answer 3

by the rule of Sarrus we get$$-a^3+\sin(x)\cos(x)-\cos(x)\sin(x)+a^2\sin^2(x)-a+a\cos^2(x)$$