Evaluate Δ=0−sinαcosαsinα0−sinβ−cosαsinβ0 Solution Expanding along R1, we get $\begin{aligned}
\Delta & =0\left|\begin{array}{cc}
0 & \sin \beta
-\sin \beta & 0
\end{array}\right|-\sin \alpha\left|\begin{array}{cc}
-\sin \alpha & \sin \beta
\cos \alpha & 0
\end{array}\right|-\cos \alpha\left|\begin{array}{cc}
-\sin \alpha & 0
\cos \alpha & -\sin \beta
\end{array}\right|
& =0-\sin \alpha(0-\sin \beta \cos \alpha)-\cos \alpha(\sin \alpha \sin \beta-0)
& =\sin \alpha \sin \beta \cos \alpha-\cos \alpha \sin \alpha \sin \beta=0
\end{aligned}
\\
Get the complete, step-by-step math solution for: "Evaluate = {ccc}0 & sin α & -cos α \\ -sin α & 0 & sin β \\ cos α & -sin β & 0 Solution Expanding along {R}_{1}, we get & =0 {cc} 0 & sin β -sin β & 0...". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Expand along the first row
To evaluate the determinant of a
3×3
matrix, we can expand along any row or column. Here, we choose to expand along the first row (
R1
). The formula for expansion along
R1
is
a11C11+a12C12+a13C13
, where
aij
are the elements and
Cij
are their cofactors. The cofactors include a sign factor
(−1)i+j
.
Step 2: Calculate the 2x2 determinants
Now, we evaluate each of the
2×2
determinants. The determinant of a
2×2
matrix
acbd
is given by ad - bc. We apply this formula to each of the three
2×2
sub-determinants.
Step 3: Simplify the expression
We simplify the terms inside the parentheses. The first term becomes
0(0+sin2β)=0
. The second term simplifies to
−sinα(−sinβcosα)
. The third term simplifies to
−cosα(sinαsinβ)
.
Step 4: Combine and find the final value
After simplifying, we get two terms:
sinαsinβcosα
and
−cosαsinαsinβ
. These two terms are identical but with opposite signs.
Step 5: Final Result
Since the two terms are identical with opposite signs, they cancel each other out, resulting in a final determinant value of
0
.