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If the matrix $P = \begin{pmatrix}1 & \alpha & 3\\1 & 3 & 3\\2 & 4 & 4\end{pmatrix}$ is the adjoint of a $3\times3$ matrix $A$ with $|A|=4$, find $\alpha$.
We use the property: $|adj(A)| = |A|^{n-1}$ for an $n\times n$ matrix. Here $n=3$, so $|P| = |A|^2 = 16$.
Compute $|P|$ by expanding along row 1 (with unknown $\alpha$):
$|P| = 1(3\cdot4-3\cdot4) - \alpha(1\cdot4-3\cdot2) + 3(1\cdot4-3\cdot2)$
$= 1(0) - \alpha(4-6) + 3(4-6) = 2\alpha - 6$
Setting $2\alpha - 6 = 16 \Rightarrow 2\alpha = 22 \Rightarrow \alpha = \mathbf{11}$
An employee works 40 hours/week at $8/hour. Bonus: 30% of hourly rate per hour of time saved. Standard time: 30 minutes per unit. Last week: 40 hours worked, 90 units produced. What were her total earnings?
Option B ($332) is correct.
Basic pay: 40 × $8 = $320
Allowed time for 90 units = 45 hours; Actual = 40 hours; Time saved = 5 hours
Bonus = 5 × (30% × $8) = 5 × $2.40 = $12
Total = $320 + $12 = $332
Find a value of $\theta$ for which $\dfrac{2+3i\sin\theta}{1-2i\sin\theta}$ is purely imaginary.
Multiply numerator and denominator by the conjugate of the denominator $(1+2i\sin\theta)$:
Numerator real part: $(2+3i\sin\theta)(1+2i\sin\theta)$ real part $= 2-6\sin^2\theta$
For purely imaginary, the real part must be zero: $2-6\sin^2\theta=0\Rightarrow\sin^2\theta=\frac{1}{3}\Rightarrow\sin\theta=\frac{1}{\sqrt{3}}$
Therefore $\theta=\sin^{-1}\!\left(\dfrac{1}{\sqrt{3}}\right)$
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