Key Concepts and Formulas

1. Matrix Multiplication: If matrix $A$ has dimensions $m \times n$ and matrix $B$ has dimensions $n \times p$, their product $AB$ will be an $m \times p$ matrix. The element $(AB)_{ij}$ is found by taking the dot product of the $i$-th row of $A$ and the $j$-th column of $B$.
2. Determinant of a $2 \times 2$ Matrix: For a matrix $M = \left[ {\begin{matrix} a & b \\ c & d \\ \end{matrix} } \right]$, its determinant is given by $\det(M) = ad - bc$.
3. Properties of Determinants: A matrix is singular if its determinant is zero. Also, if a common factor exists in any row or column of a determinant, it can be factored out.
4. Sum of Roots of a Quadratic Equation: For a quadratic equation $ax^2 + bx + c = 0$ (where $a \neq 0$), the sum of its roots is given by $-b/a$.

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Step-by-Step Solution

Step 1: Determine the Dimensions of AB and Perform Matrix Multiplication

What we are doing: We need to find the product matrix $AB$ first, as its determinant is required. Why we are doing it: The determinant property $\det(AB) = \det(A)\det(B)$ is not applicable here because $A$ and $B$ are not square matrices. Thus, we must first calculate $AB$ and then find its determinant.

Given matrices: $A = \left[ {\begin{matrix} 1 & { - 2} & \alpha \\ \alpha & 2 & { - 1} \\ \end{matrix} } \right]$ This is a $2 \times 3$ matrix.

$B = \left[ {\begin{matrix} 2 & \alpha \\ { - 1} & 2 \\ 4 & { - 5} \\ \end{matrix} } \right]$ This is a $3 \times 2$ matrix.

Since the number of columns in $A$ (3) equals the number of rows in $B$ (3), the product $AB$ is defined. The resulting matrix $AB$ will have dimensions $2 \times 2$.

Let $AB = \left[ {\begin{matrix} c_{11} & c_{12} \\ c_{21} & c_{22} \\ \end{matrix} } \right]$. We calculate each element:

• $c_{11} = (1)(2) + (-2)(-1) + (\alpha)(4) = 2 + 2 + 4\alpha = 4 + 4\alpha$ (This is the dot product of the 1st row of A and the 1st column of B)
• $c_{12} = (1)(\alpha) + (-2)(2) + (\alpha)(-5) = \alpha - 4 - 5\alpha = -4\alpha - 4$ (This is the dot product of the 1st row of A and the 2nd column of B)
• $c_{21} = (\alpha)(2) + (2)(-1) + (-1)(4) = 2\alpha - 2 - 4 = 2\alpha - 6$ (This is the dot product of the 2nd row of A and the 1st column of B)
• $c_{22} = (\alpha)(\alpha) + (2)(2) + (-1)(-5) = \alpha^2 + 4 + 5 = \alpha^2 + 9$ (This is the dot product of the 2nd row of A and the 2nd column of B)

So, the product matrix $AB$ is: $AB = \left[ {\begin{matrix} {4 + 4\alpha} & { - 4\alpha - 4} \\ {2\alpha - 6} & {{\alpha ^2} + 9} \\ \end{matrix} } \right]$

Step 2: Set the Determinant of AB to Zero

What we are doing: We are applying the determinant formula for a $2 \times 2$ matrix to $AB$ and setting it equal to zero, as given in the problem statement. Why we are doing it: The problem explicitly states that $\det(AB)=0$, which will give us an equation in terms of $\alpha$ to solve.

Using the formula $\det \left[ {\begin{matrix} a & b \\ c & d \\ \end{matrix} } \right] = ad - bc$: $\det(AB) = (4 + 4\alpha)({\alpha ^2} + 9) - (-4\alpha - 4)(2\alpha - 6) = 0$

Step 3: Simplify and Solve the Equation for $\alpha$

What we are doing: We will simplify the determinant equation to find the values of $\alpha$. Why we are doing it: This step yields the specific values of $\alpha$ that make the determinant zero. We must be careful with algebraic manipulation.

Let's simplify the equation: Notice that $4+4\alpha = 4(1+\alpha)$ and $-4\alpha-4 = -4(1+\alpha)$. Substituting these into the determinant expression: $\left| {\begin{matrix} {4(1 + \alpha)} & {-4(1 + \alpha)} \\ {2\alpha - 6} & {{\alpha ^2} + 9} \\ \end{matrix} } \right| = 0$

Now, we can factor out $4(1+\alpha)$ from the first row of the determinant. This simplifies the calculation significantly. $4(1 + \alpha) \left| {\begin{matrix} 1 & { - 1} \\ {2\alpha - 6} & {{\alpha ^2} + 9} \\ \end{matrix} } \right| = 0$

Next, calculate the determinant of the remaining $2 \times 2$ matrix: $(1)({\alpha ^2} + 9) - (-1)(2\alpha - 6) = 0$ ${\alpha ^2} + 9 + 2\alpha - 6 = 0$ ${\alpha ^2} + 2\alpha + 3 = 0$

So, the complete equation for $\alpha$ is: $4(1 + \alpha)({\alpha ^2} + 2\alpha + 3) = 0$

This equation provides two cases for the values of $\alpha$:

Case 1: From the linear factor $1 + \alpha = 0 \implies \alpha = -1$

Case 2: From the quadratic factor ${\alpha ^2} + 2\alpha + 3 = 0$ Since $\alpha \in \mathbb{C}$ (complex numbers), we solve this quadratic equation. We can use the quadratic formula $\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a=1, b=2, c=3$. The discriminant $D = b^2 - 4ac = (2)^2 - 4(1)(3) = 4 - 12 = -8$. $\alpha = \frac{-2 \pm \sqrt{-8}}{2(1)} = \frac{-2 \pm \sqrt{8}i}{2} = \frac{-2 \pm 2\sqrt{2}i}{2}$ $\alpha = -1 \pm \sqrt{2}i$ The two roots are $\alpha_1 = -1 + \sqrt{2}i$ and $\alpha_2 = -1 - \sqrt{2}i$.

Alternatively, using the sum of roots property for the quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $-b/a$. For ${\alpha ^2} + 2\alpha + 3 = 0$, the sum of roots is $-(2)/1 = -2$.

Step 4: Calculate the Sum of all values of $\alpha$ and its Absolute Value

What we are doing: We are summing all the $\alpha$ values found in Step 3 and then taking the absolute value of that sum. Why we are doing it: This is what the question asks for.

The values of $\alpha$ for which $\det(AB)=0$ are:

• $\alpha = -1$
• $\alpha = -1 + \sqrt{2}i$
• $\alpha = -1 - \sqrt{2}i$

Sum of all values of $\alpha = (-1) + (-1 + \sqrt{2}i) + (-1 - \sqrt{2}i)$ Sum of all values of $\alpha = -1 - 1 + \sqrt{2}i - 1 - \sqrt{2}i$ Sum of all values of $\alpha = -3$

Alternatively, using the sum of roots property for the quadratic: Sum of all values of $\alpha = (\text{root from linear factor}) + (\text{sum of roots from quadratic factor})$ Sum of all values of $\alpha = (-1) + (-2)$ Sum of all values of $\alpha = -3$

The absolute value of the sum is: $|-3| = 3$

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Common Mistakes & Tips

• Matrix Multiplication Errors: Be extremely careful with the "row by column" rule, especially with signs and carrying over terms. A single calculation error here will propagate through the entire problem.
• Determinant Formula: Remember the $ad-bc$ formula for $2 \times 2$ determinants and the crucial minus sign between the products.
• Algebraic Simplification: Always look for opportunities to factor common terms in determinants or equations. Factoring $4(1+\alpha)$ in Step 3 was key to avoiding a more complex cubic expansion.
• Domain of Variables: Pay attention to the specified domain of $\alpha$ (here, $\alpha \in \mathbb{C}$). If $\alpha$ were restricted to real numbers, the quadratic factor would yield no solutions.
• Sum of Roots Property: For polynomials, especially quadratics, using the sum of roots property ($-b/a$) can be a quick and efficient way to find the sum without explicitly calculating individual complex roots.

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Summary

We first calculated the product matrix $AB$ by performing matrix multiplication. Then, we set the determinant of this $2 \times 2$ matrix $AB$ to zero. We simplified the resulting equation by factoring a common term $4(1+\alpha)$ from the determinant, which led to a product of a linear and a quadratic factor. Solving these factors yielded three values for $\alpha$: one real root and two complex conjugate roots. Finally, we summed these values and took the absolute value of the sum to arrive at the final answer.

The final answer is $\boxed{3}$, which corresponds to option (A).