Key Concepts and Formulas

This problem primarily tests your understanding of fundamental properties of determinants for square matrices. For square matrices $A$ and $B$ of order $n$, and a scalar $k$:

1. Determinant of a Scalar Multiple: $\det(kA) = k^n \det(A)$. The crucial part here is $n$, the order of the matrix. If a matrix of order $n$ is multiplied by a scalar $k$, its determinant is scaled by $k^n$.
2. Determinant of a Product: $\det(AB) = \det(A) \det(B)$. This property is useful when dealing with products of matrices.
3. Determinant of a Power: $\det(A^m) = (\det A)^m$. This is a direct consequence of the product rule, applied repeatedly.

Step-by-Step Solution

We are given two square matrices $A$ and $B$ of order $n=2$. We are also given $\det(A) = 2$ and $\det(B) = 3$. Our goal is to evaluate the expression $\det \left( {(\det \,5(\det\,A)B){A^2}} \right)$ and express it in the form ${2^a}{3^b}{5^c}$ to find $a+b+c$.

Step 1: Identify the order of the matrices and the given determinant values.

• Order of matrices $A$ and $B$ is $n=2$. This is critical for applying the scalar multiplication property.
• $\det(A) = 2$
• $\det(B) = 3$

Step 2: Evaluate the innermost determinant term, $\det(5(\det A)B)$. Let's first determine the scalar multiplying matrix $B$.

• The scalar is $k_1 = 5(\det A)$.
• Substitute the given value of $\det A = 2$: $k_1 = 5 \times 2 = 10$
• Now we need to find $\det(k_1 B) = \det(10B)$.
• Since $B$ is a matrix of order $n=2$, we apply the property $\det(kX) = k^n \det(X)$: $\det(10B) = (10)^2 \det(B)$
• Substitute the given value of $\det(B) = 3$: $\det(10B) = (10)^2 \times 3 = 100 \times 3 = 300$
• So, the value of the innermost determinant is 300. This is a scalar. Let's denote this scalar as $K$. $K = \det(5(\det A)B) = 300$

Step 3: Substitute the scalar $K$ back into the main expression.

• The original expression is $\det \left( {(\det \,5(\det\,A)B){A^2}} \right)$.
• Replacing the innermost determinant with its calculated scalar value $K$: $\det(K A^2)$
• Here, $K=300$ is a scalar, and $A^2$ is a matrix. Note that $A^2$ is also a square matrix of order $n=2$.

Step 4: Apply determinant properties to evaluate $\det(K A^2)$.

• We again use the property $\det(kX) = k^n \det(X)$, where $k=K$ and $X=A^2$. Since $A^2$ is of order $n=2$: $\det(K A^2) = K^2 \det(A^2)$
• Next, we use the property $\det(A^m) = (\det A)^m$ for $\det(A^2)$: $\det(A^2) = (\det A)^2$
• Substitute this back into the expression: $\det(K A^2) = K^2 (\det A)^2$

Step 5: Substitute all known numerical values and calculate.

• We have $K = 300$ and $\det A = 2$. $\det(K A^2) = (300)^2 (2)^2$
• Calculate the values: $(300)^2 (2)^2 = (90000) (4) = 360000$

Step 6: Express the result in prime factorization form ${2^a}{3^b}{5^c}$ and find $a+b+c$.

• We need to find the prime factorization of $360000$. It's often easier to factorize the components first: $300 = 3 \times 100 = 3 \times 10^2 = 3 \times (2 \times 5)^2 = 3^1 \times 2^2 \times 5^2$
• Now substitute this into the expression $(300)^2 (2)^2$: $(300)^2 (2)^2 = (3^1 \times 2^2 \times 5^2)^2 \times 2^2$ $= (3^2 \times (2^2)^2 \times (5^2)^2) \times 2^2$ $= (3^2 \times 2^4 \times 5^4) \times 2^2$ $= 2^{4+2} \times 3^2 \times 5^4$ $= 2^6 \times 3^2 \times 5^4$
• We are given that $\det \left( {(\det \,5(\det\,A)B){A^2}} \right) = {2^a}{3^b}{5^c}$.
• Comparing our result $2^6 \times 3^2 \times 5^4$ with $2^a \times 3^b \times 5^c$: $a = 6$ $b = 2$ $c = 4$
• Finally, calculate $a+b+c$: $a+b+c = 6 + 2 + 4 = 12$

Common Mistakes & Tips

1. Incorrect Order ($n$) for Scalar Multiplication: The most common error is forgetting to use the power $n$ (the order of the matrix) when applying $\det(kA) = k^n \det(A)$. In this problem, $n=2$, so scalars are squared.
2. Confusing Scalars and Matrices: Always be clear whether an entity is a scalar (a number, like $\det A$, 300) or a matrix (like $A$, $B$, $A^2$, $10B$). Determinants of matrices are scalars.
3. Errors in Prime Factorization: Be meticulous when breaking down numbers into their prime factors. A small mistake in an exponent can lead to an incorrect final sum for $a+b+c$.

Summary

This problem required a systematic application of determinant properties. We first evaluated the innermost determinant by recognizing the scalar multiple and applying $\det(kA) = k^n \det(A)$. Then, we treated the result as a new scalar and applied the same property again, along with $\det(A^m) = (\det A)^m$, to the outer determinant. Finally, we performed prime factorization of the resulting numerical value to determine the exponents $a, b, c$ and calculated their sum. The key was correctly identifying the order of the matrices and distinguishing between scalars and matrices at each step.

The final answer is $\boxed{12}$ which corresponds to option (B).