Key Concepts and Formulas

To solve this problem, we will utilize several fundamental concepts from matrix algebra:

1. Relationship between a Matrix, its Adjoint, and its Determinant: For any square matrix $A$ of order $n$, the product of the matrix and its adjoint is equal to the determinant of the matrix multiplied by the identity matrix of the same order. This is expressed as: $A (\text{adj } A) = (\text{adj } A) A = |A| I$ where $|A|$ is the determinant of $A$ and $I$ is the identity matrix of order $n$. This identity is crucial for simplifying the left-hand side of the given equation.

2. Determinant of a $2 \times 2$ Matrix: For a $2 \times 2$ matrix M = \left[ {\matrix{ p & q \cr r & s \cr } } \right], its determinant is calculated as: $|M| = ps - qr$

3. Transpose of a Matrix: The transpose of a matrix $M$, denoted as $M^T$, is obtained by interchanging its rows and columns. For a $2 \times 2$ matrix M = \left[ {\matrix{ p & q \cr r & s \cr } } \right], its transpose is: M^T = \left[ {\matrix{ p & r \cr q & s \cr } } \right]

4. Matrix Multiplication: To multiply two matrices, say $C = AB$, the element in the $i$-th row and $j$-th column of $C$ is obtained by taking the dot product of the $i$-th row of $A$ and the $j$-th column of $B$.

5. Equality of Matrices: Two matrices are equal if and only if they have the same dimensions and their corresponding elements are equal.

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Step-by-Step Solution

We are given the matrix A = \left[ {\matrix{ {5a} & { - b} \cr 3 & 2 \cr } } \right] and the condition $A (\text{adj } A) = A A^T$. Our goal is to find the value of $5a+b$.

Step 1: Simplify the Left-Hand Side (LHS) using the fundamental identity.

• Why this step? The given equation involves $A (\text{adj } A)$ on the LHS. Directly calculating $\text{adj } A$ and then multiplying by $A$ can be tedious and prone to errors. Using the identity $A (\text{adj } A) = |A| I$ allows us to simplify the LHS into a scalar multiple of the identity matrix, which is much easier to work with.

First, let's calculate the determinant of matrix $A$: $|A| = (5a)(2) - (-b)(3)$ $|A| = 10a - (-3b)$ $|A| = 10a + 3b$

Now, substitute this into the identity $A (\text{adj } A) = |A| I$. Since $A$ is a $2 \times 2$ matrix, $I$ will be the $2 \times 2$ identity matrix, I = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]. So, the LHS becomes: A (\text{adj } A) = (10a + 3b) \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right] A (\text{adj } A) = \left[ {\matrix{ {10a + 3b} & 0 \cr 0 & {10a + 3b} \cr } } \right] \quad \text{(Equation 1)}

Step 2: Simplify the Right-Hand Side (RHS) by calculating $A A^T$.

• Why this step? The RHS of the given equation is $A A^T$. We need to explicitly calculate this matrix product to compare it with the simplified LHS. This involves finding the transpose of $A$ and then performing matrix multiplication.

First, let's find the transpose of $A$: Given A = \left[ {\matrix{ {5a} & { - b} \cr 3 & 2 \cr } } \right], we swap its rows and columns to get $A^T$: A^T = \left[ {\matrix{ {5a} & 3 \cr { - b} & 2 \cr } } \right]

Now, perform the matrix multiplication $A A^T$: A A^T = \left[ {\matrix{ {5a} & { - b} \cr 3 & 2 \cr } } \right] \left[ {\matrix{ {5a} & 3 \cr { - b} & 2 \cr } } \right]

Let's calculate each element of the resulting matrix:

• (Row 1, Column 1): $(5a)(5a) + (-b)(-b) = 25a^2 + b^2$
• (Row 1, Column 2): $(5a)(3) + (-b)(2) = 15a - 2b$
• (Row 2, Column 1): $(3)(5a) + (2)(-b) = 15a - 2b$
• (Row 2, Column 2): $(3)(3) + (2)(2) = 9 + 4 = 13$

So, the RHS becomes: A A^T = \left[ {\matrix{ {25a^2 + b^2} & {15a - 2b} \cr {15a - 2b} & {13} \cr } } \right] \quad \text{(Equation 2)}

Step 3: Equate the LHS and RHS and solve for $a$ and $b$.

• Why this step? The problem statement tells us that $A (\text{adj } A) = A A^T$. By equating the simplified forms from Equation 1 and Equation 2, we create a matrix equality. From this equality, we can derive a system of algebraic equations by comparing corresponding elements, which will allow us to find the values of $a$ and $b$.

From Equation 1 and Equation 2, we have: \left[ {\matrix{ {10a + 3b} & 0 \cr 0 & {10a + 3b} \cr } } \right] = \left[ {\matrix{ {25a^2 + b^2} & {15a - 2b} \cr {15a - 2b} & {13} \cr } } \right]

By equating corresponding elements:

1. From the (Row 1, Column 2) elements: $0 = 15a - 2b \quad \text{(Equation 3)}$ This gives us a simple linear relationship between $a$ and $b$.

2. From the (Row 2, Column 2) elements: $10a + 3b = 13 \quad \text{(Equation 4)}$ This gives us another linear equation involving $a$ and $b$.

Now we have a system of two linear equations with two variables $a$ and $b$: (3) $15a - 2b = 0$ (4) $10a + 3b = 13$

From Equation 3, we can express $b$ in terms of $a$: $2b = 15a$ $b = \frac{15}{2}a \quad \text{(Equation 5)}$

Substitute Equation 5 into Equation 4: $10a + 3\left(\frac{15}{2}a\right) = 13$ $10a + \frac{45}{2}a = 13$ To combine the terms with $a$, find a common denominator: $\frac{20a}{2} + \frac{45a}{2} = 13$ $\frac{65a}{2} = 13$ $65a = 26$ $a = \frac{26}{65}$ Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, 13: $a = \frac{2 \times 13}{5 \times 13} = \frac{2}{5}$

Now substitute the value of $a$ back into Equation 5 to find $b$: $b = \frac{15}{2} \times \frac{2}{5}$ $b = \frac{15}{5}$ $b = 3$

We have found $a = \frac{2}{5}$ and $b = 3$.

Step 4: Verify with the (Row 1, Column 1) elements (Optional but Recommended).

• Why this step? This step serves as a valuable check to ensure the correctness of our calculated values for $a$ and $b$. If they satisfy the remaining equality, it increases our confidence in the solution.

Equating the (Row 1, Column 1) elements: $10a + 3b = 25a^2 + b^2$

Substitute the values $a = \frac{2}{5}$ and $b = 3$: LHS: $10\left(\frac{2}{5}\right) + 3(3) = 4 + 9 = 13$ RHS: $25\left(\frac{2}{5}\right)^2 + (3)^2 = 25\left(\frac{4}{25}\right) + 9 = 4 + 9 = 13$

Since LHS = RHS ($13 = 13$), our values for $a$ and $b$ are consistent.

Step 5: Calculate the required expression $5a+b$.

• Why this step? This is the final step to answer the question posed in the problem, which asks for the value of the specific expression $5a+b$.

Substitute the values of $a = \frac{2}{5}$ and $b = 3$ we found: $5a + b = 5\left(\frac{2}{5}\right) + 3$ $5a + b = 2 + 3$ $5a + b = 5$

Thus, the value of $5a+b$ is $5$.

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Tips and Common Mistakes

• Master the Identity: The identity $A (\text{adj } A) = |A| I$ is fundamental. Recognizing and applying it correctly is the most important step in simplifying this type of problem.
• Identity Matrix vs. Scalar: Remember that $I$ in $|A|I$ is the identity matrix, not just the number 1. For a $2 \times 2$ matrix, I = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right].
• Careful with Matrix Multiplication: Matrix multiplication is not commutative ($AB \neq BA$ in general). Ensure you multiply matrices in the correct order as given in the problem ($A A^T$ not $A^T A$). Pay close attention to signs and arithmetic, especially when dealing with negative values.
• Systematic Approach: Break down the problem into smaller, manageable steps (simplify LHS, simplify RHS, equate, solve system of equations). This reduces the chances of errors and makes complex problems more approachable.
• Verification: If time permits, always verify your solutions by substituting the derived values back into the original equations or remaining matrix element equalities. This helps catch calculation mistakes.

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Summary and Key Takeaway

This problem effectively tests your understanding of fundamental matrix properties and operations. The key to solving it efficiently was recognizing and applying the identity $A (\text{adj } A) = |A|I$, which significantly simplified the left-hand side of the given equation. The remaining steps involved standard matrix operations: calculating a determinant, finding a transpose, performing matrix multiplication, and solving a system of linear equations obtained by equating corresponding elements of the resulting matrices. Mastering these basic operations and matrix identities is crucial for success in matrix-related JEE problems.

The final answer is $\boxed{5}$.