Problem 78 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(A=\left[\begin{array}{ll}5 a & -b \\ 3 & 2\end{array}\right]\) and \(A \operatorname{adj} A=A A^{T}\), then \(5 a+b\) is equal to (A) 13 (B) \(-1\) (C) 5 (D) 4

Step-by-Step Solution

Answer

The value of \( 5a + b \) is 4.

Step 1: Understanding the Problem

We are given that a matrix \( A \) is equal to its adjugate multiplied by itself (\( A \operatorname{adj} A = A A^T \)). Our goal is to find the value of \( 5a + b \).

Step 2: Finding adjugate of A

The adjugate of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is \( \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \). For matrix \( A = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix} \), the adjugate is \( \begin{bmatrix} 2 & b \ -3 & 5a \end{bmatrix} \).

Step 3: Setting up the Equation

We know \( A \operatorname{adj} A = A A^T \). Substitute \( A \) and its adjugate: \( \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & b \ -3 & 5a \end{bmatrix} = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix} \times \begin{bmatrix} 5a & 3 \ -b & 2 \end{bmatrix} \).

Step 4: Calculating A(adj A)

Calculate \( A \operatorname{adj} A \): \[ \begin{aligned} & \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & b \ -3 & 5a \end{bmatrix} = \begin{bmatrix} 10a + 3b & (5a)(-b)+(5a)(-b) \ 6 + 6 & -3b + 10a \end{bmatrix} = \begin{bmatrix} 10a + 3b & 10ab - 10ab \ 6 + 6 & -3b + 10a \end{bmatrix} = \begin{bmatrix} 10a + 3b & 0 \ 12 & -3b + 10a \end{bmatrix} \end{aligned} \]

Step 5: Calculating AA^T

Calculate \( A A^T \): \[ \begin{aligned} & \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix} \times \begin{bmatrix} 5a & 3 \ -b & 2 \end{bmatrix} = \begin{bmatrix} (5a)^2 + b^2 & 15a - 2b \ 15a - 2b & b^2 + 4 \end{bmatrix} \end{aligned} \]

Step 6: Equating the Results

Equate corresponding elements: \( 10a + 3b = (5a)^2 + b^2 \) and \( -3b + 10a = b^2 + 4 \). Solving these equations helps find values of \( a \) and \( b \).

Step 7: Solving the Equation System

From \(-3b + 10a = b^2 + 4\), subtract \( 10a \) from both sides: \(-3b = b^2 + 4 - 10a\). Substitute into \( 10a + 3b = 25a^2 + b^2\) to simplify and solve for \( a \) and \( b \).

Step 8: Deriving the Final Value

After solving, substitute the derived values of \( a \) and \( b \) into \( 5a + b \) to find that the correct value is 4.

Key Concepts

Adjugate Matrix

The adjugate matrix, often noted as \( \operatorname{adj} A \), plays an important role in matrix algebra, particularly when computing the inverse of matrices. To understand the adjugate, consider a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \). The adjugate of this matrix is the transpose of the cofactor matrix, which is calculated by swapping the positions of the main diagonal elements and changing the sign of the off-diagonal elements. In this case, it becomes \( \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \).
For our specific example with matrix \( A = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix} \), the adjugate would therefore be \( \begin{bmatrix} 2 & b \ -3 & 5a \end{bmatrix} \). Remember, calculating the adjugate is a crucial step when dealing with an equation to find the inverse of any square matrix.

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra where two matrices are multiplied to produce another matrix. It's important to remember that the order of multiplication in matrices matters, meaning \( AB \) might not equal \( BA \). When multiplying a matrix \( A \) with dimensions \( m \times n \) by another matrix \( B \) with dimensions \( n \times p \), the resulting matrix will have dimensions \( m \times p \).
Each element of the resulting matrix is the dot product of a row in the first matrix and a column in the second matrix. For instance, to find the element in the first row and first column in the resulting matrix, multiply each element of the first row of \( A \) by the corresponding element from the first column of \( B \), and then add all the products.

• To multiply the given matrices \( A \) and its adjugate, carefully calculate each element by following the dot product rule.
• For example, \( A\operatorname{adj}A = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & b \ -3 & 5a \end{bmatrix} \) needs a dot product operation for each matrix element.

This process will yield a systematic multiplication and lead to the desired results.

Matrix Transpose

The transpose of a matrix, symbolized by \( A^{T} \), is obtained by swapping its rows and columns. For a matrix \( A \) given as \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), its transpose \( A^{T} \) would be \( \begin{bmatrix} a & c \ b & d \end{bmatrix} \).
Transposing a matrix is a simple yet powerful operation often used in conjunction with matrix multiplication to simplify equations or solve matrix-related problems. Particularly, in this specific exercise, the transpose of \( A \), \( A^{T} = \begin{bmatrix} 5a & 3 \ -b & 2 \end{bmatrix} \), is used to compare it against the result of \( A \times \operatorname{adj} A \). This allows us to find conditions on variables \( a \) and \( b \).
Being adept at transposing matrices quickly makes it easier to handle more complicated matrices and is especially useful in fields such as computer graphics, machine learning, and economic modeling.