Problem 17 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) and \(Q=P A P^{\prime}\), then \(p^{\prime} Q^{2005} P\) is (A) \(\left[\begin{array}{cc}1 & 1 \\ 2005 & 1\end{array}\right]\) (B) \(\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]\) (C) \(\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\) (D) \(\left[\begin{array}{cc}1 & 2005 \\ 2005 & 1\end{array}\right]\)

Step-by-Step Solution

Answer

The correct answer is (B) \(\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]\).

Step 1: Find the conjugate of matrix A

We are given that \(Q = PAP'\), where \(P'\) denotes the transpose of \(P\). To find \(Q\), we first compute \(P'\) from \(P\):\[P = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}\]Thus, the transpose \(P'\) is:\[P' = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}\]Next, calculate \(PAP'\). First, calculate \(PA\):\[PA = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} + \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} - \frac{1}{2} \end{bmatrix}\]Finally, compute \(Q = (PA)P'\):\[Q = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} + \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} - \frac{1}{2} \end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}\]

Step 2: Simplify matrix Q

Carry out the matrix multiplication:- First row, first column: \(\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2} + (\frac{\sqrt{3}}{2}+\frac{1}{2})\times\frac{1}{2} = \frac{3}{4} + \frac{\sqrt{3}}{4} + \frac{1}{4} = 1\)- First row, second column: \(\frac{\sqrt{3}}{2}\times\left(-\frac{1}{2}\right) + (\frac{\sqrt{3}}{2}+\frac{1}{2})\times\frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{4} + \frac{3}{4} + \frac{\sqrt{3}}{4} = 1\)- Second row, first column: \((-\frac{1}{2})\times\frac{\sqrt{3}}{2} + (\frac{\sqrt{3}}{2}-\frac{1}{2})\times\frac{1}{2} = -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} - \frac{1}{4} = 0\)- Second row, second column: \((-\frac{1}{2})\times\left(-\frac{1}{2}\right) + (\frac{\sqrt{3}}{2}-\frac{1}{2})\times\frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} - \frac{\sqrt{3}}{4} = 1\)This yields matrix \(Q\) as the identity matrix:\[Q = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}\]

Step 3: Evaluate \(Q^{2005}\)

Since we have determined that \(Q = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}\), we know that raising an upper triangular matrix of this form to a power leads to:\[Q^{2005} = \begin{bmatrix} 1 & 2005 \ 0 & 1 \end{bmatrix}\]

Step 4: Find \(P'Q^{2005}P\)

We are tasked with computing \(P'Q^{2005}P\):- First, perform \(Q^{2005}P\):\[Q^{2005}P = \begin{bmatrix} 1 & 2005 \ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} - 1002.5 & 1003 \frac{1}{2} \ - \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}\]- Then compute \(P'Q^{2005}P\):\[P'Q^{2005}P = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} 1 & 2005 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2005 \ 0 & 1 \end{bmatrix}\]

Step 5: Conclusion: Determine the Matching Answer

Comparing our final result from Step 4 \( \begin{bmatrix} 1 & 2005 \ 0 & 1 \end{bmatrix} \) with the given options, we find it matches option (B).

Key Concepts

Matrix Transpose

Understanding the transpose of a matrix is crucial in linear algebra. When you transpose a matrix, you effectively "flip" it over its diagonal. This means that the rows of the original matrix become the columns in the transposed matrix and vice versa. For instance, if you have a matrix \[ P = \begin{bmatrix} a & b \ c & d \end{bmatrix} \],then the transpose, denoted as \( P' \), is\[ P' = \begin{bmatrix} a & c \ b & d \end{bmatrix} \].When dealing with matrix multiplication, transposes can be particularly useful. Especially, it's often needed when you wish to multiply a matrix by its transpose or find the conjugate of a matrix combination, as seen in the previous exercise with the matrix \( Q = PAP' \). Using the transpose can sometimes simplify calculations and is essential for operations like orthogonalizing matrices or finding certain solutions.

Matrix Exponentiation

Matrix exponentiation is the process of raising a matrix to a given power. Just like with numbers, raising a matrix to a power involves multiplying the matrix by itself multiple times. However, it's important to note that this is not as straightforward as it might seem due to the nature of matrix multiplication. When dealing with a matrix that has a specific form, such as an upper triangular matrix, the process becomes more predictable. For example, consider a matrix of the form:\[ Q = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \].When you raise this type of upper triangular matrix to a power, say 2005, the structure makes it easier to predict the result:\[ Q^{2005} = \begin{bmatrix} 1 & 2005 \ 0 & 1 \end{bmatrix} \].In this pattern, the top right element becomes \( n \) (the power) while the rest of the structure remains unchanged and recognizable as an identity-like matrix with a twist in its upper row.

Upper Triangular Matrix

Upper triangular matrices are a special form of matrices. They have all zeros below the main diagonal. For example, a matrix \( A \) is upper triangular if it looks like this:\[ A = \begin{bmatrix} a & b \ 0 & d \end{bmatrix} \].These matrices have special properties, which can make computation easier.

• The determinant of an upper triangular matrix is simply the product of the diagonal elements.
• The inverse of an upper triangular matrix, if it exists, will also be upper triangular.
• When it comes to exponentiation, the structure of upper triangular matrices allows us predict the layout of resultant matrices.

For example, as seen in the exercise, when a simple upper triangular matrix with ones down the diagonal and the first upper element as '1' is raised to a power, the other non-diagonal elements increase linearly. This property provides a neat shortcut when solving linear algebra problems involving such matrices, like those encountered with matrix \( Q \) in our earlier example.