Problem 19 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(A=\left|\begin{array}{ccc}5 & 5 \alpha & \alpha \\ 0 & \alpha & 5 \alpha \\\ 0 & 0 & 5\end{array}\right|\). If \(\left|A^{2}\right|=25\), then \(|\alpha|\) equals [2007] (a) \(1 / 5\) (b) 5 (c) \(5^{2}\) (d) 1

Step-by-Step Solution

Answer

The magnitude of \(\alpha\) is \(\frac{1}{5}\).

Step 1: Determine the Determinant of A

To find \(|A|\), recognize that \(A\) is an upper triangular matrix, meaning all the non-diagonal elements below the main diagonal are zero. The determinant of an upper triangular matrix is the product of its diagonal elements. Thus, for matrix \(A\), the determinant is \(|A| = 5 \cdot \alpha \cdot 5 = 25\alpha\).

Step 2: Apply Determinant Property for A Squared

The property of determinants for a matrix squared, \(|A^{2}|\), is given by \(|A^{2}| = |A|^{2}\). Substituting from Step 1, \(|A^{2}| = (25\alpha)^{2} = 625\alpha^{2}\). We are given \(|A^{2}| = 25\).

Step 3: Solve the Equation for \(\alpha\)

The equation from Step 2 is \(625\alpha^{2} = 25\). Divide both sides by 25 to simplify: \(25\alpha^{2} = 1\). Divide by 25: \(\alpha^{2} = \frac{1}{25}\). Thus, \(|\alpha| = \frac{1}{5}\).

Key Concepts

Upper Triangular Matrix

In the world of matrices, an upper triangular matrix is a special kind of matrix where all the entries below the main diagonal are zero. This means that the matrix has a simple triangular structure with all non-zero elements appearing at the top.

Understanding this property helps simplify calculations significantly. For example, when determining the determinant of an upper triangular matrix, you don't need to perform complex row reductions or other tedious operations. Instead, you just multiply the entries along the main diagonal.

Given the matrix:\[A = \left|\begin{array}{ccc}5 & 5 \alpha & \alpha \ 0 & \alpha & 5 \alpha \ 0 & 0 & 5\end{array}\right|\],you can instantly determine its determinant by multiplying the diagonal elements:

• First diagonal element: 5
• Second diagonal element: \( \alpha \)
• Third diagonal element: 5

Thus, the determinant of the matrix is \(|A| = 5 \cdot \alpha \cdot 5 = 25\alpha\). This characteristic is super handy to quickly find the determinant without much hassle!

Matrix Squared Property

The matrix squared property relates elegantly to how determinants behave when a matrix is multiplied by itself. If you have a matrix \(A\), the determinant of its square \(|A^2|\) is linked to the determinant of the original matrix by the formula:\[|A^2| = |A|^2\]This means that to find the determinant of a squared matrix, you simply square the determinant of the original matrix.

In the exercise, we first found that \(|A| = 25\alpha\). Applying the matrix squared property, the determinant of \(A^2\) becomes:\[|A^2| = (25\alpha)^2 = 625\alpha^2\]Knowing this property makes it possible to calculate the determinant of complex matrix operations without directly computing each element of \(A^2\). Instead, it leverages the simpler computation of a single determinant.

Solving for Alpha

Once you have the equation set up from previous steps, the task is now to solve for \(\alpha\). Given from the problem, \(\left|A^{2}\right| = 25\), while we determined that \(\left|A^{2}\right| = 625\alpha^{2}\).
Thus, the equation becomes:\[625\alpha^{2} = 25\]To solve for \(\alpha\), divide both sides by 25:
\[25\alpha^{2} = 1\]Next, divide by 25 on both sides:\[\alpha^{2} = \frac{1}{25}\]Keep in mind that solving for \(\alpha\) involves taking the square root. The square root of a fraction like \(\frac{1}{25}\) is calculated as:
\[\alpha = \pm \frac{1}{5}\]In this context, the absolute value \(|\alpha|\) indicates that \(\alpha\) can either be positive or negative \(\frac{1}{5}\), depending on the conditions of the problem. That's why \(|\alpha|\) is \(\frac{1}{5}\), aligning with option \(a\). This means regardless of the negative or positive sign, the size of \(\alpha\) we need is \(\frac{1}{5}\).