Problem 236 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(A=\left[\begin{array}{ll}2 & 5 \\ 1 & 6\end{array}\right]\), find \(A^{-1}\) and verify that \(A^{-1}=-\frac{1}{7} A+\frac{8}{7} I\)

Step-by-Step Solution

Answer

The inverse of matrix A is \(A^{-1} = \left[\begin{array}{ll}\frac{6}{7} & -\frac{5}{7} \ -\frac{1}{7} & \frac{2}{7}\end{array}\right]\), and it can be verified that \(A^{-1}=-\frac{1}{7} A+\frac{8}{7} I\).

Step 1: Calculate the Determinant of the Matrix

The determinant of a 2x2 matrix \(A=\left[\begin{array}{ll}a & b \ c & d\end{array}\right]\) is given by \(ad - bc\). So, calculating the determinant of A gives \(2*6 - 1*5 = 7\).

Step 2: Calculate the Inverse of the Matrix

The inverse of a 2x2 matrix \(A\) is given by \(\frac{1}{\text{{det}}(A)}*\left[\begin{array}{ll}d & -b \ -c & a\end{array}\right]\). Substituting the given values, we get \(A^{-1} = \frac{1}{7}*\left[\begin{array}{ll}6 & -5 \ -1 & 2\end{array}\right] = \left[\begin{array}{ll}\frac{6}{7} & -\frac{5}{7} \ -\frac{1}{7} & \frac{2}{7}\end{array}\right]\).

Step 3: Verify the Given Expression

Now replace \(A^{-1}\) and \(A\) in the expression \(-\frac{1}{7} A+\frac{8}{7} I\). Replacing values gives: \(-\frac{1}{7}*A+\frac{8}{7}*\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right]\) = \left[\begin{array}{ll}-\frac{2}{7} + \frac{8}{7} & -\frac{5}{7} \ -\frac{1}{7} & -\frac{6}{7} + \frac{8}{7}\end{array}\right] = \left[\begin{array}{ll}\frac{6}{7} & -\frac{5}{7} \ -\frac{1}{7} & \frac{2}{7}\end{array}\right] = A^{-1}\) This last expression is indeed the calculated inverse, so the given expression is correct.

Key Concepts

Determinant of a Matrix

The determinant is a special number that can be calculated from a square matrix. For a 2x2 matrix, the formula for the determinant is quite straightforward. Imagine you have a matrix \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\]To find the determinant, you use the formula:\[det(A) = ad - bc\]This involves multiplying the values diagonally from top-left to bottom-right and subtracting the product of the values diagonally from top-right to bottom-left.

• If the determinant is zero, the matrix does not have an inverse.
• A non-zero determinant indicates the matrix is invertible, meaning it has an inverse.

In our example, after applying the formula, we found that the determinant of the matrix \( A \) is 7. This non-zero result indicates that the matrix has an inverse, which we can proceed to calculate.

Matrix Multiplication

Matrix multiplication is a key operation when working with matrices. Understanding how to multiply matrices correctly is crucial for many applications, including finding the inverse. When multiplying matrices, it’s important to remember that the number of columns in the first matrix must match the number of rows in the second.

Here's how you multiply two matrices:

• Select a row from the first matrix and a column from the second matrix.
• Multiply each element of the row by the corresponding element of the column.
• Add up the results to get the entry in the new matrix.

It’s important to note that matrix multiplication is not commutative, meaning that \( AB \) does not necessarily equal \( BA \). However, the identity matrix plays a special role as it satisfies \( AI = A \) and \( IA = A \), where \( I \) is the identity matrix. Also, when verifying matrix expressions, correctly applying matrix multiplication ensures accurate results.

2x2 Matrices

2x2 matrices are among the simplest square matrices you can work with. They have just two rows and two columns, which makes calculations straightforward. In many cases, 2x2 matrices serve as a great introduction to more complex matrix operations.

When dealing with 2x2 matrices, some common operations include calculation of the determinant, finding the inverse, and performing matrix multiplication. These matrices often involve minimal arithmetic yet require understanding of fundamental concepts such as row and column alignment for multiplication.

• Inverses involve swapping elements of the main diagonal and changing signs on the other diagonal, followed by scaling with the reciprocal of the determinant.
• They provide a simple model for numerous real-world problems due to their ease of handling.

Mastering operations with 2x2 matrices builds a strong foundation for advancing into higher-order matrices and more complex linear algebra problems.