Problem 237 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(A=\left[\begin{array}{ll}2 & 0 \\ 3 & 1\end{array}\right]\) and \(B=\left[\begin{array}{ll}0 & 1 \\ 2 & 4\end{array}\right]\). Verify that \((A B)^{-1}=B^{-1} A^{-1}\).

Step-by-Step Solution

Answer

The given statement is proven to be true as (AB)\(^{-1}\) = B\(^{-1}\)A\(^{-1}\) = \(\left[\begin{array}{cc}-7/4 & 1/2 \ 1/2 & 0\end{array}\right]\)

Step 1: Compute the product AB

First, calculate the matrix product AB. When multiplying matrices, each element ij in the resulting matrix is the sum of the product of elements in the ith row of the first matrix and the jth column of the second matrix. So we have,AB = \(\left[\begin{array}{cc}2 & 0 \ 3 & 1\end{array}\right]\) * \(\left[\begin{array}{cc}0 & 1 \ 2 & 4\end{array}\right]\) = \(\left[\begin{array}{cc}(2*0 + 0*2) & (2*1 + 0*4) \ (3*0 + 1*2) & (3*1 + 1*4)\end{array}\right]\) = \(\left[\begin{array}{cc}0 & 2 \ 2 & 7\end{array}\right]\)

Step 2: Calculate the inverse of AB

To find the inverse of a 2x2 matrix, we can use the formula \(A^{-1} = \frac{1}{ad - bc} \left[\begin{array}{cc}d & -b \ -c & a\end{array}\right]\). For the matrix AB = \(\left[\begin{array}{cc}0 & 2 \ 2 & 7\end{array}\right]\), a = 0, b = 2, c = 2, and d = 7. So the inverse is,(AB)\(^{-1}\) = \(\frac{1}{(0*7 - 2*2)} \left[\begin{array}{cc}7 & -2 \ -2 & 0\end{array}\right]\) = \(\left[\begin{array}{cc}-7/4 & 1/2 \ 1/2 & 0\end{array}\right]\)

Step 3: Compute B\(^{-1}\) and A\(^{-1}\)

Next, find inverses of matrices B and A likewise in the above step. For B = \(\left[\begin{array}{cc}0 & 1 \ 2 & 4\end{array}\right]\), the inverse B\(^{-1}\) = \(\frac{1}{(0*4 - 1*2)} \left[\begin{array}{cc}4 & -1 \ -2 & 0\end{array}\right]\) = \(\left[\begin{array}{cc}-2 & 1/2 \ 1 & 0\end{array}\right]\)For A = \(\left[\begin{array}{cc}2 & 0 \ 3 & 1\end{array}\right]\), the inverse A\(^{-1}\) = \(\frac{1}{(2*1 - 0*3)} \left[\begin{array}{cc}1 & 0 \ -3 & 2\end{array}\right]\) = \(\left[\begin{array}{cc}1/2 & 0 \ -3/2 & 1\end{array}\right]\)

Step 4: Compute B\(^{-1}\)A\(^{-1}\)

Finally, compute the product of the inverse matrices B and A. Using the formula for multiplication of matrices, we getB\(^{-1}\)A\(^{-1}\) = \(\left[\begin{array}{cc}-2 & 1/2 \ 1 & 0\end{array}\right]\) * \(\left[\begin{array}{cc}1/2 & 0 \ -3/2 & 1\end{array}\right]\) = \(\left[\begin{array}{cc}(-2*1/2 + 1/2*-3/2) &(-2*0 + 1/2*1) \(1*1/2 + 0*-3/2) & 1*0 + 0*1\end{array}\right]\) = \(\left[\begin{array}{cc}-7/4 & 1/2 \ 1/2 & 0\end{array}\right]\)

Step 5: Verify the Solution

Comparing the results from step 2 and step 4, it is clear that (AB)\(^{-1}\) = B\(^{-1}\)A\(^{-1}\). Hence, the given statement is proven to be true.

Key Concepts

Matrix Multiplication

Matrix multiplication is an essential operation in linear algebra where two matrices are combined to produce a third matrix. Understanding how this operation works is crucial in solving complex mathematical problems efficiently.

The process of multiplying matrices involves taking each element of a row from the first matrix and multiplying it by the corresponding element of a column from the second matrix, then summing these products. This summation forms one element in the resulting matrix.

For example, to multiply matrix \( A \) by matrix \( B \), you would calculate each entry in the resulting matrix \( AB \) by following these steps:

• Select a row from matrix \( A \).
• Select a column from matrix \( B \).
• Multiply each pair of corresponding elements from that row and column.
• Add up these products to get the element for the resulting matrix \( AB \).

In essence, matrix multiplication is not simply a point-wise multiplication but a series of dot products between rows and columns, which come together to form a new matrix.

Inverse of 2x2 Matrix

The inverse of a 2x2 matrix is a matrix that, when multiplied with the original matrix, yields the identity matrix. Finding the inverse can be quite straightforward when you use the correct formula, especially for 2x2 matrices.

For a 2x2 matrix \( \left[ \begin{array}{cc} a & b \ c & d \end{array} \right] \), the inverse is given by the formula:\[A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \-c & a \end{bmatrix}\]

• Calculate the determinant \( ad - bc \). If this value is zero, the matrix is not invertible.
• Switch the positions of \( a \) and \( d \) in the matrix.
• Change the signs of \( b \) and \( c \).
• Multiply each element by \( 1/(ad - bc) \) to find the inverse matrix.

This method is quick and efficient for finding inverses of 2x2 matrices, making it a fundamental tool in linear algebra and related fields.

Linear Algebra

Linear Algebra is a branch of mathematics that focuses on vector spaces, linear mappings between these spaces, and related concepts like matrices and determinants. It provides a vital foundation for various mathematical disciplines and practical applications, from computer graphics to machine learning.

Essentially, linear algebra explores how linear equations and transformations interact within a given structure. It simplifies complex systems, making it easier to understand and solve mathematical problems in higher dimensions.

• Vectors: Objects defined by magnitude and direction, forming the basic unit in spaces.
• Matrices: Rectangular arrays of numbers representing linear transformations or system solutions.
• Determinants: Scalar values that provide information about the matrix, like its invertibility.
• Eigenvalues and Eigenvectors: Determine the essence of matrices and their behavior during transformations.

Linear algebra is foundational in various applications, including physics simulations, 3D modeling, and solving large sets of equations, making it a key mathematical tool for engineers, scientists, and mathematicians alike.