Problem 75 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(A=\left(\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right)\) and \(B=\left(\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right)\). If \(B\) is the inverse of matrix \(A\), then \(\alpha\) is (a) 5 (b) \(-1\) (c) 2 (d) \(-2\)

Step-by-Step Solution

Answer

The value of \(\alpha\) is 5.

Step 1: Understand Inverse Matrix Definition

If matrix \(B\) is the inverse of matrix \(A\), then \(AB = I_3\), where \(I_3\) is the 3x3 identity matrix.

Step 2: Write the Matrix Multiplication

Multiply matrix \(A\) by \(B\) and equate to the 3x3 identity matrix:\[ AB = \left(\begin{array}{ccc}1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1\end{array}\right)\left(\begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3\end{array}\right) = \left(\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right) \]

Step 3: Perform Matrix Multiplication

Calculate each element of the resulting product matrix by using standard matrix multiplication rules. Use each row from \(A\) and each column from \(B\).

Step 4: Equate with Identity Matrix

Compare the resulting product matrix with the identity matrix to create a system of equations. Focus particularly on the position (2,3), where the resulting matrix should be zero.

Step 5: Solve for \(\alpha\)

The element at position (2,3) of \(AB\) is derived from: \( (2)(2) + (1)(\alpha) + (-3)(3) = 0 \4 + \alpha - 9 = 0 \\alpha = 5 \) This gives \(\alpha = 5\), which is the only value that satisfies the condition for \(B\) being the inverse of \(A\).

Key Concepts

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra, involving two matrices where we systematically multiply rows by columns to produce a new matrix. Unlike simple arithmetic, matrix multiplication isn't always commutative, meaning \(AB\) may not equal \(BA\). Instead, it requires that the number of columns in the first matrix matches the number of rows in the second matrix. To multiply matrix \(A\) by matrix \(B\):

• Take each row of matrix \(A\) and multiply it with each column of matrix \(B\).
• Add the products to form a single number, setting it in the resulting matrix.

In the given exercise, we multiply a 3x3 matrix \(A\) by another 3x3 matrix \(B\). This involves using the matrix product rule to calculate each element of the resulting matrix, ensuring the layout and calculus align with matrix multiplication standards.

Identity Matrix

The identity matrix is a special type of square matrix. Think of it like the number 1 in multiplication, acting as a neutral element because multiplying any matrix by it will leave the original matrix unchanged. An identity matrix is always square, meaning it has an equal number of rows and columns. Its defining feature is that all diagonal elements are 1, while all other elements are 0. For a 3x3 identity matrix \(I_3\), it looks like this: \[I_3 = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}\]It's crucial in matrix operations because any matrix multiplied by its inverse will result in the identity matrix. Thus, in our exercise, the product \(AB\) resulted in \(I_3\), proving \(B\) as the inverse of \(A\).

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. Solving a system of equations involves finding values for the unknowns that satisfy all equations simultaneously. These are essential in linear algebra and play a key role in mechanics, physics, and economic modeling. In the context of matrices:

• Matrix multiplication often leads to creating a system of equations.
• Each position in the resultant matrix can give rise to an equation by matching with the corresponding position in the identity matrix.

In our exercise, this concept becomes pivotal as the matrix equation \(AB = I_3\) translates into several scalar equations. For example, focusing on element position (2,3) of the product, we derived an equation involving \(\alpha\), and by solving it, we confirmed \(\alpha = 5\). This verification was critical to ascertain that \(B\) was indeed the inverse of \(A\).