All Exams Test series for 1 year @ ₹349 only
Question

From the matrix equation AB = AC we can conclude. B = C provided

The correct answer is

|A| ≠ 0

Understanding the Matrix Equation AB = AC

The question asks for the condition under which, from the matrix equation \(AB = AC\), we can definitively conclude that \(B = C\). This is a fundamental concept in matrix algebra, specifically dealing with the properties of matrix multiplication and inversion.

Let's analyze the given matrix equation:

\[ AB = AC \]

Our goal is to find the condition on matrix A that allows us to 'cancel' A from both sides and get \(B = C\).

Condition for Cancelling A in the Matrix Equation

In regular algebra with numbers, if \(ab = ac\) and \(a \ne 0\), we can divide both sides by \(a\) to get \(b=c\). In matrix algebra, there is no division operation. Instead, we use the concept of the inverse matrix.

If a square matrix A has an inverse, denoted by \(A^{-1}\), then \(A^{-1}A = I\) and \(AA^{-1} = I\), where \(I\) is the identity matrix. Multiplying by the inverse is the matrix equivalent of division.

Consider the equation \(AB = AC\). If the inverse of A, \(A^{-1}\), exists, we can multiply both sides of the equation by \(A^{-1}\) from the left:

\[ A^{-1}(AB) = A^{-1}(AC) \]

Using the associative property of matrix multiplication, \((A^{-1}A)B = (A^{-1}A)C\).

Since \(A^{-1}A = I\), where I is the identity matrix, the equation becomes:

\[ IB = IC \]

Multiplying a matrix by the identity matrix \(I\) results in the original matrix (\(IB = B\) and \(IC = C\)). Therefore, we get:

\[ B = C \]

When Does the Inverse Matrix Exist?

A square matrix A has an inverse \(A^{-1}\) if and only if its determinant is non-zero. The determinant of A is denoted by \(|A|\) or \(\det(A)\).

So, the condition for \(A^{-1}\) to exist is \(|A| \ne 0\). When this condition holds, the matrix A is called a non-singular or invertible matrix.

Therefore, if we have the matrix equation \(AB = AC\), we can conclude \(B = C\) if and only if \(|A| \ne 0\).

If \(|A| = 0\), matrix A is singular and its inverse does not exist. In this case, we cannot multiply by \(A^{-1}\) and we cannot generally conclude that \(B = C\). There can be cases where \(AB = AC\) but \(B \ne C\) if \(|A| = 0\). For example, if A is a zero matrix, then \(0 \cdot B = 0 \cdot C\) is always true for any matrices B and C of compatible dimensions, even if \(B \ne C\). This shows that the condition \(|A| \ne 0\) is essential for cancelling A in the matrix equation \(AB = AC\).

Evaluating the Options

Let's look at the given options in the context of the matrix equation \(AB=AC\):

  1. \(|A| = 0\): If the determinant is zero, A is singular, and the inverse does not exist. We cannot conclude \(B=C\).
  2. \(|A| \ne 0\): If the determinant is non-zero, A is non-singular or invertible. The inverse \(A^{-1}\) exists, allowing us to multiply by \(A^{-1}\) and conclude \(B=C\).
  3. A is symmetric: A symmetric matrix is one where \(A^T = A\). This property does not guarantee that \(|A| \ne 0\). A symmetric matrix can be singular (e.g., a diagonal matrix with a zero on the diagonal).
  4. A is square: For the determinant \(|A|\) to be defined (for matrices larger than 1x1) and for the inverse \(A^{-1}\) to be in the same space as A, A must be a square matrix. However, being square alone does not guarantee \(|A| \ne 0\). A square matrix can be singular. While A must be square for \(|A|\) to be meaningful in this context, the condition for cancellation is about invertibility, not just shape.

Based on our analysis, the correct condition to conclude \(B = C\) from the matrix equation \(AB = AC\) is that A must be invertible, which means its determinant \(|A|\) must be non-zero.

This property is crucial when dealing with matrix equations and understanding when operations analogous to scalar algebra can be performed. The condition \(|A| \ne 0\) is the key to solving many linear algebra problems involving matrix equations.

Was this answer helpful?

Important Questions from Matrices

  1. A generic 3 × 3 real matrix A has eigenvalues 0, 1 and 6, and I is the 3 × 3 identity matrix. The quantity/quantities that cannot be determined from this information is/are the

  2. If \(A = \(\left[ {\begin{array}{} {coshx}&{sinhx}\\ { - sinhx}&{coshx} \end{array}} \right])\) , then trace (A 2) is equal to

  3. A tansformation T: 2 → 2 first reflects points through the vertical axis (x2 - axis ) and then reflects points through the line x2 = x1. Then the standard matrix of T is:

  4. The matrix representation of T on R2 defined by T(x, y) = (2y, 3x - y) relative to the basis {(1, 3), (2, 5)} is

  5. A skew-symmetric matrix cannot be of rank

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App