Question

Let A = $\begin{bmatrix} 1 & 2 & 1 \\ 1 & 3 & 2 \\ 2 & 4 & 1 \end{bmatrix}$ and Mij, Aij respectively denote the minor, co-factor of an element aij of matrix A, then which of the following are true?
(A) M22 = -1
(B) A23 = 0
(C) A32 = 3
(D) M23 = 1
(E) M32 = -3
Choose the correct answer from the options given below:

• A. (A) and (B) only
• B. (A), (B), (C) and (E) only
• C. (A), (D) and (E) only
• D. (A), (C) and (E) only

Hint:

Understanding the Sign of the Cofactor:

Be very careful with the sign of the cofactor. The sign is determined by \( (-1)^{i+j} \), which follows a checkerboard pattern of signs in the matrix:

\( \begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix} \)

Applying to A₂₃ and A₃₂:

For \( A_{23} \), the position (2, 3) has a \( - \) sign (since \( i + j = 2 + 3 = 5 \), and \( (-1)^5 = -1 \)).

For \( A_{32} \), the position (3, 2) also has a \( - \) sign (since \( i + j = 3 + 2 = 5 \), and \( (-1)^5 = -1 \)).

Conclusion:

When calculating cofactors, always be mindful of the checkerboard sign pattern, which is determined by \( (-1)^{i+j} \). For positions (2,3) and (3,2), the signs are both \( - \).

Answer 1

The Correct Option is A

Objective: To identify true statements by calculating specific minors and cofactors of a given 3x3 matrix.

Definitions:

The minor \( M_{ij} \) of an element \( a_{ij} \) is the determinant of the submatrix obtained by removing the \( i \)-th row and \( j \)-th column.

The cofactor \( A_{ij} \) is defined as \( A_{ij} = (-1)^{i+j} M_{ij} \).

Methodology: Calculate the minors and cofactors relevant to each given statement.

Matrix: \[ A = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 3 & 2 \\ 2 & 4 & 1 \end{bmatrix}. \]

Statement Analysis:

(A) \( M_{22} = -1 \): The minor \( M_{22} \) is calculated by removing the 2nd row and 2nd column:

\[ M_{22} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = (1)(1) - (1)(2) = 1 - 2 = -1 \] Statement (A) is true.

(B) \( A_{23} = 0 \): The cofactor \( A_{23} \) requires the minor \( M_{23} \):

\[ M_{23} = \begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} = (1)(4) - (2)(2) = 4 - 4 = 0 \] Now, calculate the cofactor:

\[ A_{23} = (-1)^{2+3} M_{23} = (-1)^5 (0) = 0 \] Statement (B) is true.

(C) \( A_{32} = 3 \): The cofactor \( A_{32} \) requires the minor \( M_{32} \):

\[ M_{32} = \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = (1)(2) - (1)(1) = 2 - 1 = 1 \] Now, calculate the cofactor:

\[ A_{32} = (-1)^{3+2} M_{32} = (-1)^5 (1) = -1 \] Statement (C) is false.

(D) \( M_{23} = 1 \): As calculated for statement (B), \( M_{23} = 0 \).

Statement (D) is false.

(E) \( M_{32} = -3 \): As calculated for statement (C), \( M_{32} = 1 \). Assuming a potential typo in the original statement, as \( M_{32} = 1 \).

Statement (E) is false.

Conclusion: Statements (A) and (B) are the only true statements.