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

Step 1: Understanding the Concept:

We need to calculate specific minors and cofactors of the given 3x3 matrix and check which of the given statements are true.

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

The cofactor \( A_{ij} \) is related to the minor by the formula \( A_{ij} = (-1)^{i+j} M_{ij} \).

Step 2: Key Formula or Approach:

We will use the above definitions and calculate the minors and cofactors for each statement.

Step 3: Detailed Explanation:

The given matrix is: \[ A = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 3 & 2 \\ 2 & 4 & 1 \end{bmatrix}. \] Let's evaluate each statement.

(A) \( M_{22} = -1 \): 
\( M_{22} \) is the minor of the element \( a_{22} = 3 \). We delete 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) \( A_{23} = 0 \): 
\( A_{23} \) is the cofactor of the element \( a_{23} = 2 \). First, find 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) \( A_{32} = 3 \): 
\( A_{32} \) is the cofactor of the element \( a_{32} = 4 \). First, find 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) \( M_{23} = 1 \): 
From our calculation for statement (B), we found that \( M_{23} = 0 \). Statement (D) is false.

(E) \( M_{32} = -3 \): 
The OCR here is likely a typo. Based on the options, let's assume it should have been \( A_{32} = -1 \), or something related. From our calculation for statement (C), we found \( M_{32} = 1 \). Statement (E) is false.

Step 4: Final Answer:

The only true statements are (A) and (B). Therefore, the correct option is (1).