Question:
If \( A = \begin{bmatrix} x & 2 & 1 -2 & y & 0 2 & 0 & -1 \end{bmatrix} \), \( x \) and \( y \) are non-zero numbers, trace of \( A = 0 \) and determinant of \( A = -6 \), then the minor of the element \( 1 \) of \( A \) is:
If \( A = \begin{bmatrix} x & 2 & 1 -2 & y & 0 2 & 0 & -1 \end{bmatrix} \), \( x \) and \( y \) are non-zero numbers, trace of \( A = 0 \) and determinant of \( A = -6 \), then the minor of the element \( 1 \) of \( A \) is:
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Always read the constraints carefully! Paying close attention to conditions like "non-zero numbers" allows you to instantly eliminate extraneous roots during quadratic factorization, keeping your calculations fast and precise.
Updated On: May 20, 2026
- \( -4 \)
- \( 4 \)
- \( 2 \)
- \( -2 \)
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The Correct Option is A
Solution and Explanation
Concept:
• Trace of a Matrix: The trace of a square matrix \( A \), denoted as \( \text{tr}(A) \), is the sum of the elements lying on its principal diagonal.
• Minor of an Element: The minor \( M_{ij} \) of an element \( a_{ij} \) in a matrix is the determinant of the submatrix left after deleting the \( i \)-th row and \( j \)-th column.
Step 1: Using the trace condition to establish a relationship between \( x \) and \( y \).
Given that the trace of matrix \( A \) is \( 0 \): \[ \text{tr}(A) = x + y + (-1) = 0 \quad \Rightarrow \quad x + y = 1 \quad \Rightarrow \quad y = 1 - x \quad \cdots (1) \]
Step 2: Using the determinant condition to find the specific values of \( x \) and \( y \).
The determinant of matrix \( A \) is given as \( -6 \): \[ |A| = \begin{vmatrix} x & 2 & 1 -2 & y & 0 2 & 0 & -1 \end{vmatrix} = -6 \] Expanding along the third column (\(C_3\)) because it contains a zero element: \[ 1 \cdot \begin{vmatrix} -2 & y 2 & 0 \end{vmatrix} - 0 + (-1) \cdot \begin{vmatrix} x & 2 -2 & y \end{vmatrix} = -6 \] \[ 1 \cdot [(-2)(0) - (y)(2)] - 1 \cdot [(x)(y) - (2)(-2)] = -6 \] \[ -2y - (xy + 4) = -6 \quad \Rightarrow \quad -2y - xy - 4 = -6 \quad \Rightarrow \quad xy + 2y = 2 \quad \cdots (2) \] Substitute equation (1) into equation (2): \[ x(1 - x) + 2(1 - x) = 2 \] \[ x - x^2 + 2 - 2x = 2 \quad \Rightarrow \quad -x^2 - x + 2 = 2 \quad \Rightarrow \quad -x^2 - x = 0 \] \[ -x(x + 1) = 0 \quad \Rightarrow \quad x = 0 \quad \text{or} \quad x = -1 \] Since the problem states that \( x \) and \( y \) are non-zero numbers, we discard \( x = 0 \). Therefore: \[ x = -1 \] Substituting \( x = -1 \) back into equation (1): \[ y = 1 - (-1) = 2 \]
Step 3: Finding the minor of the element 1.
The element \( 1 \) is located at the first row and third column position (\( a_{13} \)). To find its minor \( M_{13} \), delete the 1st row and 3rd column of matrix \( A \): \[ M_{13} = \begin{vmatrix} -2 & y 2 & 0 \end{vmatrix} = (-2)(0) - (y)(2) = -2y \] Substituting the determined value of \( y = 2 \): \[ M_{13} = -2(2) = -4 \]
• Trace of a Matrix: The trace of a square matrix \( A \), denoted as \( \text{tr}(A) \), is the sum of the elements lying on its principal diagonal.
• Minor of an Element: The minor \( M_{ij} \) of an element \( a_{ij} \) in a matrix is the determinant of the submatrix left after deleting the \( i \)-th row and \( j \)-th column.
Step 1: Using the trace condition to establish a relationship between \( x \) and \( y \).
Given that the trace of matrix \( A \) is \( 0 \): \[ \text{tr}(A) = x + y + (-1) = 0 \quad \Rightarrow \quad x + y = 1 \quad \Rightarrow \quad y = 1 - x \quad \cdots (1) \]
Step 2: Using the determinant condition to find the specific values of \( x \) and \( y \).
The determinant of matrix \( A \) is given as \( -6 \): \[ |A| = \begin{vmatrix} x & 2 & 1 -2 & y & 0 2 & 0 & -1 \end{vmatrix} = -6 \] Expanding along the third column (\(C_3\)) because it contains a zero element: \[ 1 \cdot \begin{vmatrix} -2 & y 2 & 0 \end{vmatrix} - 0 + (-1) \cdot \begin{vmatrix} x & 2 -2 & y \end{vmatrix} = -6 \] \[ 1 \cdot [(-2)(0) - (y)(2)] - 1 \cdot [(x)(y) - (2)(-2)] = -6 \] \[ -2y - (xy + 4) = -6 \quad \Rightarrow \quad -2y - xy - 4 = -6 \quad \Rightarrow \quad xy + 2y = 2 \quad \cdots (2) \] Substitute equation (1) into equation (2): \[ x(1 - x) + 2(1 - x) = 2 \] \[ x - x^2 + 2 - 2x = 2 \quad \Rightarrow \quad -x^2 - x + 2 = 2 \quad \Rightarrow \quad -x^2 - x = 0 \] \[ -x(x + 1) = 0 \quad \Rightarrow \quad x = 0 \quad \text{or} \quad x = -1 \] Since the problem states that \( x \) and \( y \) are non-zero numbers, we discard \( x = 0 \). Therefore: \[ x = -1 \] Substituting \( x = -1 \) back into equation (1): \[ y = 1 - (-1) = 2 \]
Step 3: Finding the minor of the element 1.
The element \( 1 \) is located at the first row and third column position (\( a_{13} \)). To find its minor \( M_{13} \), delete the 1st row and 3rd column of matrix \( A \): \[ M_{13} = \begin{vmatrix} -2 & y 2 & 0 \end{vmatrix} = (-2)(0) - (y)(2) = -2y \] Substituting the determined value of \( y = 2 \): \[ M_{13} = -2(2) = -4 \]
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