Question

If \[ A = \begin{bmatrix} x & 2 & 1 \\ -2 & y & 0 \\ 2 & 0 & -1 \end{bmatrix}, \] where \( x \) and \( y \) are non-zero real numbers, trace of \( A = 0 \), and determinant of \( A = -6 \), then the minor of the element 1 of \( A \) is:}

• A. \(-4\)
• B. \(4\)
• C. \(2\)
• D. \(-2\)

Hint:

The minor of an element is the determinant of the matrix after removing that element's row and column.

Answer 1

The Correct Option is A

Step 1: Use the trace condition
\[ \text{trace}(A) = x + y - 1 = 0 \implies x + y = 1 \] Step 2: Express determinant of \( A \)
\[ \det(A) = x \times \det \begin{bmatrix} y & 0
0 & -1 \end{bmatrix} - 2 \times \det \begin{bmatrix} -2 & 0
2 & -1 \end{bmatrix} + 1 \times \det \begin{bmatrix} -2 & y
2 & 0 \end{bmatrix} \] Calculate each minor: \[ \det \begin{bmatrix} y & 0 \\ 0 & -1 \end{bmatrix} = y \times (-1) - 0 = -y \] \[ \det \begin{bmatrix} -2 & 0 \\ 2 & -1 \end{bmatrix} = (-2)(-1) - 0 = 2 \] \[ \det \begin{bmatrix} -2 & y \\ 2 & 0 \end{bmatrix} = (-2)(0) - 2y = -2y \] So, \[ \det(A) = x(-y) - 2(2) + 1(-2y) = -xy - 4 - 2y \] Step 3: Use the determinant condition
\[ -xy - 4 - 2y = -6 \implies -xy - 2y = -2 \implies y(-x - 2) = -2 \] Step 4: Use \( x + y = 1 \) to substitute \( x = 1 - y \)
\[ y(- (1 - y) - 2) = -2 \implies y(-1 + y - 2) = -2 \implies y(y - 3) = -2 \] \[ y^2 - 3y + 2 = 0 \] Step 5: Solve quadratic
\[ (y - 1)(y - 2) = 0 \implies y = 1 \text{ or } y = 2 \] Step 6: Find corresponding \( x \)
If \( y=1 \), \( x = 1 - 1 = 0 \) (not allowed since \( x \neq 0 \))
If \( y=2 \), \( x = 1 - 2 = -1 \) (valid) Step 7: Minor of element 1 (first element in matrix) is
The minor of element \( a_{11} \) is determinant of the matrix formed by deleting first row and first column: \[ M_{11} = \det \begin{bmatrix} y & 0
0 & -1 \end{bmatrix} = -y = -2 \] Step 8: Correct answer
Minor of element 1 is \(-2\). Since option (1) is \(-4\), and option (4) is \(-2\), but user marked option (1) as correct, likely question refers to minor of element 1 meaning element in position (1,2) or (1,3), we verify: Minor of element in position (1,2): \[ M_{12} = \det \begin{bmatrix} -2 & 0 \ 2 &\ -1 \end{bmatrix} = (-2)(-1) - 0 = 2 \] Minor of element in position (1,3): \[ M_{13} = \det \begin{bmatrix} -2 & y \\ 2 & 0 \end{bmatrix} = (-2)(0) - 2y = -2y = -4 \] Hence, minor of element 1 (likely means element \( a_{13} \)) is \(-4\).