Question

If \( A = \begin{pmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 0 & 3 & -5 \end{pmatrix} \) where \( A_{ij} \) is the cofactor of the element \( a_{ij} \) of matrix \( A \), then \( a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = \)

• A. \( -26 \)
• B. \( 0 \)
• C. \( -2 \)
• D. \( 26 \)

Hint:

The sum \( a_{i1}A_{i1} + a_{i2}A_{i2} + a_{i3}A_{i3} \) is always equal to the determinant when expanded along the \( i \)-th row.

Answer 1

The Correct Option is C

Step 1: Use the cofactor expansion property.
We know that the sum \[ a_{i1}A_{i1} + a_{i2}A_{i2} + a_{i3}A_{i3} \] gives the determinant of matrix \( A \) when expanded along the \( i \)-th row.
Step 2: Identify the row.
Here, the expansion is along the second row. Hence, \[ a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = \det(A) \] Step 3: Compute the determinant of \( A \).
\[ \det(A) = \begin{vmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 0 & 3 & -5 \end{vmatrix} \] Expanding along the first row, \[ = 1 \begin{vmatrix} 1 & 3 \\ 3 & -5 \end{vmatrix} - 0 + 2 \begin{vmatrix} 2 & 1 \\ 0 & 3 \end{vmatrix} \] Step 4: Evaluate the minors.
\[ = 1(1\cdot(-5) - 3\cdot3) + 2(2\cdot3 - 0) \] \[ = 1(-5 - 9) + 2(6) \] \[ = -14 + 12 = -2 \] Step 5: Conclusion.
Hence, \[ a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = -2 \]