If `A=[(1,1,0),(2,1,5),(1,2,1)]` then `a_(11)A_(21)+a_(12)A_(22)+a_(13)A_(23)` is equal to

Options

A. 1
B. 0
C. -1
D. 2

Explanation

To solve the problem, we need to evaluate the expression \( a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} \) where \( A \) is given as:

\[ A = \begin{pmatrix} 1 & 1 & 0 \\ 2 & 1 & 5 \\ 1 & 2 & 1 \end{pmatrix} \]

### Step 1: Identify the elements \( a_{ij} \)

From the matrix \( A \), we can identify the elements as follows:

- \( a_{11} = 1 \) (1st row, 1st column)
- \( a_{12} = 1 \) (1st row, 2nd column)
- \( a_{13} = 0 \) (1st row, 3rd column)

### Step 2: Identify the cofactors \( A_{ij} \)

Next, we need to find the cofactors \( A_{21}, A_{22}, A_{23} \).

- **Cofactor \( A_{21} \)**: This is the determinant of the submatrix obtained by deleting the 2nd row and 1st column of \( A \):

\[ A_{21} = \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} = (1)(1) - (0)(2) = 1 \]

- **Cofactor \( A_{22} \)**: This is the determinant of the submatrix obtained by deleting the 2nd row and 2nd column of \( A \):

\[ A_{22} = \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = (1)(1) - (0)(1) = 1 \]

- **Cofactor \( A_{23} \)**: This is the determinant of the submatrix obtained by deleting the 2nd row and 3rd column of \( A \):

\[ A_{23} = \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = (1)(2) - (1)(1) = 2 - 1 = 1 \]

### Step 3: Substitute the values into the expression

Now we substitute the values we found into the expression:

\[ a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = (1)(1) + (1)(1) + (0)(1) \]

### Step 4: Calculate the result

Calculating the above expression gives:

\[ 1 + 1 + 0 = 2 \]

### Final Answer

Thus, the value of \( a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} \) is equal to \( 2 \).

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