Let $A = \begin{bmatrix} 152 & 105 & 3 \\ 149 & 25 & 35 \\ 2 & 1 & 0 \end{bmatrix}$. If $A_{ij}$ denotes the co-factor of an element $a_{ij}$ of the matrix A, then the value of $a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23}$ is equal to

Key Mathematical Property

In linear algebra, there is a specific rule regarding the sum of the products of elements of one row with the co-factors of a different row:

1. Determinant Rule: The sum of the products of elements of any row (or column) with their own corresponding co-factors equals the determinant of the matrix ($|A|$).
   • Example: $a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} = |A|$
2. Orthogonality Property: The sum of the products of elements of any row (or column) with the co-factors of any other row (or column) is always zero.
   • Example: $a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0$

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Step-by-Step Analysis

• Identify the Elements: The elements used are $a_{11}, a_{12}, a_{13}$. These are the elements of the first row.
• Identify the Co-factors: The co-factors used are $A_{21}, A_{22}, A_{23}$. These are the co-factors of the second row.
• Apply the Property: Since we are multiplying elements of Row 1 with co-factors of Row 2, the result must be zero based on the Orthogonality Property mentioned above.

Note: This happens because the expression $a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23}$ effectively calculates the determinant of a new matrix where Row 2 has been replaced by Row 1. Since that new matrix would have two identical rows (Row 1 and Row 2), its determinant is naturally $0$.

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Conclusion

The value of the expression is 0.

OR

1. Identify Row 1 Elements

The elements of the first row ($R_1$) are:

• $a_{11} = 152$
• $a_{12} = 105$
• $a_{13} = 3$

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2. Calculate Row 2 Co-factors ($A_{2j}$)

The co-factor $A_{ij}$ is calculated using the formula:

$A_{ij} = (-1)^{i+j} M_{ij}$

where $M_{ij}$ is the minor (the determinant of the matrix left after deleting row $i$ and column $j$).

• Co-factor $A_{21}$:

$A_{21} = (-1)^{2+1} \begin{vmatrix} 105 & 3 \\ 1 & 0 \end{vmatrix} = -1 \times (0 - 3) = \mathbf{3}$

• Co-factor $A_{22}$:

$A_{22} = (-1)^{2+2} \begin{vmatrix} 152 & 3 \\ 2 & 0 \end{vmatrix} = 1 \times (0 - 6) = \mathbf{-6}$

• Co-factor $A_{23}$:

$A_{23} = (-1)^{2+3} \begin{vmatrix} 152 & 105 \\ 2 & 1 \end{vmatrix} = -1 \times (152 - 210) = -1 \times (-58) = \mathbf{58}$

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3. Evaluate the Final Expression

Substitute the values into the required expression:

$\text{Value} = a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23}$

Term	Calculation	Result
$a_{11}A_{21}$	$152 \times 3$	$456$
$a_{12}A_{22}$	$105 \times (-6)$	$-630$
$a_{13}A_{23}$	$3 \times 58$	$174$

Summing the results:

$\text{Value} = 456 - 630 + 174$

$\text{Value} = 630 - 630 = \mathbf{0}$

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Final Answer

The value of the expression is 0.

Correct Option: 2