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Q3:
Using Cofactors of elements of second row, evaluate $\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$.
Solution :
Step 1: Identifying the Determinant and Elements of the Second Row
The given determinant is:
$\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$
The elements of the second row are:
- $a_{21} = 2$
- $a_{22} = 0$
- $a_{23} = 1$
Step 2: Calculating Cofactors for the Second Row
The cofactor $A_{ij}$ of an element $a_{ij}$ is defined as $A_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$ [Definition of Cofactor].
For $a_{21} = 2$:
$A_{21} = (-1)^{2+1} \begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = -1 \times [(3)(3) - (8)(2)] = -1 \times [9 - 16] = -1 \times (-7) = 7$
For $a_{22} = 0$:
$A_{22} = (-1)^{2+2} \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = 1 \times [(5)(3) - (8)(1)] = 1 \times [15 - 8] = 7$
For $a_{23} = 1$:
$A_{23} = (-1)^{2+3} \begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = -1 \times [(5)(2) - (3)(1)] = -1 \times [10 - 3] = -7$
Step 3: Evaluating the Determinant using Row 2 Cofactors
The value of a determinant is the sum of the products of elements of any row (or column) and their corresponding cofactors [Laplace Expansion Theorem]:
$\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}$
Substituting the calculated values:
$\Delta = (2)(7) + (0)(7) + (1)(-7)$
$\Delta = 14 + 0 - 7$
$\Delta = 7$
Verification via Geometric/Visual Representation
Final Solution: The value of the determinant $\Delta$ evaluated using the second row is 7.
More Questions from Class 12 Mathematics Determinants EXERCISE 4.3
- Q1(i): Write Minors and Cofactors of the elements of following determinant: (i) $\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$
- Q1(ii): Write Minors and Cofactors of the elements of following determinant: (ii) $\begin{vmatrix} a & c \\ b & d \end{vmatrix}$
- Q2(i): Write Minors and Cofactors of the elements of following determinant: (i) $\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$
- Q2(ii): Write Minors and Cofactors of the elements of following determinant: (ii) $\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}$
- Q4: Using Cofactors of elements of third column, evaluate $\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}$.
- Q5: If $\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$ and $A_{ij}$ is Cofactors of $a_{ij}$, then value of $\Delta$ is given by
CBSE Solutions for Class 12 Mathematics Determinants
Chapters in CBSE - Class 12 Mathematics
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