Question

3. Using Cofactors of elements of second row, evaluate​

Answer 1

Given:

$\Delta = \left[\begin{array}{ccc}5&3&8\\2&0&1\\1&2&3\end{array}\right]$

To find:

Evaluation of a matrix

Solution:

$\Delta = a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23}$

a{21} = 2, a{22} = 0, a{23} = 1

The co factors of the second row are:

A{21}, A{22} and A{23}

$M_{21}=\left[\begin{array}{ccc}3&8\\2&3\end{array}\right] = 3 \times 3 - 8 \times 2 = 9-16= -7$

$M_{22}=\left[\begin{array}{ccc}5&8\\1&3\end{array}\right] = 5 \times 3 - 8 \times 1 = 15-8= 7$

$M_{23}=\left[\begin{array}{ccc}5&3\\1&2\end{array}\right] = 5 \times 2 - 1 \times 3 = 10-3= 7$

Cofactor of a{21} = A{21} = (-1)^{2+1} M{21} = (-1)^3 × -7 = -1 × -7 = 7

Cofactor of a{22} = A{22} = (-1)^{2+2} M{22} = (-1)^4 × 7 = 1 × -7 = 7

Cofactor of a{23} = A{23} = (-1)^{2+3} M{23} = (-1)^5 × 7 = -1 × -7 = -7

$\Delta = a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23}$

= 2 × 7 + 0 × 7 + 1 × -7

= 14 + 0 - 7

= 7

∴ Δ = 7