Fig. 23(a)

Applying KVL in mesh I, II and II respectively

\(-2I_1-3I_2+{7I}_3=12\) … (1)

\(7I_1-4I_2-2I_3=0\) … (2)

\(-4I_1+9I_2-3I_3=0\) … (3)

\(\left[\begin{matrix}-2&-3&7\\7&-4&-2\\-4&9&-3\\\end{matrix}\right]\left[\begin{matrix}I_1\\I_2\\I_3\\\end{matrix}\right]=\left[\begin{matrix}12\\0\\0\\\end{matrix}\right]\)

Solving above equation by Cramer’s rule

\(I_1=\frac{\left[\begin{matrix}12&-3&7\\0&-4&-2\\0&9&-3\\\end{matrix}\right]}{\left[\begin{matrix}-2&-3&7\\7&-4&-2\\-4&9&-3\\\end{matrix}\right]}=\frac{12(12+18)}{182}=\frac{360}{182}=1.978\ Amp\).

\(I_2=\frac{\left[\begin{matrix}-2&12&7\\7&0&-2\\-4&0&-3\\\end{matrix}\right]}{\left[\begin{matrix}-2&-3&7\\7&-4&-2\\-4&9&-3\\\end{matrix}\right]}=\frac{-12(-21-8)}{182}=\frac{348}{182}=1.912\ Amp\).

\(I_3=\frac{\left[\begin{matrix}-2&-3&12\\7&-4&0\\-4&9&0\\\end{matrix}\right]}{\left[\begin{matrix}-2&-3&7\\7&-4&-2\\-4&9&-3\\\end{matrix}\right]}=\frac{12(63-16)}{182}=\frac{564}{182}=3.098Amp\).

Hence current in different branches are 1.978 A, 1.921 A and 3.098 A.