$A = \left( {\begin{array}{*{20}{c}}
1&2&3\\
1&3&5\\
2&5&a
\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}
6\\
9\\
b
\end{array}} \right)$
Since, system is consistent and has infonitely many solutions
$\therefore $ (adj.$A$) $B=0$
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
{3a - 25}&{15 - 2a}&1\\
{10 - a}&{a - 6}&{ - 2}\\
{ - 1}&{ - 1}&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
6\\
9\\
b
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
0\\
0\\
0
\end{array}} \right)$
$ \Rightarrow - 6 - 9 + b = 0 \Rightarrow b = 15$
and $6(10-a)+9(a-6)-2(b)=0$
$ \Rightarrow 60 - 6a + 9a - 54 - 30 = 0$
$ \Rightarrow 3a = 24 \Rightarrow a = 8$
Hence, $a=8,b=15$.
વિધાન $-I$ : ${A^{ - 1}} = \frac{1}{7}\left( {5I - A} \right).$
વિધાન $-II$ : બહુપદી $A^3 - 2A^2 - 3A + I$ ને $5\, (A - 4I)$ સ્વરૂપમાં દર્શાવી શકાય .
$x+y+3 z=0$
$x+3 y+k^{2} z=0$
$3 x+y+3 z=0$
માટે શૂન્યેતર ઉકેલ $(x, y, z)$ જ્યાં $k \in R$ હોય તો $x +\left(\frac{ y }{ z }\right)$ ની કિમત મેળવો