અહી શ્રેણિક $A=\left[a_{i j}\right]_{3 \times 3}$ કે જ્યાં
$a_{i j}=J_{6+i, 3}-J_{i+3,3}, \quad i \leq j$
$\quad\quad\quad\quad\quad\quad0 , \quad\quad\quad i>j$.
તો $\left|\operatorname{adj} A^{-1}\right|$ મેળવો.
$\mathrm{J}_{6 + i, 3}-\mathrm{J}_{i+3,3} ; \mathrm{i} \leq \mathrm{j}$
$\Rightarrow \int_{0}^{\frac{1}{2}} \frac{\mathrm{x}^{6+\mathrm{i}}}{\mathrm{x}^{3}-1}-\int_{0}^{\frac{1}{2}} \frac{\mathrm{x}^{\mathrm{i}+3}}{\mathrm{x}^{3}-1}$
$\Rightarrow \int_{0}^{1 / 2} \frac{\mathrm{x}^{\mathrm{i}+3}\left(\mathrm{x}^{3}-1\right)}{\mathrm{x}^{3}-1}$
$\Rightarrow \frac{\mathrm{x}^{3+\mathrm{i}+1}}{3+\mathrm{i}+1}=\left(\frac{\mathrm{x}^{4+\mathrm{i}}}{4+\mathrm{i}}\right)^{1 / 2}$
$\mathrm{a}_{\mathrm{ij}}=\mathrm{j}_{6+\mathrm{i}, 3}-\mathrm{j}_{\mathrm{i}+3,3}=\frac{\left(\frac{1}{2}\right)^{4+\mathrm{i}}}{4+\mathrm{i}}$
$\mathrm{a}_{11}=\frac{\left(\frac{1}{2}\right)^{5}}{5}=\frac{1}{5.2^{5}}$
$\mathrm{a}_{12}=\frac{1}{5.2^{5}}$
$\mathrm{a}_{13}=\frac{1}{5.2^{5}}$
$\mathrm{a}_{22}=\frac{1}{6.2^{6}}$
$\mathrm{a}_{23}=\frac{1}{6.2^{6}}$
$\mathrm{a}_{33}=\frac{1}{7.2^{7}}$
$\mathrm{A}=\left[\begin{array}{ccc}\frac{1}{5.2^{5}} & \frac{1}{5.2^{5}} & \frac{1}{5.2^{5}} \\ 0 & \frac{1}{6.2^{6}} & \frac{1}{6.2^{6}} \\ 0 & 0 & \frac{1}{7.2^{7}}\end{array}\right]$
$|\mathrm{A}|=\frac{1}{5.2^{5}}\left[\frac{1}{6.2^{6}} \times \frac{1}{7.2^{7}}\right]$
$|A|=\frac{1}{210.2^{18}}$
$\left|\operatorname{adj} A^{-1}\right|=\left|A^{-1}\right|^{n-1}=\left|A^{-1}\right|^{2}=\frac{1}{(|A|)^{2}}$
$\Rightarrow\left(210.2^{18}\right)^{2}$
$(105)^{2} \times 2^{38}$
$x+y+z=5$ ; $x+2 y+3 z=\mu$ ; $x+3 y+\lambda z=1$
ને બનાવમાં આવે છે.જો $\mathrm{p}$ એ સમીકરણ સંહતિને એકાકી ઉકેલ હોય તેની સંભાવના દર્શાવે છે અને $\mathrm{q}$ એ સમીકરણ સંહતિનો ઉકેલગણ ખાલીગણ છે તેની સંભાવના દર્શાવે છે તો