જો $A = \left[ {\begin{array}{*{20}{c}}2&4&5\\4&8&{10}\\{ - 6}&{ - 12}&{ - 15}\end{array}} \right]$. તો $A$ નો રેન્ક મેળવો.
A$0$
B$1$
C$2$
D$3$
Medium
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B$1$
b (b) $A = {\left[ {\begin{array}{*{20}{c}}2&4&5\\4&8&{10}\\{ - 6}&{ - 12}&{ - 15}\end{array}} \right]_{3 \times 3}}$
$|A|\, = 0$, then rank cannot be $ 3.$
Considering a $2 \times 2$ minor, $\left[ {\begin{array}{*{20}{c}}2&4\\4&8\end{array}} \right]$ its determinant is zero.
Similarly considering
$\left[ {\begin{array}{*{20}{c}}4&5\\8&{10}\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}4&8\\{ - 6}&{ - 12}\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}8&{10}\\{ - 12}&{15}\end{array}} \right],\,\left[ {\begin{array}{*{20}{c}}2&5\\4&{10}\end{array}} \right],\,$
$\left[ {\begin{array}{*{20}{c}}4&{10}\\{ - 6}&{ - 15}\end{array}} \right]$ their determinants is zero. Each rank can not be $2$. Thus rank =$ 1.$
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જો $x = cy + bz,\,\,y = az + cx,\,\,z = bx + ay\ ($કે જ્યાં $ x, y, z $ બધા શૂન્ય ન હોય$)$ તો $x = 0, y = 0, z = 0$ સિવાય નો ઉકેલ હોય તો $ a, b $ અને $c$ વચ્ચેનો સંબંધ મેળવો.
$-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ અંતરાલમાં $\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$ ના વાસ્તવિક ભિન્ન બીજની સંખ્યા મેળવો.