$\left| {\begin{array}{*{20}{c}}{a - 1}&a&{bc}\\{b - 1}&b&{ca}\\{c - 1}&c&{ab}\end{array}} \right| = $

  • A
    $0$
  • B
    $(a - b)(b - c)(c - a)$
  • C
    ${a^3} + {b^3} + {c^3} - 3abc$
  • D
    આમાંથી કોઈ નહીં

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Similar Questions

નિશ્ચાયક $\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$ નું મૂલ્ય શું છે?

જો $\left| {\begin{array}{*{20}{c}} {a - b}&{b - c}&{c - a} \\ {b - c}&{c - a}&{a - b} \\ {c - a + 1}&{a - b}&{b - c} \end{array}} \right| = 0$,જ્યાં $a, b, c \in R - \{0\}$,તો:

નિશ્ચાયકના ગુણધર્મોનો ઉપયોગ કરીને સાબિત કરો કે:
$\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|=1$

જો $A$ અને $B$ બે ચોરસ શ્રેણિકો હોય,જ્યાં $\det(A) = 5$ અને $\det(B^T \cdot A^T) = -15$ હોય,તો $\det(B)$ ની કિંમત શોધો.

સાબિત કરો કે $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=abc+bc+ca+ab$.

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