If a, b, c are all different from zero and left| {begin{array}{*{20}{c}} {1 + a}&1&1 1

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3 and 4 .Determinants and Matrices

normal

If $a, b, c$ are all different from zero and $\left| {\begin{array}{*{20}{c}} {1 + a}&1&1\\ 1&{1 + b}&1\\ 1&1&{1 + c} \end{array}} \right| = 0$ , then the value of $a^{-1} + b^{-1} + c^{-1}$ is

$abc$

$a^{-1}\, b^{-1}\, c^{-1}$

$-a-b-c$

$-1$

Solution

$C_1 \rightarrow C_1 – C_2 \, \,and\, \, C_2 \rightarrow C_2 – C_3 \, \,and\, \,$ then open by $R_1$ to get $ab + abc + ac + bc = 0;$ divided by $abc$

Standard 12

Mathematics

Prove that $\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|=4 a b c$

medium

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{31}&{37}&{92}\\{31}&{58}&{71}\\{31}&{105}&{24}\end{array}\,} \right|$ is

easy

Let $P=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]$ and $I$ be the identity matrix of order $3$ . If $\left.Q=q_{i j}\right]$ is a matrix such that $P^{50}-Q=I$, then $\frac{q_{31}+q_{32}}{q_{21}}$ equals

hard

(IIT-2016)

If $A =$ $\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$ (where $bc \ne 0$) satisfies the equations $x^2 + k = 0$, then