If a^2 + b^2 + c^2 = - 2 and f (x) = left| {,begin{array}{*{20}{c}}{1 + {a^2}x}&{(1 + {b^2})x}&a

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3 and 4 .Determinants and Matrices

normal

If $a^2 + b^2 + c^2 = - 2$ and $f (x) = $ $\left| {\,\begin{array}{*{20}{c}}{1 + {a^2}x}&{(1 + {b^2})x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{1 + {b^2}x}&{(1 + {c^2})x}\\ {(1 + {a^2})x}&{(1 + {b^2})x}&{1 + {c^2}x}\end{array}\,} \right|$ then $f (x)$ is a polynomial of degree

$0$

$1$

$2$

$3$

Solution

$C_1 \rightarrow C_1 + C_2 + C_3$

$\left| {\,\begin{array}{*{20}{c}} {1 + 2x + x({a^2} + {b^2} + {c^2})}&{(1 + {b^2})x}&{(1 + {c^2})x}\\ {1 + 2x + x({a^2} + {b^2} + {c^2})}&{1 + {b^2}x}&{(1 + {c^2})x}\\ {1 + 2x + x({a^2} + {b^2} + {c^2})}&{(1 + {b^2})x}&{1 + {c^2}x} \end{array}\,} \right|$

$\left| {\,\begin{array}{*{20}{c}}1&{(1 + {b^2})x}&{(1 + {c^2})x}\\1&{1 + {b^2}x}&{(1 + {c^2})x}\\1&{(1 + {b^2})x}&{1 + {c^2}x} \end{array}\,} \right|$

$R_2 \rightarrow R-2 – R_1\, and\, R_3 \rightarrow R_3 – R_1$

$\left| {\,\begin{array}{*{20}{c}}1&{(1 + {b^2})x}&{(1 + {c^2})x}\\0&{1 – x}&0\\ 0&0&{1 – x}\end{array}\,} \right|$

$f (x) = (1 – x)^2 = 1 – 2x + x^2$

Standard 12

Mathematics

The value of $\left| {\,\begin{array}{*{20}{c}}1&1&1\\{bc}&{ca}&{ab}\\{b + c}&{c + a}&{a + b}\end{array}\,} \right|$is

medium

If $A, B, C$ are the angles of a triangle and $\left| {\begin{array}{*{20}{c}}1&1&1\\{1 + \sin A}&{1 + \sin B}&{1 + \sin C}\\{\sin A + {{\sin }^2}A}&{\sin B + {{\sin }^2}B}&{\sin C + {{\sin }^2}C} \end{array}} \right|$ $= 0$, then the triangle is

normal

Evaluate $\left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right|$

medium

$\left| {\,\begin{array}{*{20}{c}}{a – b – c}&{2a}&{2a}\\{2b}&{b – c – a}&{2b}\\{2c}&{2c}&{c – a – b}\end{array}\,} \right| = $

hard

Evaluate $\left|\begin{array}{ccc}102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|$

easy