If mathrm{a, b, c} are positive and unequal, show that value of determinant Delta=left|begin{array}{

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3 and 4 .Determinants and Matrices

hard

If $\mathrm{a, b, c}$ are positive and unequal, show that value of the determinant $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is negative.

Option A

Option B

Option C

Option D

Solution

Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}$ to the given determinant, we get

$\Delta = \left| {\begin{array}{*{20}{l}}
{a + b + c}&b&c \\
{a + b + c}&c&a \\
{a + b + c}&a&b
\end{array}} \right| = (a + b + c)\left| {\begin{array}{*{20}{c}}
1&b&c \\
1&c&a \\
1&a&b
\end{array}} \right|$

${ = (a + b + c)\left| {\begin{array}{*{20}{c}}
1&b&c \\
0&{c – b}&{a – c} \\
0&{a – b}&{b – c}
\end{array}} \right|}$ ${{\text{(Applying }}{{\text{R}}_2} \to {{\text{R}}_2} – {{\text{R}}_1},{\text{ and }}{{\text{R}}_3} \to {{\text{R}}_3} – {{\text{R}}_1})}$

$ = (a + b + c)[(c – b)(b – c) – (a – c)(a – b)]$ ${\text{(Expanding along }}{{\text{C}}_1}{\text{ ) }}$

$ = (a + b + c)\left( { – {a^2} – {b^2} – {c^2} + ab + bc + ca} \right)$

which is negative (since $\left.a+b+c>0 \text { and }(a-b)^{2}+(b-c)^{2}+(c-a)^{2}>0\right)$

Standard 12

Mathematics

Evaluate $\left|\begin{array}{ccc}1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y\end{array}\right|$

medium

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$

easy

The value of $\left| {\,\begin{array}{*{20}{c}}{{5^2}}&{{5^3}}&{{5^4}}\\{{5^3}}&{{5^4}}&{{5^5}}\\{{5^4}}&{{5^5}}&{{5^7}}\end{array}\,} \right|$ is

easy

Give the correct order of initials $T$ or $F$ for following statements. Use $T$ if statement is true and $F$ if it is false.

Statement $-1$ : If the graphs of two linear equations in two variables are neither parallel nor the same, then there is a unique solution to the system. Statement $-2$ : If the system of equations $ax + by = 0, cx + dy = 0$ has a non-zero solution, then it has infinitely many solutions.

Statement $-3$ : The system $x + y + z = 1, x = y, y = 1 + z$ is inconsistent. Statement $-4$ : If two of the equations in a system of three linear equations are inconsistent, then the whole system is inconsistent.

normal

If $a, b, c,$ are non zero complex numbers satisfying $a^2 + b^2 + c^2 = 0$ and $\left| {\begin{array}{*{20}{c}}
{{b^2} + {c^2}}&{ab}&{ac}\\
{ab}&{{c^2} + {a^2}}&{bc}\\
{ac}&{bc}&{{a^2} + {b^2}}
\end{array}} \right| = k{a^2}{b^2}{c^2},$ then $k$ is equal to

hard

(AIEEE-2012)

If $\mathrm{a, b, c}$ are positive and unequal, show that value of the determinant $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is negative.

Solution

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The value of $\left| {\,\begin{array}{*{20}{c}}{{5^2}}&{{5^3}}&{{5^4}}\\{{5^3}}&{{5^4}}&{{5^5}}\\{{5^4}}&{{5^5}}&{{5^7}}\end{array}\,} \right|$ is

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Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$

If $a, b, c,$ are non zero complex numbers satisfying $a^2 + b^2 + c^2 = 0$ and $\left| {\begin{array}{*{20}{c}}
{{b^2} + {c^2}}&{ab}&{ac}\\
{ab}&{{c^2} + {a^2}}&{bc}\\
{ac}&{bc}&{{a^2} + {b^2}}
\end{array}} \right| = k{a^2}{b^2}{c^2},$ then $k$ is equal to