Left| {,begin{array}{*{20}{c}}1&a&{{a^2} - bc}1&b&{{b^2} - ac}1&c&{{c^2}

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3 and 4 .Determinants and Matrices

medium

$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2} - bc}\\1&b&{{b^2} - ac}\\1&c&{{c^2} - ab}\end{array}\,} \right| = $

$0$

${a^3} + {b^3} + {c^3} - 3abc$

$3abc$

${(a + b + c)^3}$

(IIT-1988)

Solution

(a) $\left| \,\begin{matrix}
1 & a & {{a}^{2}}-bc \\
1 & b & {{b}^{2}}-ac \\
1 & c & {{c}^{2}}-ab \\
\end{matrix}\, \right|=\left| \,\begin{matrix}
0 & a-b & (a-b)\,(a+b+c) \\
0 & b-c & (b-c)\,\,(a+b+c) \\
1 & c & {{c}^{2}}-ab \\
\end{matrix}\, \right|$

by $\left\{ \begin{array}{l}{R_1} \to {R_1} – {R_2}\\{R_2} \to {R_2} – {R_3}\end{array} \right.$

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

$\{\because\,\,{R_1} \equiv {R_2}\} $

Standard 12

Mathematics

If $ 5$ is one root of the equation $\left| {\,\begin{array}{*{20}{c}}x&3&7\\2&x&{ – 2}\\7&8&x\end{array}\,} \right| = 0$, then other two roots of the equation are

medium

Let $A =$ $\left[ {\begin{array}{*{20}{c}}{1 + {x^2} – {y^2} – {z^2}}&{2(xy + z)}&{2(zx – y)}\\{2(xy – z)}&{1 + {y^2} – {z^2} – {x^2}}&{2(yz + x)}\\{2(zx + y)}&{2(yz – x)}&{1 + {z^2} – {x^2} – {y^2}}\end{array}} \right]$ then det. $A$ is equal to

normal

The system of equations $-k x+3 y-14 z=25$ $-15 x+4 y-k z=3$ $-4 x+y+3 z=4$ is consistent for all $k$ in the set

medium

(JEE MAIN-2022)

If $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha – b}\\b&c&{b\alpha – c}\\2&1&0\end{array}\,} \right| = 0$ and $\alpha \ne \frac{1}{2},$ then

medium

Consider the system of linear equations ${a_1}x + {b_1}y + {c_1}z + {d_1} = 0$, ${a_2}x + {b_2}y + {c_2}z + {d_2} = 0$ and ${a_3}x + {b_3}y + {c_3}z + {d_3} = 0$. Let us denote by $\Delta (a,b,c)$ the determinant $\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ if $\Delta (a,b,c) \ne 0$, then the value of $x$ in the unique solution of the above equations is

medium

$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2} - bc}\\1&b&{{b^2} - ac}\\1&c&{{c^2} - ab}\end{array}\,} \right| = $

Solution

Similar Questions

If $ 5$ is one root of the equation $\left| {\,\begin{array}{*{20}{c}}x&3&7\\2&x&{ – 2}\\7&8&x\end{array}\,} \right| = 0$, then other two roots of the equation are

Let $A =$ $\left[ {\begin{array}{*{20}{c}}{1 + {x^2} – {y^2} – {z^2}}&{2(xy + z)}&{2(zx – y)}\\{2(xy – z)}&{1 + {y^2} – {z^2} – {x^2}}&{2(yz + x)}\\{2(zx + y)}&{2(yz – x)}&{1 + {z^2} – {x^2} – {y^2}}\end{array}} \right]$ then det. $A$ is equal to

The system of equations $-k x+3 y-14 z=25$ $-15 x+4 y-k z=3$ $-4 x+y+3 z=4$ is consistent for all $k$ in the set

If $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha – b}\\b&c&{b\alpha – c}\\2&1&0\end{array}\,} \right| = 0$ and $\alpha \ne \frac{1}{2},$ then

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