Using properties of determinants, prove that: left|begin{array}{ccc}α & α^{2} & β+gamma ?

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3 and 4 .Determinants and Matrices

hard

Using properties of determinants, prove that:

$\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta\end{array}\right|=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$

Option A

Option B

Option C

Option D

Solution

$\Delta=\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta\end{array}\right|$

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1},$ we have:

$\Delta=\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ \beta-\alpha & \beta^{2}-\alpha^{2} & \alpha-\beta \\ \gamma-\alpha & \gamma^{2}-\alpha^{2} & \alpha-\gamma\end{array}\right|$

$=(\beta-\alpha)(\gamma-\alpha)\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ 1 & \beta+\alpha & -1 \\ 1 & \gamma+\alpha & -1\end{array}\right|$

Applying $R_{3} \rightarrow R_{3}-R_{2},$ we have:

$\Delta=(\beta-\alpha)(\gamma-\alpha)\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ 1 & \beta+\alpha & -1 \\ 0 & \gamma-\beta & 0\end{array}\right|$

Expanding along $R_{3},$ we have:

$\Delta=(\beta-\alpha)(\gamma-\alpha)[-(\gamma-\beta)(-\alpha-\beta-\gamma)]$

$=(\beta-\alpha)(\gamma-\alpha)(\gamma-\beta)(\alpha+\beta+\gamma)$

$=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)$

Hence, the given result is proved.

Standard 12

Mathematics

By using properties of determinants, show that:

$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$

hard

Using the property of determinants and without expanding, Prove that

$\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$

easy

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

medium

By using properties of determinants, show that:

$\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a+b)(b-c)(c-a)$

medium

If $a, b$ and $c$ are real numbers, and

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

Show that either $a+b+c=0$ or $a=b=c$.

hard

Using properties of determinants, prove that: $\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta\end{array}\right|=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$

Solution

Similar Questions

By using properties of determinants, show that: $\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$

Using the property of determinants and without expanding, Prove that $\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$

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

By using properties of determinants, show that: $\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a+b)(b-c)(c-a)$

If $a, b$ and $c$ are real numbers, and $\Delta=\left|\begin{array}{lll} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right|=0$ Show that either $a+b+c=0$ or $a=b=c$.

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Using properties of determinants, prove that:

$\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta\end{array}\right|=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$

By using properties of determinants, show that:

$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$

Using the property of determinants and without expanding, Prove that

$\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$

By using properties of determinants, show that:

$\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a+b)(b-c)(c-a)$

If $a, b$ and $c$ are real numbers, and

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

Show that either $a+b+c=0$ or $a=b=c$.