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

English

Gujarati

Hindi

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

Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a b c+b c+c a+a b$

hard

By using properties of determinants, show that:

$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$

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

By using properties of determinants, show that:

$\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|=\left(1-x^{3}\right)^{2}$

hard

Let $P=\left[a_{\|}\right]$be $a \times 3$ matrix and let $Q=\left[b_1\right]$, where $b_1=2^{1+j} a_{\|}$for $1 \leq i, j \leq 3$. If the determinant of $P$ is $2$ , then the determinant of the matrix $Q$ is

hard

(IIT-2012)

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

Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a b c+b c+c a+a b$

By using properties of determinants, show that: $\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$

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

By using properties of determinants, show that: $\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|=\left(1-x^{3}\right)^{2}$

Let $P=\left[a_{\|}\right]$be $a \times 3$ matrix and let $Q=\left[b_1\right]$, where $b_1=2^{1+j} a_{\|}$for $1 \leq i, j \leq 3$. If the determinant of $P$ is $2$ , then the determinant of the matrix $Q$ is

Start a Free Trial Now

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:

$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$

By using properties of determinants, show that:

$\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|=\left(1-x^{3}\right)^{2}$