First slide

Introduction to Determinants

Question

Let $Δ = |\begin{array}{ccc} - bc & b^{2} + bc & c^{2} + bc \\ a^{2} + ac & - ac & c^{2} + ac \\ a^{2} + ab & b^{2} + ab & - ab \end{array}|$ and the equation ${px}^{3} + {qx}^{2} + rx + s = 0$ has roots a, b, c, where $a, b, c \in R^{+}$ .

Moderate

Question

The value of $Δ$ is

A

$r^{2} / p^{2}$
B

$r^{3} / p^{3}$
C

- s/p
D

none of these

Solution

Multiplying R₁, R₂, R₃ by a, b, c, respectively, and then taking a, b, c common from C₁, C₂, and C₃, we get
$Δ = |\begin{array}{ccc} - bc & ab + ac & ac + ab \\ ab + bc & - ac & bc + ab \\ ac + bc & bc + ac & - ab \end{array}|$
Now, using $C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{1}$ , and then taking ( ab + bc + ca) common from C₂ and C₃, we get
$Δ = |\begin{array}{ccc} - bc & 1 & 1 \\ ab + bc & - 1 & 0 \\ ac + bc & 0 & - 1 \end{array}| \times (ab + bc + ca)^{2}$
Now, applying $R_{2} \to R_{2} + R_{1}$ , we get
$Δ = |\begin{array}{ccc} - bc & 1 & 1 \\ ab & 0 & 1 \\ ac + bc & 0 & - 1 \end{array}| (ab + bc + ca)^{2}$
Expanding along C₂, we get
$\begin{array}{l} Δ = (ab + bc + ca)^{2} [ac + bc + ab] \\ = (ab + bc + ca)^{3} \\ = (r / p)^{3} = r^{3} / p^{3} \end{array}$
Now given a, b, c are all positive, then
$A.M. \geq G.M.$
$\Rightarrow \frac{ab + bc + ac}{3} \geq (ab \times bc \times ac)^{1 / 3}$
$\begin{array}{l} or (ab + bc + ac)^{3} \geq 27 a^{2} b^{2} c^{2} \\ or (ab + bc + ac)^{3} \geq 27 (s^{2} / p^{2}) \end{array}$
If $Δ = 27$ , then ab + bc + ca = 3, and given that a² + b² + c² = 3, from (a + b + c)² = a² + b² + c² + 2(ab + bc + ca), we have
$a + b + c = \pm 3$
$\begin{array}{l} \Rightarrow a + b + c = 3 (∵ a, b, c \in R^{+}) \\ \Rightarrow 3 p + q = 0 \end{array}$

Question

The value of $∆$ is

A

$\leq 9 r^{2} / p^{2}$
B

$\geq 27 s^{2} / p^{2}$
C

$\leq 27 s^{3} / p^{3}$
D

none of these

Solution

Multiplying R₁, R₂, R₃ by a, b, c, respectively, and then taking a, b, c common from C₁, C₂, and C₃, we get
$Δ = |\begin{array}{ccc} - bc & ab + ac & ac + ab \\ ab + bc & - ac & bc + ab \\ ac + bc & bc + ac & - ab \end{array}|$
Now, using $C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{1}$ , and then taking ( ab + bc + ca) common from C₂ and C₃, we get
$Δ = |\begin{array}{ccc} - bc & 1 & 1 \\ ab + bc & - 1 & 0 \\ ac + bc & 0 & - 1 \end{array}| \times (ab + bc + ca)^{2}$
Now, applying $R_{2} \to R_{2} + R_{1}$ , we get
$Δ = |\begin{array}{ccc} - bc & 1 & 1 \\ ab & 0 & 1 \\ ac + bc & 0 & - 1 \end{array}| (ab + bc + ca)^{2}$
Expanding along C₂, we get
$\begin{array}{l} Δ = (ab + bc + ca)^{2} [ac + bc + ab] \\ = (ab + bc + ca)^{3} \\ = (r / p)^{3} = r^{3} / p^{3} \end{array}$
Now given a, b, c are all positive, then
$A.M. \geq G.M.$
$\Rightarrow \frac{ab + bc + ac}{3} \geq (ab \times bc \times ac)^{1 / 3}$
$\begin{array}{l} or (ab + bc + ac)^{3} \geq 27 a^{2} b^{2} c^{2} \\ or (ab + bc + ac)^{3} \geq 27 (s^{2} / p^{2}) \end{array}$
If $Δ = 27$ , then ab + bc + ca = 3, and given that a² + b² + c² = 3, from (a + b + c)² = a² + b² + c² + 2(ab + bc + ca), we have
$a + b + c = \pm 3$
$\begin{array}{l} \Rightarrow a + b + c = 3 (∵ a, b, c \in R^{+}) \\ \Rightarrow 3 p + q = 0 \end{array}$

Question

If $Δ = 27$ , and a² + b² + c² = 3, then

A

3p + 2q = 0
B

4p + 3q = 0
C

3p + q = 0
D

none of these

Solution

Multiplying R₁, R₂, R₃ by a, b, c, respectively, and then taking a, b, c common from C₁, C₂, and C₃, we get
$Δ = |\begin{array}{ccc} - bc & ab + ac & ac + ab \\ ab + bc & - ac & bc + ab \\ ac + bc & bc + ac & - ab \end{array}|$
Now, using $C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{1}$ , and then taking ( ab + bc + ca) common from C₂ and C₃, we get
$Δ = |\begin{array}{ccc} - bc & 1 & 1 \\ ab + bc & - 1 & 0 \\ ac + bc & 0 & - 1 \end{array}| \times (ab + bc + ca)^{2}$
Now, applying $R_{2} \to R_{2} + R_{1}$ , we get
$Δ = |\begin{array}{ccc} - bc & 1 & 1 \\ ab & 0 & 1 \\ ac + bc & 0 & - 1 \end{array}| (ab + bc + ca)^{2}$
Expanding along C₂, we get
$\begin{array}{l} Δ = (ab + bc + ca)^{2} [ac + bc + ab] \\ = (ab + bc + ca)^{3} \\ = (r / p)^{3} = r^{3} / p^{3} \end{array}$
Now given a, b, c are all positive, then
$A.M. \geq G.M.$
$\Rightarrow \frac{ab + bc + ac}{3} \geq (ab \times bc \times ac)^{1 / 3}$
$\begin{array}{l} or (ab + bc + ac)^{3} \geq 27 a^{2} b^{2} c^{2} \\ or (ab + bc + ac)^{3} \geq 27 (s^{2} / p^{2}) \end{array}$
If $Δ = 27$ , then ab + bc + ca = 3, and given that a² + b² + c² = 3, from (a + b + c)² = a² + b² + c² + 2(ab + bc + ca), we have
$a + b + c = \pm 3$
$\begin{array}{l} \Rightarrow a + b + c = 3 (∵ a, b, c \in R^{+}) \\ \Rightarrow 3 p + q = 0 \end{array}$

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