QBManager

Matrix and Determinant

79028 An equilateral triangle has each side equal to ' $a$ '. If the coordinates of its vertices are $(x_{1}, y_{1})$ , $(x_{2}, y_{2}), (x_{3}, y_{3})$ , then what is the square of the determinant $| \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} |$ equal to ?

1

3 a^{4}

2

3 a^{4} / 4

3

3 a^{4} / 2

4

2 a^{4}

Explanation:

(B) : Area of equilateral triangle $= \frac{\sqrt{3}}{4} (side)^{2} = \frac{\sqrt{3}}{4} a^{2}$
Area $= \frac{1}{2} | \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} | = {(\frac{\sqrt{3} a^{2}}{2})}^{2}$
Square of det. $= \frac{3 a^{4}}{4}$
$| \begin{array}{lll} 0 & 1 & 0 \end{array} |$

SCRA-2010

Matrix and Determinant

79029 If the system of equations $α x + y + z = 5, x + 2 y$ $+ 3 z = 4, x + 3 y + 5 z = β$ , has infinitely many solutions, then the ordered pair $(α, β)$ is equal to :

1

(1, - 3)

2

(- 1, 3)

3

(1, 3)

4

(- 1, - 3)

Explanation:

(C) : : Given,
$α x + y + z = 5$
$x + 2 y + 3 z = 4$
$x + 3 y + 5 z = β$
For infinitely many solutions,
$Δ = Δ_{1} = Δ_{2} = Δ_{3} = 0$
$Δ = | \begin{array}{lll} α & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{array} |$
$= α (10 - 9) - 1 (5 - 3) + 1 (3 - 2)$
$= α - 2 + 1 = α - 1$
$α - 1 = 0$
$α = 1 Δ_{1} = | \begin{array}{lll} 5 & 1 & 1 \\ 4 & 2 & 3 \\ b & 3 & 5 \end{array} | = 0 5 (10 - 9) - 1 (20 - 3 β) + 1 (12 - 2 β) = 0 β = 3 (α, β) = (1, 3)$

JEE Main-2022-26.06.2022

Matrix and Determinant

79030 Let $A = (\begin{array}{cc} 4 & - 2 \\ α & β \end{array})$ If $A^{2} + γ A + 18 I = 0$ , then $\det$ (A) is equal to

1 -18

2 18

3 -50

4 50

Explanation:

(B) : The characteristic equation for $A$ is $| A - λ I | = 0$
$\Rightarrow | \begin{array}{cc} 4 - λ & - 2 \\ α & β - λ \end{array} | = 0$
$\Rightarrow (4 - λ) (β - λ) + 2 α = 0$
$\Rightarrow$ Put $λ = A$
$A^{2} - (β + 4) A + (4 β + 2 α) I = 0$
On comparison
$- (β + 4) = γ & 4 β + 2 α = 18$
And $| A | = 4 β + 2 α = 18$

JEE Main-2022-27.07.2022

Matrix and Determinant

79031 Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{T} = α A + I$ , where $α \in R - {- 1, 1}$ . If det $(A^{2} - A) = 4$ , then the sum of all possible values of $α$ is equal to

1 0

2

\frac{3}{2}

3

\frac{5}{2}

4 2

Explanation:

(C)Given,
$A^{T} = α A + I$
$A = α A^{T} + I$
$A = α (α A + I) + I$
$A = α^{2} A + (α + 1) I$
$A (1 - α^{2}) = (α + 1) I$
$A = \frac{I}{1 - α}$
$| A | = \frac{I}{(1 - α)^{2}}$
$| A^{2} - A | = | A | | A - I |$
$A - I = \frac{I}{1 - α} - I = \frac{α}{1 - α} I$
$| A - I | = {(\frac{α}{1 - α})}^{2}$
Now, $| A^{2} - A | = 4$
$| A | | A - I | = 4$
$\begin{array}{ll} \Rightarrow & \frac{1}{(1 - α)^{2}} \frac{α^{2}}{(1 - α)^{2}} = 4 \\ \Rightarrow & \frac{α}{(1 - α)^{2}} = \pm 2 \\ \Rightarrow & 2 (1 - α)^{2} = \pm α \end{array}$
$(C_{1}) 2 (1 - α)^{2} = α$
$(C_{2}) 2 (1 - α)^{3} = - α$
$2 α^{2} - 5 α + 2 = 0$
$2 α^{2} - 3 α + 2 = 0$
$α_{1} + α_{2} = \frac{5}{2}$
$α \notin R$
Sum of value of $α = \frac{5}{2}$

JEE Main-2023-11.04.2023

Matrix and Determinant

79028 An equilateral triangle has each side equal to ' $a$ '. If the coordinates of its vertices are $(x_{1}, y_{1})$ , $(x_{2}, y_{2}), (x_{3}, y_{3})$ , then what is the square of the determinant $| \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} |$ equal to ?

1

3 a^{4}

2

3 a^{4} / 4

3

3 a^{4} / 2

4

2 a^{4}

Explanation:

(B) : Area of equilateral triangle $= \frac{\sqrt{3}}{4} (side)^{2} = \frac{\sqrt{3}}{4} a^{2}$
Area $= \frac{1}{2} | \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} | = {(\frac{\sqrt{3} a^{2}}{2})}^{2}$
Square of det. $= \frac{3 a^{4}}{4}$
$| \begin{array}{lll} 0 & 1 & 0 \end{array} |$

SCRA-2010

Matrix and Determinant

79029 If the system of equations $α x + y + z = 5, x + 2 y$ $+ 3 z = 4, x + 3 y + 5 z = β$ , has infinitely many solutions, then the ordered pair $(α, β)$ is equal to :

1

(1, - 3)

2

(- 1, 3)

3

(1, 3)

4

(- 1, - 3)

Explanation:

(C) : : Given,
$α x + y + z = 5$
$x + 2 y + 3 z = 4$
$x + 3 y + 5 z = β$
For infinitely many solutions,
$Δ = Δ_{1} = Δ_{2} = Δ_{3} = 0$
$Δ = | \begin{array}{lll} α & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{array} |$
$= α (10 - 9) - 1 (5 - 3) + 1 (3 - 2)$
$= α - 2 + 1 = α - 1$
$α - 1 = 0$
$α = 1 Δ_{1} = | \begin{array}{lll} 5 & 1 & 1 \\ 4 & 2 & 3 \\ b & 3 & 5 \end{array} | = 0 5 (10 - 9) - 1 (20 - 3 β) + 1 (12 - 2 β) = 0 β = 3 (α, β) = (1, 3)$

JEE Main-2022-26.06.2022

Matrix and Determinant

79030 Let $A = (\begin{array}{cc} 4 & - 2 \\ α & β \end{array})$ If $A^{2} + γ A + 18 I = 0$ , then $\det$ (A) is equal to

1 -18

2 18

3 -50

4 50

Explanation:

(B) : The characteristic equation for $A$ is $| A - λ I | = 0$
$\Rightarrow | \begin{array}{cc} 4 - λ & - 2 \\ α & β - λ \end{array} | = 0$
$\Rightarrow (4 - λ) (β - λ) + 2 α = 0$
$\Rightarrow$ Put $λ = A$
$A^{2} - (β + 4) A + (4 β + 2 α) I = 0$
On comparison
$- (β + 4) = γ & 4 β + 2 α = 18$
And $| A | = 4 β + 2 α = 18$

JEE Main-2022-27.07.2022

Matrix and Determinant

79031 Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{T} = α A + I$ , where $α \in R - {- 1, 1}$ . If det $(A^{2} - A) = 4$ , then the sum of all possible values of $α$ is equal to

1 0

2

\frac{3}{2}

3

\frac{5}{2}

4 2

Explanation:

(C)Given,
$A^{T} = α A + I$
$A = α A^{T} + I$
$A = α (α A + I) + I$
$A = α^{2} A + (α + 1) I$
$A (1 - α^{2}) = (α + 1) I$
$A = \frac{I}{1 - α}$
$| A | = \frac{I}{(1 - α)^{2}}$
$| A^{2} - A | = | A | | A - I |$
$A - I = \frac{I}{1 - α} - I = \frac{α}{1 - α} I$
$| A - I | = {(\frac{α}{1 - α})}^{2}$
Now, $| A^{2} - A | = 4$
$| A | | A - I | = 4$
$\begin{array}{ll} \Rightarrow & \frac{1}{(1 - α)^{2}} \frac{α^{2}}{(1 - α)^{2}} = 4 \\ \Rightarrow & \frac{α}{(1 - α)^{2}} = \pm 2 \\ \Rightarrow & 2 (1 - α)^{2} = \pm α \end{array}$
$(C_{1}) 2 (1 - α)^{2} = α$
$(C_{2}) 2 (1 - α)^{3} = - α$
$2 α^{2} - 5 α + 2 = 0$
$2 α^{2} - 3 α + 2 = 0$
$α_{1} + α_{2} = \frac{5}{2}$
$α \notin R$
Sum of value of $α = \frac{5}{2}$

JEE Main-2023-11.04.2023

Matrix and Determinant

79028 An equilateral triangle has each side equal to ' $a$ '. If the coordinates of its vertices are $(x_{1}, y_{1})$ , $(x_{2}, y_{2}), (x_{3}, y_{3})$ , then what is the square of the determinant $| \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} |$ equal to ?

1

3 a^{4}

2

3 a^{4} / 4

3

3 a^{4} / 2

4

2 a^{4}

Explanation:

(B) : Area of equilateral triangle $= \frac{\sqrt{3}}{4} (side)^{2} = \frac{\sqrt{3}}{4} a^{2}$
Area $= \frac{1}{2} | \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} | = {(\frac{\sqrt{3} a^{2}}{2})}^{2}$
Square of det. $= \frac{3 a^{4}}{4}$
$| \begin{array}{lll} 0 & 1 & 0 \end{array} |$

SCRA-2010

Matrix and Determinant

79029 If the system of equations $α x + y + z = 5, x + 2 y$ $+ 3 z = 4, x + 3 y + 5 z = β$ , has infinitely many solutions, then the ordered pair $(α, β)$ is equal to :

1

(1, - 3)

2

(- 1, 3)

3

(1, 3)

4

(- 1, - 3)

Explanation:

(C) : : Given,
$α x + y + z = 5$
$x + 2 y + 3 z = 4$
$x + 3 y + 5 z = β$
For infinitely many solutions,
$Δ = Δ_{1} = Δ_{2} = Δ_{3} = 0$
$Δ = | \begin{array}{lll} α & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{array} |$
$= α (10 - 9) - 1 (5 - 3) + 1 (3 - 2)$
$= α - 2 + 1 = α - 1$
$α - 1 = 0$
$α = 1 Δ_{1} = | \begin{array}{lll} 5 & 1 & 1 \\ 4 & 2 & 3 \\ b & 3 & 5 \end{array} | = 0 5 (10 - 9) - 1 (20 - 3 β) + 1 (12 - 2 β) = 0 β = 3 (α, β) = (1, 3)$

JEE Main-2022-26.06.2022

Matrix and Determinant

79030 Let $A = (\begin{array}{cc} 4 & - 2 \\ α & β \end{array})$ If $A^{2} + γ A + 18 I = 0$ , then $\det$ (A) is equal to

1 -18

2 18

3 -50

4 50

Explanation:

(B) : The characteristic equation for $A$ is $| A - λ I | = 0$
$\Rightarrow | \begin{array}{cc} 4 - λ & - 2 \\ α & β - λ \end{array} | = 0$
$\Rightarrow (4 - λ) (β - λ) + 2 α = 0$
$\Rightarrow$ Put $λ = A$
$A^{2} - (β + 4) A + (4 β + 2 α) I = 0$
On comparison
$- (β + 4) = γ & 4 β + 2 α = 18$
And $| A | = 4 β + 2 α = 18$

JEE Main-2022-27.07.2022

Matrix and Determinant

79031 Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{T} = α A + I$ , where $α \in R - {- 1, 1}$ . If det $(A^{2} - A) = 4$ , then the sum of all possible values of $α$ is equal to

1 0

2

\frac{3}{2}

3

\frac{5}{2}

4 2

Explanation:

(C)Given,
$A^{T} = α A + I$
$A = α A^{T} + I$
$A = α (α A + I) + I$
$A = α^{2} A + (α + 1) I$
$A (1 - α^{2}) = (α + 1) I$
$A = \frac{I}{1 - α}$
$| A | = \frac{I}{(1 - α)^{2}}$
$| A^{2} - A | = | A | | A - I |$
$A - I = \frac{I}{1 - α} - I = \frac{α}{1 - α} I$
$| A - I | = {(\frac{α}{1 - α})}^{2}$
Now, $| A^{2} - A | = 4$
$| A | | A - I | = 4$
$\begin{array}{ll} \Rightarrow & \frac{1}{(1 - α)^{2}} \frac{α^{2}}{(1 - α)^{2}} = 4 \\ \Rightarrow & \frac{α}{(1 - α)^{2}} = \pm 2 \\ \Rightarrow & 2 (1 - α)^{2} = \pm α \end{array}$
$(C_{1}) 2 (1 - α)^{2} = α$
$(C_{2}) 2 (1 - α)^{3} = - α$
$2 α^{2} - 5 α + 2 = 0$
$2 α^{2} - 3 α + 2 = 0$
$α_{1} + α_{2} = \frac{5}{2}$
$α \notin R$
Sum of value of $α = \frac{5}{2}$

JEE Main-2023-11.04.2023

Matrix and Determinant

79028 An equilateral triangle has each side equal to ' $a$ '. If the coordinates of its vertices are $(x_{1}, y_{1})$ , $(x_{2}, y_{2}), (x_{3}, y_{3})$ , then what is the square of the determinant $| \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} |$ equal to ?

1

3 a^{4}

2

3 a^{4} / 4

3

3 a^{4} / 2

4

2 a^{4}

Explanation:

(B) : Area of equilateral triangle $= \frac{\sqrt{3}}{4} (side)^{2} = \frac{\sqrt{3}}{4} a^{2}$
Area $= \frac{1}{2} | \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} | = {(\frac{\sqrt{3} a^{2}}{2})}^{2}$
Square of det. $= \frac{3 a^{4}}{4}$
$| \begin{array}{lll} 0 & 1 & 0 \end{array} |$

SCRA-2010

Matrix and Determinant

79029 If the system of equations $α x + y + z = 5, x + 2 y$ $+ 3 z = 4, x + 3 y + 5 z = β$ , has infinitely many solutions, then the ordered pair $(α, β)$ is equal to :

1

(1, - 3)

2

(- 1, 3)

3

(1, 3)

4

(- 1, - 3)

Explanation:

(C) : : Given,
$α x + y + z = 5$
$x + 2 y + 3 z = 4$
$x + 3 y + 5 z = β$
For infinitely many solutions,
$Δ = Δ_{1} = Δ_{2} = Δ_{3} = 0$
$Δ = | \begin{array}{lll} α & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{array} |$
$= α (10 - 9) - 1 (5 - 3) + 1 (3 - 2)$
$= α - 2 + 1 = α - 1$
$α - 1 = 0$
$α = 1 Δ_{1} = | \begin{array}{lll} 5 & 1 & 1 \\ 4 & 2 & 3 \\ b & 3 & 5 \end{array} | = 0 5 (10 - 9) - 1 (20 - 3 β) + 1 (12 - 2 β) = 0 β = 3 (α, β) = (1, 3)$

JEE Main-2022-26.06.2022

Matrix and Determinant

79030 Let $A = (\begin{array}{cc} 4 & - 2 \\ α & β \end{array})$ If $A^{2} + γ A + 18 I = 0$ , then $\det$ (A) is equal to

1 -18

2 18

3 -50

4 50

Explanation:

(B) : The characteristic equation for $A$ is $| A - λ I | = 0$
$\Rightarrow | \begin{array}{cc} 4 - λ & - 2 \\ α & β - λ \end{array} | = 0$
$\Rightarrow (4 - λ) (β - λ) + 2 α = 0$
$\Rightarrow$ Put $λ = A$
$A^{2} - (β + 4) A + (4 β + 2 α) I = 0$
On comparison
$- (β + 4) = γ & 4 β + 2 α = 18$
And $| A | = 4 β + 2 α = 18$

JEE Main-2022-27.07.2022

Matrix and Determinant

79031 Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{T} = α A + I$ , where $α \in R - {- 1, 1}$ . If det $(A^{2} - A) = 4$ , then the sum of all possible values of $α$ is equal to

1 0

2

\frac{3}{2}

3

\frac{5}{2}

4 2

Explanation:

(C)Given,
$A^{T} = α A + I$
$A = α A^{T} + I$
$A = α (α A + I) + I$
$A = α^{2} A + (α + 1) I$
$A (1 - α^{2}) = (α + 1) I$
$A = \frac{I}{1 - α}$
$| A | = \frac{I}{(1 - α)^{2}}$
$| A^{2} - A | = | A | | A - I |$
$A - I = \frac{I}{1 - α} - I = \frac{α}{1 - α} I$
$| A - I | = {(\frac{α}{1 - α})}^{2}$
Now, $| A^{2} - A | = 4$
$| A | | A - I | = 4$
$\begin{array}{ll} \Rightarrow & \frac{1}{(1 - α)^{2}} \frac{α^{2}}{(1 - α)^{2}} = 4 \\ \Rightarrow & \frac{α}{(1 - α)^{2}} = \pm 2 \\ \Rightarrow & 2 (1 - α)^{2} = \pm α \end{array}$
$(C_{1}) 2 (1 - α)^{2} = α$
$(C_{2}) 2 (1 - α)^{3} = - α$
$2 α^{2} - 5 α + 2 = 0$
$2 α^{2} - 3 α + 2 = 0$
$α_{1} + α_{2} = \frac{5}{2}$
$α \notin R$
Sum of value of $α = \frac{5}{2}$

JEE Main-2023-11.04.2023

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