Determinants Test

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Determinants Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

If $$\triangle_{r}=\begin{vmatrix} { 2 }^{ r-1 } & 2.{ 3 }^{ r-1 } & 4.{ 5 }^{ r-1 } \\ x & y & z \\ { 2 }^{ n-1 } & { 3 }^{ n-1 } & { 5 }^{ n-1 } \end{vmatrix}$$, then $$\sum_{r=1}^{n}(\triangle_{r})$$ is equal to

A
$$xyz$$

B
$$1$$

C
$$-1$$

D
$$0$$

Solution
Question 2
1 / -0

if $$a^{2},b^{2}+c^{2}+ab+bc+ca \le 0 \forall a,b,c \epsilon R$$, then value of the determinant $$\left| \begin{matrix} \left( a^{ 2 }+b^{ 2 }+c^{ 2 } \right) ^{ 2 } & a^{ 2 }+b^{ 2 } & 1 \\ 1 & (b+c+2) & b^{ 2 }+c^{ 2 } \\ c^{ 2 }+a^{ 2 } & 1 & (c+a+2)^{ 2 } \end{matrix} \right| $$ equals

A
$$65$$

B
$$a^{2}+b^{2}+c^{2}+31$$

C
$$4(a^{2}+b^{2}+c^{2})$$

D
$$0$$

Solution
Question 3
1 / -0

Adj $$\begin{bmatrix} 1 & 0 & 2 \\ -1 & 5 & -2 \\ 0 & 2 & 1 \end{bmatrix}=\begin{bmatrix} 9 & a & -2 \\ -1 & 1 & 0 \\ -2 & 2 & b \end{bmatrix}\Rightarrow \left[ \begin{matrix} a & b \end{matrix} \right] =$$

A
$$\left[ \begin{matrix} -4 & 5 \end{matrix} \right]$$

B
$$\left[ \begin{matrix} -4 & -1 \end{matrix} \right]$$

C
$$\left[ \begin{matrix} 4 & 1 \end{matrix} \right]$$

D
$$\left[ \begin{matrix} 4 & -1 \end{matrix} \right]$$

Solution

$$\begin{array}{l} A=\left( { \begin{array} { *{ 20 }{ c } }1 & 0 & 2 \\ { -1 } & 5 & { -2 } \\ 0 & 2 & 1 \end{array} } \right) \\ \Rightarrow cofactor\, \, of\, \, A \\ \Rightarrow { C_{ 11 } }=9,\, \, { C_{ 12 } }=-1 ,\, \, { C_{ 13 } }=-2 \\\Rightarrow{C_{21}}=-4,\ {C_{22}}=1,\ {C_{23}}=2\\ \Rightarrow { C_{ 31 } }=-10,\, \, { c_{ 32 } }=0.\, \, { c_{ 33 } }=5 \\ \Rightarrow adJ\left( { \begin{array} { *{ 20 }{ c } }1 & 0 & 2 \\ { -1 } & 5 & { -2 } \\ 0 & 2 & 1 \end{array} } \right) =\left( { \begin{array} { *{ 20 }{ c } }9 & -4 & { -10 } \\ -1 & 1 & 0 \\ { -2 } & { 2 } & 5 \end{array} } \right) =\left( { \begin{array} { *{ 20 }{ c } }9 & a & { -10 } \\ -1 & 1 & 0 \\ { -2 } & { 2 } & b \end{array} } \right) \end{array}$$
hence by compearing $$a = -4,\,\,b = 5$$
Question 4
1 / -0

$$\Delta = \left| \matrix{ 1 + {a^2} + {a^4}\;\;1 + ab + {a^2}{b^2}\;\;1 + ac + {a^2}{c^2} \hfill \cr 1 + ab + {a^2}{b^2}\;\;1 + {b^2} + {b^4}\;\;1 + bc + {b^2}{c^2} \hfill \cr 1 + ac + {a^2}{c^2}\;\;1 + bc + {b^2}{c^2}\;\;1 + {c^2} + {c^4} \hfill \cr} \right|is\;equal\;to$$

A
$${\left( {a + b + c} \right)^6}$$

B
$${\left( {a - b} \right)^2}{\left( {b - c} \right)^2}{\left( {c - a} \right)^2}$$

C
4 (a-b)(b-c)(c-a)

D
None of these

Solution
Question 5
1 / -0

Let $$A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \, and \, U_1, U_2, U_3$$ be column matrices satisfying $$AU_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , AU_2 = \begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix} , AU_3 = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$$. If U is $$3 \times 3$$ matrix, whose columns are $$U_1, U_2, U_3$$. then |U| is

A
$$-11$$

B
$$-3$$

C
$$\dfrac{3}{2}$$

D
$$2$$

Solution
Question 6
1 / -0

Matrix $$A = \left| {\begin{array}{*{20}{c}}x & 3 & 2\\1 & y & 4\\2 & 2 & z\end{array}} \right|$$, if $$xyz=60$$ and $$8x+4y+3z=20$$, then $$a(adjA)$$ is equal to

A
$$\left| {\begin{array}{*{20}{c}}
{64} & 0 & 0\\
0 & {64} & 0\\
0 & 0 & {64}
\end{array}} \right|$$

B
$$\left| {\begin{array}{*{20}{c}}
{88} & 0 & 0\\
0 & {88} & 0\\
0 & 0 & {88}
\end{array}} \right|$$

C
$$\left| {\begin{array}{*{20}{c}}
{68} & 0 & 0\\
0 & {68} & 0\\
0 & 0 & {68}
\end{array}} \right|$$

D
$$\left| {\begin{array}{*{20}{c}}
{34} & 0 & 0\\
0 & {34} & 0\\
0 & 0 & {34}
\end{array}} \right|$$

Solution
Question 7
1 / -0

$$\begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix}=$$

A
$$xyz\left( \dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } \right) $$

B
$$xyz$$

C
$$1+\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z }$$

D
$$\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z }$$

Solution
Question 8
1 / -0

The number of distinct values of a $$2 \times 2$$ determinant whose entries are from set $$\{-1, 0, 1\}$$ is

A
$$4$$

B
$$6$$

C
$$5$$

D
$$3$$

Solution
Question 9
1 / -0

$$f\left( x \right) = \left| {\begin{array}{*{20}{c}}{x - 2}&{{{\left( {x - 1} \right)}^2}}&{{x^3}}\\{x - 1}&{{x^2}}&{{{\left( {x + 1} \right)}^3}}\\x&{{{\left( {x + 1} \right)}^2}}&{{{\left( {x + 2} \right)}^3}}\end{array}} \right|$$

A
$$0$$

B
$$2$$

C
$$-2$$

D
None of these

Solution
Question 10
1 / -0

If $$\theta \varepsilon R,$$ then the determinant $$\Delta =\begin{vmatrix} \sin { \theta } & \cos { \theta } & \sin { 2\theta } \\ \sin { \left( \theta +\dfrac { 2\pi }{ 3 } \right) } & \cos { \left( \theta +\dfrac { 2\pi }{ 3 } \right) } & \sin { \left( 2\theta +\dfrac { 4\pi }{ 3 } \right) } \\ \sin { \left( \theta -\dfrac { 2\pi }{ 3 } \right) } & \cos { \left( \theta -\dfrac { 2\pi }{ 3 } \right) } & \sin { \left( 2\theta -\dfrac { 4\pi }{ 3 } \right) } \end{vmatrix}=$$

A
$$-\sin {\theta} - \cos {\theta}$$

B
$$\sin {2\theta}$$

C
$$1+\sin {2\theta} - \cos {2\theta}$$

D
$$None\ of\ these$$