Determinants Test

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Determinants Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

If $$a, b$$ and $$c$$ are in $$AP$$, then determinant $$\begin{vmatrix}x + 2 & x + 3 & x + 2a\\ x + 3 & x + 4 & x + 2b\\ x + 4 & x + 5 & x + 2c\end{vmatrix}$$ is

A
$$0$$

B
$$1$$

C
$$x$$

D
$$2x$$

Solution

Let $$\triangle = \begin{vmatrix}x + 2 & x + 3 & x + 2a\\ x + 3 & x + 4 & x + 2b\\ x + 4 & x + 5 & x + 2c\end{vmatrix}$$

$$= \dfrac {1}{2} \begin{vmatrix}x + 2 & x + 3 & x + 2a\\ 0 & 0 & 2(2b - a - c)\\ x + 4 & x + 5 & x + 2c\end{vmatrix}$$ ...... (using $$R_{2}\rightarrow 2R_{2} - R_{1} - R_{2})$$

But $$a, b$$ and $$c$$ are in $$AP$$ using $$2b = a + c$$, we get
$$\triangle = \dfrac {1}{2} \begin{vmatrix}x + 2 & x + 3 & x + 2a\\ 0 & 0 & 0\\ x + 4 & x + 5 & x + 2c\end{vmatrix}$$
$$=0$$ ........ (Since, all elements of $$R_{2}$$ are zero.)
Hence, $$\begin{vmatrix}x + 2 & x + 3 & x + 2a\\ x + 3 & x + 4 & x + 2b\\ x + 4 & x + 5 & x + 2c\end{vmatrix}=0$$
Question 2
1 / -0

If $$\triangle (r) = \begin{vmatrix}r & r^{3}\\ 1 & n(n + 1)\end{vmatrix}$$, then $$\displaystyle \sum_{r = 1}^{n} \triangle (r)$$ is equal to

A
$$\displaystyle \sum_{r = 1}^{n} r^{2}$$

B
$$\displaystyle \sum_{r = 1}^{n} r^{3}$$

C
$$\displaystyle \sum_{r = 1}^{n} r$$

D
$$\displaystyle \sum_{r = 1}^{n} r^{4}$$

Solution

$$\triangle (r) = \begin{vmatrix}r & r^{3}\\ 1 & n(n + 1)\end{vmatrix}$$
$$\implies \displaystyle \sum_{r = 1}{n} \triangle (r) = \begin{vmatrix}\displaystyle \sum_{r = 1}^{n}r &\displaystyle \sum_{r = 1}^{n}r^{3} \\ 1 & n(n + 1)\end{vmatrix}$$

$$= \begin{vmatrix} \dfrac {n(n + 1)}{2}& \dfrac {[n(n + 1)]^{2}}{4}\\ 1 & n(n + 1)\end{vmatrix}$$

$$= \dfrac {[n(n + 1)]^{2}}{2} - \dfrac {[n(n + 1)]^{2}}{4}$$

$$= \dfrac {[n(n + 1)]^{2}}{4} = \displaystyle \sum_{r = 1}^{n} r^{3}$$
Question 3
1 / -0

If the points $$(2,5),(4,6)$$ and $$(a,a)$$ are collinear, then the value of $$a$$ is equal to

A
$$-8$$

B
$$4$$

C
$$-4$$

D
$$8$$

Solution

Consider the given points.
$$(2, 5), (4, 6)$$ and $$(a, a)$$

Since, these points are collinear means that the area of triangle must me zero.

So,

$$\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=0$$

where $$(x_1,y_1),(x_2,y_2),(x_3,y_3)$$ are the points

Therefore,
$$2(6-a)+4(a-5)+a(5-6)=0$$

$$12-2a+4a-20+5a-6a=0$$

$$a-8=0$$

$$a=8$$

Hence, this is the answer.
Question 4
1 / -0

If A is an invertible matrix, then what is $$det$$ $$(A^{-1})$$ equal to?

A
$$det A$$

B
$$\displaystyle\frac{1}{det A}$$

C
$$1$$

D
None of the above

Solution

We know
$$A{ A }^{ -1 }=I$$
$$\Rightarrow \left| A{ A }^{ -1 } \right| =\left| I \right| $$
$$\Rightarrow \left| A \right| \left| { A }^{ -1 } \right| =1$$
$$\Rightarrow \left| { A }^{ -1 } \right| =\cfrac { 1 }{ \left| A \right| } $$
$$\Rightarrow \left| { A }^{ -1 } \right| =\left| A \right| ^{ -1 }$$
Option B
Question 5
1 / -0

If $$ A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} $$ and $$ B = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$$ then what is determinant of AB ?

A
$$0$$

B
$$1$$

C
$$10$$

D
$$20$$

Solution

Given $$A=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$$ and $$B=\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$$

$$A\times B=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}\times \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 1 \times 1+2 \times 0 & 1 \times 1+2 \times 0 \\ 2 \times 1+3 \times 0 & 2 \times 1+3 \times 0 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix}\\$$
$$ \left| A\times B \right| =\begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix}=2 \times 1-2 \times 1=0$$
Question 6
1 / -0

If $$\begin{vmatrix} 8 & -5 & 1 \\ 5 & x & 3\\ 6 & 3 & 1 \end{vmatrix} = 2 $$ then what is the value of $$x$$ ?

A
$$44$$

B
$$55$$

C
$$61$$

D
$$84$$

Solution

$$\begin{vmatrix} 8 & -5 & 1 \\ 5 & x & 3 \\ 6 & 3 & 1 \end{vmatrix}=2$$
$$8\left( x-9 \right) +5\left( 5-18 \right) +1\left( 15-6x \right) =2$$
$$8x-72-65+15-6x=2$$
$$2x=122$$
$$x=61$$
Question 7
1 / -0

The cofactor of the element $$4$$ in the determinant
$$\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 5 & 8 & 9 \end{vmatrix}$$
is

A
$$2$$

B
$$4$$

C
$$6$$

D
$$-6$$

Solution

Cofactor of element 4 in $$\left| \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} \right| $$ is $${(-1)}^3\left| \begin{matrix} 2 & 3 \\ 8 & 9 \end{matrix} \right| =6$$

We got the matrice $$\left| \begin{matrix} 2 & 3 \\ 8 & 9 \end{matrix} \right| $$ after deleting column 1 and row 2.
Question 8
1 / -0

If $$\begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1\\ 20 & 3 & i\end{vmatrix} =x+iy,$$ where $$i=\sqrt{-1}$$, then what is x equal to?

A
$$3$$

B
$$2$$

C
$$1$$

D
$$0$$

Solution

$$\begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix}$$

$${ R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 2 }$$

$$\begin{vmatrix} 6i+4 & 0 & 0 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix}$$

$$6i+4(-3+3)=0$$

$$\Rightarrow 0+i(0)=x+i4$$

$$\Rightarrow x=0$$
Question 9
1 / -0

If $$\begin{vmatrix} a & b & 0 \\ 0 & a & b \\ b & 0 & a\end{vmatrix}=0$$, then which one of the following is correct?

A
$$\displaystyle\frac{a}{b}$$ is one of the cube roots of unity

B
$$\displaystyle\frac{a}{b}$$ is one of the cube roots of $$-1$$.

C
a is one of the cube roots of unity

D
b is one of the cube roots of unity.

Solution

$$\begin{vmatrix} a & b & 0 \\ 0 & a & b \\ b & 0 & a \end{vmatrix}=0$$
$$\Rightarrow a\left( { a }^{ 2 } \right) -b\left( -{ b }^{ 2 } \right) =0$$
$$\Rightarrow { a }^{ 3 }=-{ b }^{ 3 }$$
$$=\cfrac { a }{ b } =\sqrt [ 3 ]{ (-1) } $$
Option B
Question 10
1 / -0

Which of the following are correct in respect of the system of equations $$x + y + z = 8, x - y + 2z = 6$$ and $$3x - y + 5z = k$$?
1. They have no solution, if $$k = 15$$.
2. They have infinitely many solutions, if $$k = 20$$.
3. They have unique solution, if $$k = 25$$.
Select the correct answer using the code given below

A
1 and 2 only

B
2 and 3 only

C
1 and 3 only

D
1, 2 and 3

Solution

First we need to know if the above equations are linearly dependent or no,
In order to figure that out lets find the determinant $$D$$ which is;
$$D=\left| \begin{matrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ 3 & -1 & 5 \end{matrix} \right| =0$$
as $$D=0$$ the equations are linearly dependent, which means the system of equations can either be inconsistent or have infinitely many solutions. That shall depend on the value of k.
For there to be inifinitely many solutions , values of $${ D }_{ 1 }, { D }_{ 2 }, { D }_{ 3 }$$ should be $$0$$ , defined as;
$${ D }_{ 1 }=\left| \begin{matrix} 8 & 1 & 1 \\ 6 & -1 & 2 \\ k & -1 & 5 \end{matrix} \right| \\ { D }_{ 2 }=\left| \begin{matrix} 1 & 8 & 1 \\ 1 & 6 & 2 \\ 3 & k & 5 \end{matrix} \right| \\ { D }_{ 3 }=\left| \begin{matrix} 1 & 1 & 8 \\ 1 & -1 & 6 \\ 3 & -1 & k \end{matrix} \right| $$
which on solving you get;
$${ D }_{ 1 }=\left| \begin{matrix} 8 & 1 & 1 \\ 6 & -1 & 2 \\ k & -1 & 5 \end{matrix} \right| =3k-60=3(k-20)\\ { D }_{ 2 }=\left| \begin{matrix} 1 & 8 & 1 \\ 1 & 6 & 2 \\ 3 & k & 5 \end{matrix} \right| =k-20\\ { D }_{ 3 }=\left| \begin{matrix} 1 & 1 & 8 \\ 1 & -1 & 6 \\ 3 & -1 & k \end{matrix} \right| =40-2k=-2(k-20)$$
Now for all the Determinants to be zero , its evident that $$k=20$$
Hence statement 2 is correct.
Subsequently for there to be no solution $$k\neq 20$$, hence statement 1 is correct as well,
As $$D$$ is zero the system of equations CANNOT have a unique solution hence, statement 3 is WRONG.

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Determinants Test - 45

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Determinants Test - 45

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SHARING IS CARING

Question 1

$$0$$

$$1$$

$$x$$

$$2x$$

Solution

Question 2

$$\displaystyle \sum_{r = 1}^{n} r^{2}$$

$$\displaystyle \sum_{r = 1}^{n} r^{3}$$

$$\displaystyle \sum_{r = 1}^{n} r$$

$$\displaystyle \sum_{r = 1}^{n} r^{4}$$

Solution

Question 3

$$-8$$

$$4$$

$$-4$$

$$8$$

Solution

Question 4

$$det A$$

$$\displaystyle\frac{1}{det A}$$

$$1$$

None of the above

Solution

Question 5

$$0$$

$$1$$

$$10$$

$$20$$

Solution

Question 6

$$44$$

$$55$$

$$61$$

$$84$$

Solution

Question 7

$$2$$

$$4$$

$$6$$

$$-6$$

Solution

Question 8

$$3$$

$$2$$

$$1$$

$$0$$

Solution

Question 9

$$\displaystyle\frac{a}{b}$$ is one of the cube roots of unity

$$\displaystyle\frac{a}{b}$$ is one of the cube roots of $$-1$$.

a is one of the cube roots of unity

b is one of the cube roots of unity.

Solution

Question 10

1 and 2 only

2 and 3 only

1 and 3 only

1, 2 and 3

Solution

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