Determinants Test

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Determinants Test - 69

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Weekly Quiz Competition

Question 1
1 / -0

A= \begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix} then Adj(A)=

A
$${A^T}$$

B
$$3{A^T}$$

C
$${A^{ - 1}}$$

D
$$ - {A^T}$$

Solution
Question 2
1 / -0

If $$A=\begin{bmatrix} -4 & -1 \\ 3 & 1 \end{bmatrix}$$ then the determinant of the matrix $$\left( {A}^{2016}-2{A}^{2015}-{A}^{2014} \right) $$ is

A
$$-2016$$

B
$$-25$$

C
$$2016$$

D
$$-175$$
Question 3
1 / -0

There are $$12$$ points in a plane. The number of the straight lines joining any two of them when $$3$$ of them are collinear is.

A
$$60$$

B
$$62$$

C
$$64$$

D
$$66$$

Solution
Question 4
1 / -0

If adj B = A, |P| = |Q| = 1, then adj $$\left( { Q }^{ -1 }{ BP }^{ -1 } \right) $$ is

A
PQ

B
QAP

C
PAQ

D
$${ PA }^{ -1 }Q$$

Solution
Question 5
1 / -0

If $$A$$ is $$4\times 4$$ matrix and if $$\left| \left| A \right| adj\left( \left| A \right| A \right) \right| ={ \left| A \right| }^{ n }$$, then $$n$$ is

A
$$11$$

B
$$13$$

C
$$17$$

D
$$19$$
Question 6
1 / -0

If $$A=\begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$$ and $$A(adj\, A)=A{A}^{T}$$ then $$5a+3b$$ is equal to

A
$$5$$

B
$$4$$

C
$$11$$

D
$$-1$$
Question 7
1 / -0

Two straight lines intersects at a point O. Points $$A_1, A_2,....A_n $$ are taken on one line and $$B_1,B_2,....B_n $$ on the other. If the point O is not to be used, the number of triangles that can be drawn using these points as vertices, is:

A
$$n(n-1)$$

B
$$n(n-1)^2$$

C
$$n^2(n-1)$$

D
$$n^2(n-1)^2$$

Solution
Question 8
1 / -0

If A and B are square matrices of order 3 such that $$\left | A \right | $$= -1,$$\left | B \right | $$=3, then $$\left | 3AB \right | $$ equals

A
-9

B
-81

C
-27

D
81

Solution

For any square matrix $$X$$ of order $$n$$, $$det(kX)=k^n(det X)$$ where $$k$$ is any constant.
Here, $$A$$ and $$B$$ are square matrices of order 3 such that $$|A|=-1$$ and $$|B|=3$$.
Now, $$|3AB|=3^3|AB|$$ (Since $$A$$ and $$B$$ are both of order 3 thus $$AB$$ is also a matrix of order 3)
$$\Rightarrow |3AB| =27|A||B|$$ (Since for any square matrices $$A$$ and $$B$$, $$|AB|=|A||B|$$)
$$\Rightarrow |3AB| =27(-1)(3)=-81$$

Thus, $$|3AB|=-81$$.
Question 9
1 / -0

If $$f'(x)=\begin{vmatrix} mx & mx-p & mx+p \\ n & n+p & n-p \\ mx+2n & mx+2n+p & mx+2n-p \end{vmatrix}$$, then $$y=f(x)$$ represents

A
a straight line parallel to x-axis

B
a straight line parallel to y-axis

C
parabola

D
a straight line with negative slope

Solution

Given $${ f }^{ ' }\left( x \right) =\left| \begin{matrix} mx & mx-p & mx+p \\ n & n+p & n-p \\ mx+2n & mx+2n+p & mx+2n-p \end{matrix} \right| $$

$${ R }_{ 3 }-2{ R }_{ 2 }\rightarrow { R }_{ 3 }$$

$$\therefore { f }^{ ' }\left( x \right) =\left| \begin{matrix} mx & mx-p & mx+p \\ n & n+p & n-p \\ mx+2n-2n & mx+2n+p-2\left( n+p \right) & mx+2n-p-2\left( n-p \right) \end{matrix} \right| $$

$$\therefore { f }^{ ' }\left( x \right) =\left| \begin{matrix} mx & mx-p & mx+p \\ n & n+p & n-p \\ mx & mx+2n+p-2n-2p & mx+2n-p-2n+2p \end{matrix} \right| $$

$$\therefore { f }^{ ' }\left( x \right) =\left| \begin{matrix} mx & mx-p & mx+p \\ n & n+p & n-p \\ mx & mx-p & mx+p \end{matrix} \right| $$

Now, $${ R }_{ 1 }={ R }_{ 3 }$$

$$\therefore { f }^{ ' }\left( x \right) =0$$

$$\therefore { f }\left( x \right) =c$$

$$\therefore y=c$$

This is the equation of straight line parallel to x axis.
Question 10
1 / -0

Let $$ A= \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 0 & 5 \\ 0 & 2 & 1 \end{matrix} \right] $$ and $$ B = \left[ \begin{matrix} 0 \\ -3 \\ 1 \end{matrix} \right] $$ which of the following is true ?

A
$$ AX = B $$ has a unique solution

B
$$ AX = B $$ has exactly three solution

C
$$ AX =B $$ has infinitely many solutions

D
$$ AX = B $$ is inconsistent

Solution