Mathematics Test

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Mathematics Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

If matrix A such that A²= 2A − I , where II is the identity matrix, then for n ≥ 2, where n ∈ N ; Aⁿis equal to :

A
nA − (n − 1) I

B
n A I

C
2ⁿ⁻¹A − (n − 1) I

D
2ⁿ⁻¹A − I

Solution

Given:
$A^{2}=2 A-I$
$A^{3}=A^{2} A=(2 A-I) A=2 A^{2}-I A=2 A^{2}-A=2(2 A-I)-A$
$\Rightarrow A^{3}=3 A-2 I$
$A^{4}=\left(A^{2}\right)\left(A^{2}\right)=(2 A-I)(2 A-I)=4 A^{2}-4 A I+I=4(2 A-I)-4 A+I$
$\Rightarrow A^{4}=4 A-3 I$
$\ldots$ and so on...
According to the pattern, we can say that $A^{n}=n A-(n-1) I$
Hence, option $\mathrm{A}$.
Question 2
1 / -0

lf the order of A is 4×3, the order of B is 4 × 5 and the order of C is 7 × 3, then the order of (A^TB)^TC^Tis :

A
4 x 5

B
3 x 7

C
4x 3

D
5x 7

Solution

By operation of matrixes (4), order of $A=4 \times 3$
So, By property of traspose (8) Order of $A^{T}=3 \times 4$
So, order of $A^{T} B=3 \times 5$ order of $\left(A^{T} B\right)^{T}=5 \times 3$
and order of $C^{T}=3 \times 7$. . So, order of $\left(A^{T} B\right)^{T} C^{T} 5 \times 7$
Hence, the correct option is (D)
Question 3
1 / -0

If $A=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1 & i+1 \\ i-1 & 1\end{array}\right]$, then
$A\left(A^{T}\right)$ equals :

A
0

B
I

C
−I

D
2 I

Solution

$A=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1 & i+1 \\ i-1 & 1\end{array}\right]$
$A^{T}=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1 & i-1 \\ i+1 & 1\end{array}\right]$
$\overline{A^{T}}=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1 & -i-1 \\ 1-i & 1\end{array}\right]$
Now, $A \overline{A^{T}}=\frac{1}{3}\left[\begin{array}{cc}1 & i+1 \\ i-1 & 1\end{array}\right]\left[\begin{array}{cc}1 & -i-1 \\ 1-i & 1\end{array}\right]$
$\begin{aligned}=& \frac{1}{3}\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right] \\=& I \\ \Rightarrow A \overline{A^{T}} &=I \end{aligned}$
Hence, the correct option is (B)
Question 4
1 / -0

$$
\left|\begin{array}{lll}
a+x & a-x & a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|=0 \text { then the }
$$
non-zero value of $\mathrm{x}=\ldots \ldots \ldots$

A
a

B
3a

C
2a

D
4a

Solution

By operation of matrix (5), $A=\left|\begin{array}{lll}a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x\end{array}\right|$
$=(a+x)\left[(a+x)^{2}-(a-x)^{2}\right]-(a-x)\left[a^{2}-x^{2}-(a-x)^{2}\right]+(a-x)\left[(a-x)^{2}-\left(a^{2}-x^{2}\right)\right]$
$=(a+x)(4 a x)-(a-x)\left(a^{2}-x^{2}\right)+2(a-x)^{3}-(a-x)\left(a^{2}-x^{2}\right)$
$=(a+x)(4 a x)+2(a-x)\left[a^{2}+x^{2}-2 a x-a^{2}+x^{2}\right]$
$=(a+x) 4 a x+2(a-x)\left(2 x^{2}-2 a x\right)$
$A=4 a^{2} x+4 a x^{2}+4 a x^{2}-4 a^{2} x-a x^{3}+4 a x^{2}$
$A=8 a x^{2}+4 a x^{2}-4 x^{3}$
$A=12 a x^{2}-4 x^{3}$
Given, $A=12 a x^{2}-4 x^{3}=0$
$x^{2}(12 a-x)=0$
$x=3 a$
Hence, the correct option is (B)
Question 5
1 / -0

The value of a for which the equations 3x − y + az =1, 2x + y + z = 2, x + 2y − az = −1 fail to have unique solution is :

A
7/2

B
-7/2

C
2/7

D
-2/7

Solution

For fail to have unique solu. $\left|\begin{array}{ccc}3 & -1 & a \\ 2 & 1 & 1 \\ 1 & 2 & -a\end{array}\right|=0$
$3(-a-2)+1(-2 a-1)+a(4-1)=0$
$-3 a-6-2 a-1+4 a-a=0$
$2 a+7=0$
$a=-7 / 2$
Hence, the correct option is (B)
Question 6
1 / -0

If $A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1\end{array}\right]$ and $A^{-1}=\left[\begin{array}{ccc}1 / 2 & 1 / 2 & 1 / 2 \\ -4 & 3 & c \\ 5 / 2 & -3 / 2 & 1 / 2\end{array}\right]$, then the values of $a$ and $c$ are equal to :

A
1,1

B
1,-1

C
1,2

D
-1,1

Solution

$A=\left[\begin{array}{ccc}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1\end{array}\right]$ and $\quad A^{-1}=\left[\begin{array}{ccc}1 / 2 & 1 / 2 & 1 / 2 \\ -4 & 3 & c \\ 5 / 2 & -3 / 2 & 1 / 2\end{array}\right]$
Here, $|A|=-1(-8)+2(a-6)=2 a-4$
adj $A=C^{T}=\left[\begin{array}{ccc}2-3 a & 8 & a-6 \\ 2 a-1 & -6 & 3 \\ -1 & 2 & -1\end{array}\right]^{T}$
$\Rightarrow a d j A=\left[\begin{array}{ccc}2-3 a & 2 a-1 & -1 \\ 8 & -6 & 2 \\ a-6 & 3 & -1\end{array}\right]$
Hence, $A^{-1}=\frac{1}{2 a-4}\left[\begin{array}{ccc}2-3 a & 2 a-1 & -1 \\ 8 & -6 & 2 \\ a-6 & 3 & -1\end{array}\right]$
Comparing with given $A^{-1}$, $-\frac{1}{2 a-4}=\frac{1}{2}$
$\Rightarrow a=1$
Also, $\frac{2}{2 a-4}=c$
$\Rightarrow c=-1$
Thus the values of $a$ and $c$ are 1 and -1 respectively.
Hence, the correct option is (B)
Question 7
1 / -0

The value of a third order determinant is 11, then the value of the square of the determinant formed by the cofactors will be :

A
11

B
121

C
1331

D
14641

Solution

Given, determinant of order 3 is
11
$\Delta=11$
Then, the value of determinant formed by cofactors is $\Delta^{\prime}=(\Delta)^{n-1}$
So, $\Delta^{\prime}=121$
Required value $=\left(\Delta^{\prime}\right)^{2}=(121)^{2}=14641$
Hence, the correct option is (D)
Question 8
1 / -0

If A is a square matrix of order n such that its elements are polynomials in x and its r-rows become identical for x = α then :

A
(x − α) is a factor of |A|

B
(x − α)^r−1is a factor of |A|

C
(x − α)^r+1|A|

D
(x − α)^r is a factor of A.

Solution

$r$ rows of the matrix become
identical when $x=\alpha$.
If we use row transformations
on $r-1$ of the identical rows such that the remaining identical row is subtracted from the other rows, then we will get $(x-\alpha)$ as a common factor in each of the $r-1$ rows. $(x-\alpha,$ will be a factor since all the entries are polynomials.) Hence, $(x-\alpha)^{r-1}$ will be a common factor of $|A|$.
Hence, the correct option is (B)
Question 9
1 / -0

lf A and B are two square matrices such that B=−A⁻¹BA then (A + B)²= ?

A
0

B
A² + B²

C
A²+ 2AB + B²

D
A + B

Solution

Pre multiply by A $A B=-A A^{-1} B A$
$\therefore A B=-B A$
$\therefore(A+B)^{2}=A^{2}+B^{2}+A B+B A$
$=A^{2}+B^{2}$
Hence, the correct option is (B)
Question 10
1 / -0

The rank of the matrix
$A=\left\{\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right\} i s$

A
3

B
2

C
1

D
0

Solution

The determinant of matrix
$A_{3 \times 3}$ is o
Now, take any $2 \times 2$ matrix and calculating its determinant value its also comes out to be o And if we take single element of A we found that it is non-zero
$\therefore$ Rank of given matrix $=1$
Hence, the correct option is (C)

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Mathematics Test - 46

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Mathematics Test - 46

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Question 1

nA − (n − 1) I

n A I

2n−1A − (n − 1) I

2n−1A − I

Solution

Question 2

4 x 5

3 x 7

4x 3

5x 7

Solution

Question 3

0

I

−I

2 I

Solution

Question 4

a

3a

2a

4a

Solution

Question 5

7/2

-7/2

2/7

-2/7

Solution

Question 6

1,1

1,-1

1,2

-1,1

Solution

Question 7

11

121

1331

14641

Solution

Question 8

(x − α) is a factor of |A|

(x − α)r−1is a factor of |A|

(x − α)r+1|A|

(x − α)r is a factor of A.

Solution

Question 9

0

A2 + B2

A2+ 2AB + B2

A + B

Solution

Question 10

3

2

1

0

Solution

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2ⁿ⁻¹A − (n − 1) I

2ⁿ⁻¹A − I

(x − α)^r−1is a factor of |A|

(x − α)^r+1|A|

(x − α)^r is a factor of A.

A² + B²

A²+ 2AB + B²