Matrices Test

Result

Matrices Test - 22

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$A = \begin{bmatrix} 2& 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\end{bmatrix}$$, then $$A^6 =$$

A
$$6A$$

B
$$12A$$

C
$$16A$$

D
$$32A$$

Solution

$$A = \begin{bmatrix}2&0&0\\0&2&0\\0&0&2 \end{bmatrix} = 2 \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = 2I$$

$$\therefore A^6 = (2I)^6 = 2^6\times I^6 = 2^6I$$

$$A^6 = 2^6I = 2^5(2I) = 2^5A = 32A$$
Question 2
1 / -0

Least number of changes for the expression $$ax^{2} + bxy + cy^{2} + dx + ey + f$$ to be symmetric in x and y is

A
$$a = b, c = d$$

B
$$b = c, e = f$$

C
$$a = c, d = e$$

D
$$a = f, b = e, c = d$$

Solution

$$ax^2+bxy+cy^2+dx+ey+f$$.......(1)
Replace by $$x$$ by $$y$$ and $$y$$ by $$x$$
$$\Rightarrow ay^2+byx+cx^2+dy+ex+f$$.......(2)
if (1) and (2) are equal then
$$\boxed{a=c\,and\,d=e}$$
Question 3
1 / -0

If $$A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$, $$B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$, then $$BA =$$

A
$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

B
$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$

C
$$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$

D
$$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

Solution

$$A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, B=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$

$$BA=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$

$$=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$
Question 4
1 / -0

If $$A$$ and $$B$$ are square matrices such that $$AB = I$$ and $$BA = I$$, then $$B$$ is

A
Unit matrix

B
Null matrix

C
Multiplicative inverse matrix of $$A$$

D
$$- A$$

Solution

$$AB=I$$ $$\&$$ $$BA=I$$ then $$B$$ is the multiplicative inverse of $$A.$$
Hence, the answer is multiplicative inverse matrix of $$A$$.
Question 5
1 / -0

Consider the following statements in respect of the matrix $$A = \begin{bmatrix} 0 & 1 & 2\\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix}$$ :
1. The matrix A is skew-symmetric.
2. The matrix A is symmetric.
3. The matrix A is invertible.
Which of the above statements is/are correct ?

A
1 only

B
3 only

C
1 and 3

D
2 and 3

Solution

For a matrix to be skew-symmetric
$$ { A }^{ T }=-A$$
While for a matrix to be symmetric
$$ { A }^{ T }= A$$
and for a matrix to be invertible
$$ \left| A \right| \neq 0$$
$$\Rightarrow $$so for $$ \begin{bmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix}$$
$$ { A }^{ T }=-\begin{bmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix}$$
and $$\left| A \right| =0$$
Hence it is skew symmetric but not invertible
Question 6
1 / -0

If the sum of the matrices $$\begin{bmatrix} x \\ x \\ y \end{bmatrix},\begin{bmatrix} y \\ y \\ z \end{bmatrix}$$ and $$\begin{bmatrix} z \\ 0 \\ 0 \end{bmatrix}$$ is the matrix $$\begin{bmatrix} 10 \\ 5 \\ 5 \end{bmatrix}$$, then what is the value of $$y$$?

A
$$-5$$

B
$$0$$

C
$$5$$

D
$$10$$

Solution

Since, we have
$$\left[ \begin{matrix} x \\ x \\ y \end{matrix} \right] +\left[ \begin{matrix} y \\ y \\ z \end{matrix} \right] +\left[ \begin{matrix} z \\ 0 \\ 0 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \\ 5 \end{matrix} \right]$$
$$\therefore x+y+z=10,$$ $$x+y=5$$ and $$ y+z=5$$
Replacing $$x+y=5$$ in $$x+y+z=10$$
We have, $$z=5$$
Also, $$y+z=5$$
$$\therefore y=5-z=0$$
Option B is correct.
Question 7
1 / -0

If $$[2\ 3\ 4] \begin{bmatrix}1 & x &3 \\ 2 & 4 & 5\\ 3 & 2 &x \end{bmatrix} \begin{bmatrix} x\\ 2 \\  0 \end{bmatrix} = 0$$, then $$x =$$ ________.

A
$$\dfrac {7}{3}$$

B
$$\dfrac {5}{3}$$

C
$$-\dfrac {5}{3}$$

D
$$-\dfrac {7}{3}$$

Solution

Given, $$[2\ 3\ 4] \begin{bmatrix}1 & x &3 \\ 2 & 4 & 5\\ 3 & 2 &x \end{bmatrix} \begin{bmatrix} x\\ 2 \\  0 \end{bmatrix} = 0$$
$$\begin {bmatrix} 20 &20+2x&21+4x \end {bmatrix} \begin{bmatrix} x\\ 2 \\  0 \end{bmatrix} = 0$$
$$\Rightarrow 20x+40+4x=0$$
$$\Rightarrow 24x + 40 = 0$$
$$\Rightarrow x = -\dfrac {5}{3}$$
Question 8
1 / -0

If $$A$$ is any matrix, then the product $$AA$$ is defined only when A is a matrix of order $$m \times n$$ where :

A
$$m > n $$

B
$$m < n$$

C
$$m=n $$

D
$$m \leq n $$

Solution

Let two matrix be $$A$$ and $$B$$ of order $$m\times n$$ and $$l\times k$$ respectively.
They can be multiplied if $$n=l$$
So, if a matrix $$A$$ of order $$m\times n $$ is to be multiplied by $$A$$
Then, $$m=n$$
Question 9
1 / -0

Consider the following statements:
1. The product of two non-zero matrices can never be identity matrix.
2. The product of two non-zero matrices can never be zero matrix.
Which of the above statements is/are correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

Product of a matrix and it's inverse is always an identity matrix.
Example :-
$$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$
And both are non-zero.
Thus, statement $$1$$ is incorrect.
Product of two non-zero matrices can be a zero matrix. Example,
$$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$
Thus, statement $$2$$ is incorrect.
Hence, option D is correct.
Question 10
1 / -0

What is $$\begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} a& h & g\\ h & b & f\\ g & f & c\end{bmatrix}$$ equal to?

A
$$\begin{bmatrix}ax + hy + gz & h + b + f & g + f + c\end{bmatrix}$$

B
$$\begin{bmatrix}a & h & g\\ hx & by & fz\\ g & f & c\end{bmatrix}$$

C
$$\begin{bmatrix}ax & hy & gz\\ hx & by & fz\\ gx & fy & cz\end{bmatrix}$$

D
$$\begin{bmatrix} ax + hy + gz & hx + by + fz & gx + fy + cz\end{bmatrix}$$

Solution

Order of matrix $$\begin{bmatrix} x& y & z \end{bmatrix}$$ is $$(1\times 2)$$
Order of matrix $$\begin{bmatrix} a & h & g\\ h & b & f\\ g & f & c\end{bmatrix} $$ is $$(3\times 3)$$

So, the product $$\begin{bmatrix} x& y& z \end{bmatrix}$$$$\begin{bmatrix} a & h & g\\ h & b & f\\ g & f & c\end{bmatrix} =\begin{bmatrix} ax+hy+gz & hx+by+fz & gx+fy+cz\end{bmatrix}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Matrices Test - 22

Result

Matrices Test - 22

Score

-

Rank

-

SHARING IS CARING

Question 1

$$6A$$

$$12A$$

$$16A$$

$$32A$$

Solution

Question 2

$$a = b, c = d$$

$$b = c, e = f$$

$$a = c, d = e$$

$$a = f, b = e, c = d$$

Solution

Question 3

$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$

$$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$

$$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

Solution

Question 4

Unit matrix

Null matrix

Multiplicative inverse matrix of $$A$$

$$- A$$

Solution

Question 5

1 only

3 only

1 and 3

2 and 3

Solution

Question 6

$$-5$$

$$0$$

$$5$$

$$10$$

Solution

Question 7

$$\dfrac {7}{3}$$

$$\dfrac {5}{3}$$

$$-\dfrac {5}{3}$$

$$-\dfrac {7}{3}$$

Solution

Question 8

$$m > n $$

$$m < n$$

$$m=n $$

$$m \leq n $$

Solution

Question 9

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 10

$$\begin{bmatrix}ax + hy + gz & h + b + f & g + f + c\end{bmatrix}$$

$$\begin{bmatrix}a & h & g\\ hx & by & fz\\ g & f & c\end{bmatrix}$$

$$\begin{bmatrix}ax & hy & gz\\ hx & by & fz\\ gx & fy & cz\end{bmatrix}$$

$$\begin{bmatrix} ax + hy + gz & hx + by + fz & gx + fy + cz\end{bmatrix}$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.