Matrices Test

Result

Matrices Test - 27

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

lf $$\mathrm{A}=\left\{\begin{array}{lll}
1 & 1 & 3\\
5 & 2 & 6\\
-2 & -1 & -3
\end{array}\right\},$$ then $$\mathrm{A}^{3}$$ is a/an

A
diagonal matrix

B
square matrix

C
null matrix

D
unit Matrix

Solution

$$A^2$$$$=$$$$\left\{\begin{array}{lll} 1 & 1 & 3\\ 5 & 2 & 6\\ -2 & -1 & -3 \end{array}\right\}\times\left\{\begin{array}{lll} 1 & 1 & 3\\ 5 & 2 & 6\\ -2 & -1 & -3 \end{array}\right\}$$$$=$$$$\left\{\begin{array}{lll} 0 & 0 & 0\\ 3 & 3 & 9\\ -1 & -1 & -3 \end{array}\right\}$$
$$A^3$$$$=$$$${A^2}\times{A}$$
$$A^3$$$$=$$$$\left\{\begin{array}{lll} 0 & 0 & 0\\ 3 & 3 & 9\\ -1 & -1 & -3 \end{array}\right\}\times\left\{\begin{array}{lll} 1 & 1 & 3\\ 5 & 2 & 6\\ -2 & -1 & -3 \end{array}\right\}$$$$=$$$$\left\{\begin{array}{lll} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right\}$$
Question 2
1 / -0

$$A=\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}$$ then $$A^{3}-35A=$$

A
$$\mathrm{A}$$

B
$$2\mathrm{A}$$

C
$$3\mathrm{A}$$

D
$$4\mathrm{A}$$

Solution

$$A=\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}$$

$${ A }^{ 2 }=\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}=\begin{bmatrix} 12 & 12 & 12 \\ 12 & 12 & 12 \\ 12 & 12 & 12 \end{bmatrix}$$

$${ A }^{ 3 }=\begin{bmatrix} 12 & 12 & 12 \\ 12 & 12 & 12 \\ 12 & 12 & 12 \end{bmatrix}\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}=\begin{bmatrix} 72 & 72 & 72 \\ 72 & 72 & 72 \\ 72 & 72 & 72 \end{bmatrix}$$

$${ A }^{ 3 }-35A=\begin{bmatrix} 72 & 72 & 72 \\ 72 & 72 & 72 \\ 72 & 72 & 72 \end{bmatrix}-35\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}$$

$$=\begin{bmatrix} 72 & 72 & 72 \\ 72 & 72 & 72 \\ 72 & 72 & 72 \end{bmatrix}-\begin{bmatrix} 70 & 70 & 70 \\ 70 & 70 & 70 \\ 70 & 70 & 70 \end{bmatrix}$$

$$=\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}$$
Question 3
1 / -0

Let $$\left[\begin{array}{ll}
2 & -2\\-2 & \ 5
\end{array}\right]=\left[\begin{array}{ll}
1 & 0\\
-1 & 1
\end{array}\right]\left[\begin{array}{ll}
2 & 0\\
0 & x
\end{array}\right]\left[\begin{array}{ll}
1 & -1\\
0 & 1
\end{array}\right]$$, then the value of $$x$$ is

A
$$-3$$

B
$$-2$$

C
$$0$$

D
$$3$$

Solution

Consider, $$\begin{bmatrix}
1 & 0\\
-1 & 1
\end{bmatrix}\begin{bmatrix}
2 & 0\\
0 & x
\end{bmatrix}\begin{bmatrix}
1 & -1\\
0 & 1
\end{bmatrix}$$

$$=\begin{bmatrix}
2 & 0\\
-2 & x
\end{bmatrix}\begin{bmatrix}
1 & -1\\
0 & 1
\end{bmatrix}$$

By given, we have
$$\begin{bmatrix}
2 & -2\\
-2 & 2+x
\end{bmatrix}=\begin{bmatrix}
2 & -2\\
-2 & 5
\end{bmatrix}$$
By equality of above matrices, we have
$$2+x=5$$
$$\therefore x =3$$
Question 4
1 / -0

If $$A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$, then additive inverse of A is

A
$$A^{T}$$

B
$$A^{-1}$$

C
$$-A$$

D
$$-A^{-1}$$

Solution
Question 5
1 / -0

If A and B are two matrices such that $$AB = B$$ and $$BA = A$$, then $$A^2 + B^2$$ is equal to

A
$$2AB$$

B
$$2BA$$

C
$$A + B$$

D
$$AB$$

Solution

$$A^{2}+B^{2}=AA+BB$$
$$=A(BA)+B(AB)$$ (Given $$AB=B \, and \, BA =A$$)
$$=(AB)A+(BA)B$$
$$=BA+AB$$
$$=A+B$$
Question 6
1 / -0

If $$A \times \begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix} = [1 \ \ 2],$$ then A =

A
$$\left[\dfrac {1}{2} \ \ \ 1\right]$$

B
$$\begin{bmatrix} 1 & 2\\ 1 & 0 \end{bmatrix}$$

C
$$\begin{bmatrix} 2 & 1\\ 1 & 5 \end{bmatrix}$$

D
$$\left[1 \ \ \ \dfrac {1}{2} \right]$$

Solution
Question 7
1 / -0

$$\begin{bmatrix}a\ \
b\end{bmatrix}$$ x $$\begin{bmatrix}x\\y \end{bmatrix} =$$

A
$$\begin{bmatrix}ax+ay

+bx+by

\end{bmatrix}$$

B
$$\begin{bmatrix}ax

\\ by

\end{bmatrix}$$

C
$$\begin{bmatrix}ax

+by

\end{bmatrix}$$

D
$$\begin{bmatrix}ax\ \

by

\end{bmatrix}$$

Solution
Question 8
1 / -0

If $$A = \begin{bmatrix} 1 & -1\\ 2 & -1 \end{bmatrix}; B = \begin{bmatrix} 1 & 1\\ 4 & -1 \end{bmatrix},$$ then $$A^2 + B^2 =$$

A
$$2 I$$

B
$$4 I$$

C
$$\begin{bmatrix} 7 & 0\\ 0 & 7 \end{bmatrix}$$

D
$$\begin{bmatrix} 1 & -1\\ 0 & 5 \end{bmatrix}$$

Solution
Question 9
1 / -0

Multiplication of two matrices $$A$$ and $$ B$$ i.e. $$AB,$$ is possible if and only if

A
rows of $$A=$$ columns of $$B$$

B
$$A=B$$

C
column of $$A =$$ rows of $$B$$

D
none of these

Solution

Multiplication of two matrices $$A$$ and $$ B$$ is possible if and only if the number of columns in the first matrix is same as the number of rows in the second matrix.
Question 10
1 / -0

If $$P=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ and if
$$PQ=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$,then $$Q=$$

A
$$\begin{bmatrix} -4 & 2 \\ 3 & -1 \end{bmatrix}$$

B
$$\begin{bmatrix} 4 & -2 \\ 3 & 1 \end{bmatrix}$$

C
$$\begin{bmatrix} 5 & 2 \\ 3 & 1 \end{bmatrix}$$

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

Solution

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Matrices Test - 27

Result

Matrices Test - 27

Score

-

Rank

-

SHARING IS CARING

Question 1

diagonal matrix

square matrix

null matrix

unit Matrix

Solution

Question 2

$$\mathrm{A}$$

$$2\mathrm{A}$$

$$3\mathrm{A}$$

$$4\mathrm{A}$$

Solution

Question 3

$$-3$$

$$-2$$

$$0$$

$$3$$

Solution

Question 4

$$A^{T}$$

$$A^{-1}$$

$$-A$$

$$-A^{-1}$$

Solution

Question 5

$$2AB$$

$$2BA$$

$$A + B$$

$$AB$$

Solution

Question 6

$$\left[\dfrac {1}{2} \ \ \ 1\right]$$

$$\begin{bmatrix} 1 & 2\\ 1 & 0 \end{bmatrix}$$

$$\begin{bmatrix} 2 & 1\\ 1 & 5 \end{bmatrix}$$

$$\left[1 \ \ \ \dfrac {1}{2} \right]$$

Solution

Question 7

$$\begin{bmatrix}ax+ay+bx+by \end{bmatrix}$$

$$\begin{bmatrix}ax\\ by\end{bmatrix}$$

$$\begin{bmatrix}ax+by \end{bmatrix}$$

$$\begin{bmatrix}ax\ \by \end{bmatrix}$$

Solution

Question 8

$$2 I$$

$$4 I$$

$$\begin{bmatrix} 7 & 0\\ 0 & 7 \end{bmatrix}$$

$$\begin{bmatrix} 1 & -1\\ 0 & 5 \end{bmatrix}$$

Solution

Question 9

rows of $$A=$$ columns of $$B$$

$$A=B$$

column of $$A =$$ rows of $$B$$

none of these

Solution

Question 10

$$\begin{bmatrix} -4 & 2 \\ 3 & -1 \end{bmatrix}$$

$$\begin{bmatrix} 4 & -2 \\ 3 & 1 \end{bmatrix}$$

$$\begin{bmatrix} 5 & 2 \\ 3 & 1 \end{bmatrix}$$

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

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.

$$\begin{bmatrix}ax+ay

+bx+by

\end{bmatrix}$$

$$\begin{bmatrix}ax

\\ by

\end{bmatrix}$$

$$\begin{bmatrix}ax

+by

\end{bmatrix}$$

$$\begin{bmatrix}ax\ \

by

\end{bmatrix}$$