If $$A=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$, then $$A^{4}=$$

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

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

MCQ Questions for Class 12 Commerce Maths Matrices Quiz 1

Let

$X$ and

$Y$ be two arbitrary,

$3\times 3$ , non-zero, skew-symmetric matrices and

$Z$ be an arbitrary

$3\times 3$ , non-zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric?

0%
$Y^3Z^4-Z^4Y^3$
0%
$X^{44}+Y^{44}$
0%
$X^4Z^3-Z^3X^4$
0%
$X^{23}+Y^{23}$

Explanation

Given:

$X^T = -X, Y^T = -Y$ and

$Z^T = Z$
Using Properties of transpose:

$(A+B)^T = A^T + B^T$ and

$(AB)^T = B^TA^T$

Option A:

$(Y^3Z^4 - Z^4Y^3)^T = (Y^3Z^4)^T - (Z^4Y^3)^T$

$\Rightarrow (Z^4)^T(Y^3)^T - (Y^3)^T(Z^4)^T = (Z^T)^4(Y^T)^3 - (Y^T)^3(Z^T)^4 = Y^3Z^4 - Z^4Y^3$
Its a symmetric.

Option B:

$(X^{44} + Y^{44})^T = (X^T)^{44} + (Y^T)^{44} = X^{44} + Y^{44}$
Its a symmetric.

Option C:

$(X^4Z^3 - Z^3X^4)^T = (X^4Z^3)^T - (Z^3X^4)^T$

$\Rightarrow (Z^3)^T(X^4)^T - (X^4)^T(Z^3)^T = (Z^T)^3(X^T)^4 - (X^T)^4(Z^T)^3 = Z^3X^4 - X^4Z^3 = - (X^4Z^3 - Z^3X^4)$
Its a skew symmetric.

Option D:

$(X^{23} + Y^{23})^T = (X^T)^{23} + (Y^T)^{23} = -(X^{23} + Y^{23})$
Its a skew symmetric.

Hence, option C,D.

$\displaystyle A=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \end{matrix} \right]$ and

$\displaystyle I=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]$ , then the correct statement is:

0%
$\displaystyle { A }^{ 2 }+5A-7I=O$
0%
$\displaystyle -{ A }^{ 2 }+5A+7I=O$
0%
$\displaystyle { A }^{ 2 }-5A+7I=O$
0%
$\displaystyle { A }^{ 2 }+5A+7I=O$

Explanation

Now,

$\displaystyle { A }^{ 2 }=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \end{matrix} \right] \left[ \begin{matrix} 3 & 1 \\ -1 & 2 \end{matrix} \right]$

$\displaystyle =\left[ \begin{matrix} 8 & 5 \\ -5 & 3 \end{matrix} \right]$

$\displaystyle \therefore \quad { A }^{ 2 }-5A+7I=\left[ \begin{matrix} 8 & 5 \\ -5 & 3 \end{matrix} \right] -\left[ \begin{matrix} 15 & 5 \\ -5 & 10 \end{matrix} \right] +\left[ \begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix} \right]$

$\displaystyle =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right]$

If AB = AC then

0%
B = C
0%
$B\neq C$
0%
B need not be equal to C
0%
B = -C

$\mathrm{A}^{2}=\mathrm{A},\ \mathrm{B}^{2}=\mathrm{B},\ \mathrm{A}\mathrm{B}=\mathrm{B}\mathrm{A}=O$ (Null Matrix), then

$(\mathrm{A}+\mathrm{B})^{2}=$

0%
$\mathrm{A}-\mathrm{B}$
0%
$\mathrm{A}+\mathrm{B}$
0%
$\mathrm{A}^{2}-\mathrm{B}^{2}$
0%
$0$

Explanation

$(A + B)^{2} = A^{2} + B^{2} + AB + BA$

$= A + B$

$I$

$A=\left[\begin{array}{ll} 0 & 1\\ 1 & 0 \end{array}\right]$ ,

$A^{4}=$
(

$I$ is an identity matrix.)

0%
$I$
0%
$0$
0%
$\mathrm{A}$
0%
$4{I}$

Explanation

Given,

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

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

Now,

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

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

$\Rightarrow A^{2}=I$

$\Rightarrow A^{4}=I$

$\mathrm{A}= \left[\begin{array}{lll} o & c & -b\\ -c & o & a\\ b & -a & o \end{array}\right]\mathrm{a}\mathrm{n}\mathrm{d}$

$\mathrm{B}=\left[\begin{array}{lll} a^{2} & ab & ac\\ ab & b^{2} & bc\\ ac & bc & c^{2} \end{array}\right],$ then

$\mathrm{A}\mathrm{B}=$

0%
$\mathrm{A}$
0%
$\mathrm{B}$
0%
$I$
0%
$\mathrm{O}$

Explanation

Multiplying the two matrices we get,

$\begin{bmatrix} 0+abc-abc & 0+c{ b }^{ 2 }-c{ b }^{ 2 } & 0+b{ c }^{ 2 }-b{ c }^{ 2 } \\ -{ a }^{ 2 }c+0+{ a }^{ 2 }c & -abc+0+abc & -a{ c }^{ 2 }+0+a{ c }^{ 2 } \\ { a }^{ 2 }b-{ a }^{ 2 }b+0 & a{ b }^{ 2 }-a{ b }^{ 2 }+0 & abc-abc+0 \end{bmatrix}$

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

$= O$

Hence the answer is option D

lIf

$\mathrm{A} =\left[\begin{array}{ll} a & 0\\ a & 0 \end{array}\right],\ \mathrm{B}=\left[\begin{array}{ll} 0 & 0\\ b & b \end{array}\right],$ then

$\mathrm{A}\mathrm{B}=$

0%
$O$
0%
$\mathrm{B}\mathrm{A}$
0%
$\mathrm{A}\mathrm{B}$
0%
$\mathrm{A}\mathrm{B}\mathrm{A}\mathrm{B}$

Explanation

$\displaystyle AB = \begin{bmatrix} a&0\\a&0 \end{bmatrix} \begin{bmatrix} 0&0\\b&b \end{bmatrix}$

$=\begin{bmatrix} a\times 0+0\times b&a\times 0+0\times b\\a\times 0+0\times b&a\times 0+0\times b \end{bmatrix}$

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

$A=\left[\begin{array}{lll} 1 & -2 & 3\\ -4 & 2 & 5 \end{array}\right]$ and

$B=\left[\begin{array}{ll} 2 & 3\\ 4 & 5\\ 2 & 1 \end{array}\right],$ then

0%
$\mathrm{A}\mathrm{B},\ \mathrm{B}\mathrm{A}$ exist and equal
0%
$\mathrm{A}\mathrm{B},\ \mathrm{B}\mathrm{A}$ exist and are not equal
0%
$\mathrm{A}\mathrm{B}$ exists and $\mathrm{B}\mathrm{A}$ does not exist
0%
$\mathrm{A}\mathrm{B}$ does not exist and $\mathrm{B}\mathrm{A}$ exists

Explanation

Since, we have

$A = 2 \times 3 , \: B = 3 \times 2$

So, both

$AB$ and

$BA$ exist.
But both are not equal to each other because bith are if different order

$\because AB \rightarrow 2 \times 2$ and

$BA \rightarrow 3 \times 3.$

$\left[\begin{array}{ll} x & 0\\ 0 & y \end{array}\right]\left[\begin{array}{ll} a & b\\ c & d \end{array}\right]=$

0%
$\left[\begin{array}{ll} ax & b_{X}\\ yc & dy \end{array}\right]$
0%
$\left[\begin{array}{ll} ax & 0\\ 0 & dy \end{array}\right]$
0%
$\left[\begin{array}{ll} ay & cy\\ bx & dy \end{array}\right]$
0%
$\left[\begin{array}{ll} 0& ax\\ dy & 0 \end{array}\right]$

Explanation

The value of

$\begin{bmatrix} x&0\\0&y \end{bmatrix} \begin{bmatrix} a&b\\c&d \end{bmatrix}$

$=\begin{bmatrix} x\times a+0\times c&x\times b+0\times d\\ 0\times a+y\times c&0\times b+y\times d \end{bmatrix}$

$=\begin{bmatrix} ax&bx\\cy&dy \end{bmatrix}$

$\mathrm{A}=\left[\begin{array}{lll} 1 & -3 & -4\\ -1 & 3 & 4\\ 1 & -3 & -4 \end{array}\right]$ , then

$\mathrm{A}^{2}=$

0%
$A$
0%
$- A$
0%
Null matrix
0%
$2A$

Explanation

$A^2=A.A$

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

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

$A=[x,y], B=\left[\begin{array}{ll} a & h\\ h & b \end{array}\right], C=\left[\begin{array}{l} x\\ y \end{array}\right]$ ,
then

$\mathrm{A}\mathrm{B}\mathrm{C}=$

0%
$(ax+hy+bxy)$
0%
$(ax^{2}+2hxy+by^{2})$
0%
$(ax^{2}-2hxy+by^{2})$
0%
$(bx^{2}-2hxy+ay^{2})$

Explanation

Since,

$A = \begin{bmatrix} x &y \end{bmatrix}$

$B = \begin{bmatrix} a &h \\ h &b \end{bmatrix}$

$C =\begin{bmatrix} x\\ y \end{bmatrix}$

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

$=\begin{bmatrix} xa+yh &xh+yb \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}$

$=\begin{bmatrix} x^{2}a+xyh+xyh+y^{2}b \end{bmatrix}$

$=\begin{bmatrix} x^{2}a+2xyh+y^{2}b \end{bmatrix}$

$\mathrm{A}$ is skew-symmetric matrix and

$\mathrm{n}$ is even positive integer, then

$\mathrm{A}^{\mathrm{n}}$ is

0%
a symmetric matrix
0%
skew-symmetric matrix
0%
diagonal matrix
0%
triangular matrix

Explanation

If A is skew symmetric or symmetric matrix then

$A^{2}$ is a symmetric matrix.

$A^{n} = (A^{2})^{n/2}$
n is even.

$\therefore A^{n}$ is symmetric matrix.

$\because A$ is a skew symmetric matrix.

$A=[1\ \ 2\ \ 3\ \ 4]$ and

$AB = [3 \ \ 4\ \ -1],$ then the order of
matrix

$B$ is

0%
$2 \times 3$
0%
$3\times 3$
0%
$4 \times 3$
0%
$1 \times 3$

Explanation

Given matrix A

$=\left [ 1 \ \ \ 2 \ \ 3\ \ 4 \right ]$ of order 1

$\times 4$
and AB =

$\left [ 3 \ \ 4 \ \ -1 \right ]$ of order 1

$\times 3$
by multiplication of matrix

$\left ( AB \right )_{m\times n}=\left ( A \right )_{m\times k}\left ( B \right )_{k\times n}$
i.e Number of columns of A has to be equal to number of row of B

$\therefore$ Order of B is

$4 \times 3$

$A=\begin{bmatrix}x& -7\\ 7& y\end{bmatrix}$ is a skew-symmetric matrix,
then (x,y) =

0%
(1,-1)
0%
(7,-7)
0%
(0,0)
0%
(14,-14)

Explanation

$A=\begin{bmatrix}x& -7\\ 7& y\end{bmatrix}$
Now

$A^T=\begin{bmatrix}x& 7\\ -7& y\end{bmatrix}$
Now for

$A$ to be skew symmetric matrix,

$A+A^T=O$

$\Rightarrow \begin{bmatrix}x& -7\\ 7& y\end{bmatrix}+\begin{bmatrix}x& 7\\ -7& y\end{bmatrix}=\begin{bmatrix}0& 0\\ 0& 0\end{bmatrix}$

$\Rightarrow \begin{bmatrix}2x& 0\\ 0& 2y\end{bmatrix}=\begin{bmatrix}0& 0\\ 0& 0\end{bmatrix}$
Now equating the corresponding elements we get,

$(x,y)=(0,0)$

$A=\left[\begin{array}{lll} 0 & 1 & -2\\1 & 0 & 3\\2 &-3 & 0 \end{array}\right]$ then

$\mathrm{A}+\mathrm{A}^{\mathrm{T}}=$

0%
$\left[\begin{array}{lll} 0 & 2 & 0\\ 2 & 0 & 0\\ 0 & 0 & 0 \end{array}\right]$
0%
$\left[\begin{array}{lll} 1 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 4 \end{array}\right]$
0%
$\left[\begin{array}{lll} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{array}\right]$
0%
$\left[\begin{array}{lll} 2 & 0 & 2\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{array}\right]$

Explanation

Given,

$A=\begin{bmatrix} 0 & 1 & -2\\ 1 & 0& 3\\ 2 & -3 & 0 \end{bmatrix}$
So, By property of transpose (1),

$A^{T}=\begin{bmatrix} 0 & 1 & 2\\ 1 & 0 & -3\\ -2 & 3 & 0 \end{bmatrix}$
So, By operation of matrixes (2),

$A+A^{T}=\begin{bmatrix} 0 & 2 & 0\\ 2 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$

The order of

$[\mathrm{x} \space \mathrm{y} \space\mathrm{z}] \left[\begin{array}{lll} \mathrm{a} & \mathrm{h} & \mathrm{g}\\ \mathrm{h} & \mathrm{b} & \mathrm{f}\\ \mathrm{g} & \mathrm{f} & \mathrm{c} \end{array}\right]\left[\begin{array}{l} \mathrm{x}\\ \mathrm{y}\\ \mathrm{z} \end{array}\right]$ is

0%
$3\mathrm{x}1$
0%
$1\mathrm{x}1$
0%
$1\mathrm{x}3$
0%
$3\mathrm{x}3$

Explanation

$[x y z]_{1 \times 3}\begin{Bmatrix} a & h & g\\ h & b & f\\ g & f & c \end{Bmatrix}_{3 \times 3}\begin{Bmatrix} x\\ y\\ z \end{Bmatrix}_{3 \times 1}$

$\Rightarrow [p \space q \space r]_{1 \times 3}\begin{Bmatrix} x\\ y\\ z \end{Bmatrix}_{3 \times 1}$

$\Rightarrow [v]_{1 \times 1}$

$A$ and

$B$ are two matrices such that

$A + B$ and

$AB$ are both defined, then

0%
$A$ and $B$ are two matrices not necessarily of same order
0%
$A$ and $B$ are square matrices of same order
0%
$A$ and $B$ are matrices of same type
0%
$A$ and $B$ are rectangular matrices of same order

Explanation

Addition of matrix is define when both matrixes are of same order.
If A and B are two matrices such that

$A+B$ is define, then

$A$ and

$B$ are matrices of same order.
So, If order of

$A=a \times b$
then order of

$B=a \times b$
and

$AB$ is defined, [by operation of matrixes] then the number of columns of first matrix is equal to the number of rows of second matrix.

$\Rightarrow$

$b=a.$
So,

$A$ and

$B$ both are square matrix of same order.

If the transpose of a matrix is equal to the additive
inverse, then matrix is called _________
matrix.

0%
symmetric
0%
skew symmetric
0%
identity
0%
inverse

$I = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix},$ then find

$I^3$

0%
$1$
0%
$I$
0%
$0$
0%
does not exist

Explanation

$I^n$ ( for any natural

$n$ )

$= I.$

$I$ is known as the identity matrix.

$A= \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}$ and

$B= \begin{bmatrix} 1\\ 0\\ 5\end{bmatrix},$ then

$AB =$

0%
$\begin{bmatrix} 1 & 0 & 15 \end{bmatrix}$
0%
$\begin{bmatrix} 4 & 0 & 30 \end{bmatrix}$
0%
$\begin{bmatrix} 16\\ 34\end{bmatrix}$
0%
$\begin{bmatrix} 16 & 34 \end{bmatrix}$

Explanation

$AB= \begin{bmatrix} 1 &2 &3 \\ 4 &5 &6 \end{bmatrix}\begin{bmatrix} 1\\ 0\\ 5\end{bmatrix}=\begin{bmatrix} 1+ 0+ 15\\ 4+ 0+30 \end{bmatrix}=\begin{bmatrix} 16\\ 34 \end{bmatrix}$
Thus, C will be correct answer.

$A = \begin{bmatrix}a & b\end{bmatrix},\space B = \begin{bmatrix}-b & -a \end{bmatrix}$ and

$C = \begin{bmatrix}a \\ -a\end{bmatrix}$ , then the correct statement is

0%
$A = -B$
0%
$A+B = A-B$
0%
$AC = BC$
0%
$CA = CB$

Explanation

Given

$A = \begin{bmatrix}a & b\end{bmatrix}, B = \begin{bmatrix}-b & -a \end{bmatrix}$ and

$C = \begin{bmatrix}a \\ -a\end{bmatrix}$
We will check by options.
Clearly ,

$A \ne -B$ as their corresponding elements are different only.

$A+B= \begin{bmatrix}a-b & b-a\end{bmatrix}$

$A-B=\begin{bmatrix}a+b & b+a\end{bmatrix}$
So,

$A+B \ne A-B$

Now,

$AC=\begin{bmatrix}a & b\end{bmatrix}\begin{bmatrix}a \\ -a\end{bmatrix}$

$\Rightarrow AC=\begin{bmatrix}a^{2}-ab\end{bmatrix}$

$BC=\begin{bmatrix}-b & -a \end{bmatrix}\begin{bmatrix}a \\ -a\end{bmatrix}$

$\Rightarrow BC=\begin{bmatrix}a^{2}-ab\end{bmatrix}$
Hence,

$AC=BC$
Option

$C$ is correct

Now,

$CA=\begin{bmatrix}a \\ -a\end{bmatrix}\begin{bmatrix}a & b\end{bmatrix}$

$\Rightarrow CA=\begin{bmatrix}a^{2}&ab\\-a^{2}&-ab \end{bmatrix}$

$CB=\begin{bmatrix}a \\ -a\end{bmatrix}\begin{bmatrix}-b & -a \end{bmatrix}$

$\Rightarrow CB=\begin{bmatrix}-ab&-a^{2}\\ab & a^{2}\end{bmatrix}$
Hence,

$CA\ne CB$

$\displaystyle A = \begin{bmatrix} 1 & -2 & 4 \\ 2 & 3 & 2 \\ 3 & 1 & 5 \end{bmatrix}$ and

$\displaystyle B = \begin{bmatrix} 0 & -2 & 4 \\ 1 & 3 & 2 \\ -1 & 1 & 5 \end{bmatrix}$ , then

$A + B$ is

0%
$\displaystyle \begin{bmatrix} 1 & -2 & 4 \\ 3 & 3 & 2 \\ 2 & 1 & 5 \end{bmatrix}$
0%
$\displaystyle \begin{bmatrix} 1 & -2 & 8 \\ 3 & 3 & 4 \\ 2 & 1 & 10 \end{bmatrix}$
0%
$\displaystyle \begin{bmatrix} 1 & -4 & 8 \\ 3 & 6 & 4 \\ 2 & 2 & 10 \end{bmatrix}$
0%
none of these

Explanation

Given,

$\displaystyle A = \begin{bmatrix} 1 & -2 & 4 \\ 2 & 3 & 2 \\ 3 & 1 & 5 \end{bmatrix}$ and

$\displaystyle B = \begin{bmatrix} 0 & -2 & 4 \\ 1 & 3 & 2 \\ -1 & 1 & 5 \end{bmatrix}$

Therefore,

$A+B=\begin {bmatrix} 1+0 & -2-2 & 4+4 \\2+1 & 3+3 & 2+2 \\3-1 & 1+1 & 5+5\end {bmatrix}$

$=\begin{bmatrix} 1 & -4 & 8 \\ 3 & 6 & 4 \\ 2 & 2 & 10 \end{bmatrix}$

$A=\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}$ , then

$A^{3}-A^{2}=$

0%
$2A$
0%
$2I$
0%
$A$
0%
$I$

Explanation

$A^{2}=A\times A=\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}$

$A^{3}=A^{2}\times A=\begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix}$

$A^{3}-A^{2}=\begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix}-\begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}=\begin{bmatrix} -2 & 0 \\ 0 & 4 \end{bmatrix}$

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

$\begin{bmatrix}3 & -1 \\ 2 & 5\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}4 \\ -3\end{bmatrix},$ find

$x$ and

$y$

0%
$x=3,\space y = -1$
0%
$x=2,\space y = 5$
0%
$x=1,\space y = -1$
0%
$x=-1,\space y = 1$

Explanation

$\begin{bmatrix}3 & -1 \\ 2 & 5\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}4 \\ -3\end{bmatrix}$

$\Rightarrow \begin{bmatrix} 3x-y \\ 2x+5y \end{bmatrix}=\begin{bmatrix} 4 \\ -3 \end{bmatrix}$

$\Rightarrow 3x-y=4, 2x+5y=-3$
On solving we get

$x=1, y=-1$

$\displaystyle \:A= \left [ \begin{matrix}4 &x+2 \\2x-3 &x+1 \end{matrix} \right ]$ is symmetric, then x=

0%
3
0%
5
0%
2
0%
4

Explanation

Given

$\displaystyle \:A= \left [ \begin{matrix}4 &x+2 \\2x-3 &x+1 \end{matrix} \right ]$ is symmteric.

$A^{T}=A$

$\Rightarrow \left[ \begin{matrix} 4 & 2x-3 \\ x+2 & x+1 \end{matrix} \right] =\left [ \begin{matrix}4 &x+2 \\2x-3 &x+1 \end{matrix} \right ]$

$\Rightarrow x+2=2x-3$

$\Rightarrow x=5$

$\begin{bmatrix} 1 & 2 & 3 \end{bmatrix} B=\begin{bmatrix} 3 & 4 \end{bmatrix}$ , then the order of the matrix

$B$ is

0%
$3\times 1$
0%
$1\times 3$
0%
$2\times 3$
0%
$3\times 2$

Explanation

Let order of the matrix

$B$ be

$m\times n$

$[\array{1 & 2&3]}_{1\times 3}[B]_{m\times n}=\array{[3&4]}_{1\times 2}$

For the above multiplication to be valid,

$m$ must be equal to

$3$ i.e.,

$m=3$

Order of the resultant matrix is gievn by

$1\times n$ .

Comparing this with the order of the given matrix we get,

$n=2$

Therefore, order of the matrix

$B$ is

$3 \times 2$

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

$A^{4}=$

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

Explanation

Given,

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

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

$A^{4}=A^{2}\times A^{2}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Given that

$\displaystyle M=\begin{bmatrix}3 &-2 \\-4 &0 \end{bmatrix}\:and\:N=\begin{bmatrix}-2 &2 \\5 &0 \end{bmatrix}$ , then

$M+N$ is a

0%
null matrix
0%
unit matrix
0%
$\displaystyle \begin{bmatrix} 1 & 0 \\1 &0 \end{bmatrix}$
0%
$\displaystyle \begin{bmatrix} 0 &1 \\1 &1 \end{bmatrix}$

Explanation

Given:

$\displaystyle M=\begin{bmatrix}3 &-2 \\-4 &0 \end{bmatrix}\:and\:N=\begin{bmatrix}-2 &2 \\5 &0 \end{bmatrix}$

Adding

$M$ and

$N$ , we get

$M+N=\begin{bmatrix} 3 & -2 \\ -4 & 0 \end{bmatrix}+\begin{bmatrix} -2 & 2 \\ 5 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$

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

$B=\begin{bmatrix} 1 & 3 & 2 \\ 2 & 3 & 4 \end{bmatrix}$ , then

$AB$ equal to

0%
$\begin{bmatrix} 8 & 15 & 16 \\ 5 & 9 & 10 \end{bmatrix}$
0%
$\begin{bmatrix} 8 & 5 \\ 15 & 9 \\ 16 & 10 \end{bmatrix}$
0%
$\begin{bmatrix} 8 & 5 \\ 15 & 9 \end{bmatrix}$
0%
None of these

Explanation

$AB=\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 1 & 3 & 2 \\ 2 & 3 & 4 \end{bmatrix}=\begin{bmatrix} 2.1+2.2 & 2.3+3.3 & 2.2+3.4 \\ 1.1+2.2 & 1.3+2.3 & 1.2+2.4 \end{bmatrix}=\begin{bmatrix} 8 & 15 & 16 \\ 5 & 9 & 10 \end{bmatrix}$

Ans: A

$A = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{bmatrix}$ , then

$A^2$ is equal to

0%
$A$
0%
$-A$
0%
null matrix
0%
$I$

Explanation

Given,

$A = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{bmatrix}$

$A^{2}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix}$

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

$\Rightarrow A^{2}=I$

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 1 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 1 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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