MCQ Questions for Class 12 Commerce Maths Matrices Quiz 11

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

$B = \begin{bmatrix}1 & -2 & y\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$ and

$AB = I_{3}$ , then

$x + y=$ _____.

0%
$0$
0%
$-1$
0%
$2$
0%
None of these

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

$A^{2}$ is equal to

0%
null matrix
0%
unit matrix
0%
$-A$
0%
$A$

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

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

Explanation

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

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

$\Rightarrow A = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}(-1)\begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix}$

$= (-1)\begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix}=-\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$

$\begin{bmatrix} 1/25 & 0 \\ x & 1/25 \end{bmatrix}\quad =\quad \begin{bmatrix} 5 & 0 \\ -a & 5 \end{bmatrix}^{ -2 }$ , then the value of x is

0%
$a / 125$
0%
$2a / 125$
0%
$2a / 25$
0%
None of these

Explanation

Let

$A = \begin{bmatrix} 5 & 0 \\ -a & 5 \end{bmatrix}$

$\Rightarrow (A)= \begin{bmatrix} 5 & 0 \\ -a & 5 \end{bmatrix}$

$\Rightarrow A^{-1} = \frac {1}{ |A| } \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} = \frac {1}{25} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix}$

$\Rightarrow A^{-2} =(A^{-1})^2 = \frac {1}{25} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} \frac {1}{25} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix}$

$=\frac {1}{625} \begin{bmatrix}25 & 0 \\10a&25 \end{bmatrix}$

Now ,

$\begin{bmatrix} 1/2 & 0 \\ x & 1/25 \end{bmatrix}=\begin{bmatrix} \frac {1}{25} &0 \\ \frac {2a}{125} & \frac {1}{25} \end{bmatrix}$

$\Rightarrow x = 2a /125$

$AB = A$ and

$BA =B,$ then

0%
$A^2 B = A^2$
0%
$B^2A =B^2$
0%
$ABA = A$
0%
$BAB = B$

Explanation

We have,

$AB = A$ and

$BA =B$

Now, checking each option:

Option A:

$A^2B=A(AB)=AA=A^2$

Option B:

$B^2A=B(BA)=BB=B^2$

Option C:

$ABA=A(BA)=AB=A$ and

Option D:

$BAB = B(AB)=BA=B.$

Hence, all options are correct.

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

$B$ is a square matrix of order

$2$ such that

$AB = I$ then

$B = ?$

0%
$\begin{bmatrix}1 &2 \\ 2 & 3\end{bmatrix}$
0%
$\begin{bmatrix}1 &\dfrac {1}{2} \\ 2 & \dfrac {3}{2}\end{bmatrix}$
0%
$\begin{bmatrix}1 &2 \\ \dfrac {1}{2} & \dfrac {3}{2}\end{bmatrix}$
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None of these

Find the value of

$x,$ if

$\left [ 1\ x\ 1 \right ]$

$\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}$

$\begin{bmatrix}1 \\ 2 \\ x \end{bmatrix}=O.$

0%
$-2$
0%
$2$
0%
$-14$
0%
9

Explanation

Given,

$\left [ 1\ x\ 1 \right ]_{1 \times 3}$

$\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}_{ _3 \times_ 3}$

$\begin{bmatrix}1 \\ 2 \\ x \end{bmatrix}_{_3 \times _1}= O$

$\Rightarrow$

$[1+ 2x + 15\,\,\, 3 + 5x +3 \,\,\, 2 + x + 2] \begin{bmatrix}1 \\ 2 \\ x \end{bmatrix}= O$

$\Rightarrow$

$[16 + 2x + 15\,\,\, 3+5x + 3\,\,\, 2 + x +2]$

$\begin{bmatrix}1 \\ 2 \\ x \end{bmatrix} = O$

$\Rightarrow$

$[(16 +2x) \cdot 1 + (6 + 5x ) \cdot 2 + (4 + x ) \cdot x] = 0$
=

$\Rightarrow$

$(16 + 2x ) + (12 + 10x ) + (4x + x2) = 0$

$\Rightarrow$

$x^2 + 16x + 28 = 0$

$\Rightarrow$

$(x + 14 ) (x+ 2) = 0$

$\Rightarrow$

$x + 14 = 0$ or

$x + 2 = 0$
Hence,

$x = -14$ or

$x = -2.$

The matrix

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix}$ is a

0%
Identity matrix
0%
Symmetric matrix
0%
Skew symmetric matrix
0%
None of these

Explanation

Let

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

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

So, the given matrix is a symmetric matrix.

[Since, in a square

$A$ . If

$A'=A$ , then

$A$ is called symmetric matrix]

If matrix

$A=[a_{ij}]_{2\times 2}$ , where

$a_{ij} *1$ if

$1*j$ and

$0$ if

$i=j$ then

$A^2$ is equal to

0%
$I$
0%
$A$
0%
$0$
0%
$None\ of\ these$

Explanation

We have,

$A=[a_{ij}]_{2\times 2}$ . where

$a_{ij}=1$ if

$1\neq j$ and if

$i=j$

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

And

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

$AA^{'}$ is always a symmetric matrix for any matrix

$A$ .

0%
True
0%
False

Explanation

It is true,

Because,

$[AA^{'}]^{'}=(A^{'})^{'}A^{'}=[AA^{'}]$

So,

$AA'$ is a symmetric matrix for any matrix

$A$

$A$ and

$B$ are square matrices of the same order, then

$(A+B)(A-B)$ is equal to

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$A^2-B^2$
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$A^2-BA-AB-B^2$
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$A^2-B^2+BA-AB$
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$A^2-BA+B^2+AB$

Explanation

$(A+B)(A-B)=A(A-B)+B(A-B)$ .........because matrix product is distributive.

$=A^2-AB+BA-B^2$

And as matrix product is not commutative, so

$AB\neq BA$

Hence option

$C$ is correct.

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

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

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Only $AB$ is defined
0%
only $BA$ is defined
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$AB$ and $BA$ both are defined
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$AB$ and $BA$ both are not defined

Explanation

Let

$A=[a_{ij}]_{2\times 3}, B=[b_{ij}]_{3\times 2}$ .

For,

$AB$ columns of

$A=3$ is equal to rows of

$B=3$ . So,

$AB$ is defined.

For,

$BA$ columns of

$B=2$ is equal to rows of

$A=2$ . So,

$BA$ is defined.

So, Both

$AB$ and

$BA$ are defined.

$(C)$ is the correct answer.

State true/false:

$A$ and

$B$ are two matrices of orders

$3 \times 2$ and

$2 \times 3$ respectively, then their sum

$A + B$ is possible.

0%
True
0%
False

Explanation

The Sum of matrices

$A+B$ is possible only when the order of both the matrices

$A$ and

$B$ are same.

If A=

$\displaystyle \begin{vmatrix} a & b &c \\ x & y & z \\ l & m & n \end{vmatrix}$ is a skew-symmetric matrix then which of the following is equal to x+y+z?

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$a+b+c$
0%
$l+m+n$
0%
$-b-m$
0%
$c-l-n$

Explanation

$A$ is a skew symmetric matrix, then

${A}^{T} = -A$

Now,

${A}^{T} = \begin{vmatrix} a & x & l \\ b & y & m \\ c & z & n \end{vmatrix}\quad$

So,

${A}^{T} = -A$

$=> \begin{vmatrix} a & x & l \\ b & y & m \\ c & z & n \end{vmatrix}\quad = \begin{vmatrix} -a & -b & -c \\ -x & -y & -z \\ -l & -m & -n \end{vmatrix}$

$=> x = -b ; y = -y$ or

$y = 0 ; z = - m$

So,

$x + y + z = -b + 0 -m = -b-m$

Given

$A =\begin{bmatrix} 2& 1\\2 &1 \end{bmatrix} .$ Find all possible matrix

$X$ for which

$AX = A.$

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$X =\begin{bmatrix} a&b \\ 2 -2a&1 -2b\end{bmatrix}$ for a, $b\in R$
0%
$X =\begin{bmatrix} b&a \\ 2 -2b&1 -2a\end{bmatrix}$ for a, $b\in R$
0%
$X =\begin{bmatrix} b&a \\ 1 -2a&2 -2b\end{bmatrix}$ for a, $b\in R$
0%
none of these

Explanation

Given

$A =\begin{bmatrix} 2& 1\\2 &1 \end{bmatrix}$
We will check by options

Option A:
Consider,

$AX=\begin{bmatrix} 2& 1\\2 &1 \end{bmatrix} \begin{bmatrix} a&b \\ 2 -2a&1 -2b\end{bmatrix}$

$=\begin{bmatrix} 2a+2-2a & 2b+1-2b \\ 2a+2-2a & 2b+1-2b \end{bmatrix}$

$\Rightarrow AX=\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}$

$\Rightarrow AX=A$
Hence, this matrix X satisfies given condition.

Option B:
Consider

$AX=\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} b & a \\ 2-2b & 1-2a \end{bmatrix}$

$=\begin{bmatrix} 2b+2-2b & 2a+1-2a \\ 2b+2-2b & 2a+1-2a \end{bmatrix}$

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

$\Rightarrow AX=A$
Hence, this matrix X also satisfies the given equation.

Option C:
Consider,

$AX=\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} b & a \\ 1-2a & 2-2b \end{bmatrix}$

$=\begin{bmatrix} 2b+1-2a & 2a+2-2b \\ 2b+1-2a & 2a+2-2b \end{bmatrix}$

$\Rightarrow AX\ne A$
Hence, option C is not possible matrix X.

$\alpha ,\beta ,\gamma$ are three real numbers and

$A=\begin{bmatrix} 1&\cos (\alpha -\beta ) &\cos (\alpha -\gamma ) \\\cos (\beta -\alpha ) &1 & \cos (\beta -\gamma ) \\ \cos (\gamma -\alpha ) & \cos (\gamma -\beta ) &1 \end{bmatrix}$ then

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$A$ is symmetric
0%
$A$ is orthogonal
0%
$A$ is singular
0%
$A$ is not invertible.

Explanation

$\cos(-x)=\cos x$

we can write

$\cos(\alpha-\beta)=\cos(\beta-\alpha)$

So clearly

$A=A^T$

$A$ is symmetric

We can write

$A$ as

$A=\begin{bmatrix} \cos \alpha &\sin \alpha &0 \\ \cos \beta & \sin \beta &0 \\\cos \gamma &\sin \gamma &0 \end{bmatrix}\begin{bmatrix} \cos \alpha &\cos \beta &\cos \gamma \\ \sin \alpha & \sin \beta &\sin \gamma \\0 &0 &0 \end{bmatrix}$

$|A|=0$

means

$A$ is singular and non-invertible.

Let

$\alpha =\pi /5$ and

$A=\begin{bmatrix}\cos \alpha & \sin \alpha \\-\sin \alpha &\cos \alpha \end{bmatrix}$ and

$B = A + A^2 + A^3 + A^4$ , then

0%
singular
0%
non-singular
0%
skew-symmetric
0%
$|B|=1$

Explanation

Given

$A=\begin{bmatrix}\cos \alpha & \sin \alpha \\-\sin \alpha &\cos \alpha \end{bmatrix}$ and

$\alpha =\pi /5$

$A^2=\begin{bmatrix}\cos\! 2\alpha & \sin\! 2\alpha \\-\sin\! 2\alpha &\cos\! 2\alpha \end{bmatrix}$ ,

$A^3=\begin{bmatrix}\cos\! 3\alpha & \sin\! 3\alpha \\-\sin\! 3\alpha &\cos\! 3\alpha \end{bmatrix}$
and

$A^4=\begin{bmatrix}\cos\! 4\alpha & \sin\! 4\alpha \\-\sin\! 4\alpha &\cos\! 4\alpha \end{bmatrix}$
We have

$\cos \alpha +\cos 2\alpha +\cos 3\alpha +\cos 4\alpha$

$=\cos \alpha +\cos 2\alpha +\cos(\pi -2\alpha)+\cos(\pi -\alpha) [\:\because 5\alpha=\pi]$

$=\cos \alpha +\cos 2\alpha -\cos 2\alpha-\cos \alpha=0$
and

$\sin \alpha+\sin 2\alpha+\sin 3\alpha+\sin 4\alpha$

$=\sin \alpha+\sin 2\alpha+\sin(\pi -2\alpha)+\sin(\pi -\alpha)$

$=2 [\sin \alpha+\sin 2\alpha]$

$\displaystyle =2\left \{ 2\sin \left [ \frac{3\alpha }{2} \right ]\cos \frac{\alpha }{2} \right \}=4\sin \left [ \frac{3\pi }{10} \right ]\cos \frac{\pi }{10}$

$\displaystyle =4\sin \left [ \frac{\pi }{2}-\frac{\pi }{5} \right ]\cos \frac{\pi }{10}$

$\displaystyle =4\cos \frac{\pi }{5}\cos \frac{\pi }{10}=a$ (say)
Thus,

$B=\begin{bmatrix} 0&a \\-a &0 \end{bmatrix}$

$\therefore B$ is skew-symmetric.
Also,

$\displaystyle |B|=a^2 = 16\cos^2 \frac{\pi }{5}\cos^2 \frac{\pi }{10}>0$

$\therefore B$ is non-singular.
Hence, options B and C.

If the matrix

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

$A^n=\begin{bmatrix}a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a\end{bmatrix}. n \in N$ where

0%
$a = 2n, b = 2^n$
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$a = 2^n, b = 2n$
0%
$a = 2^n, b = n2^{n-1}$
0%
$a = 2^n, b = n2^n$

Explanation

Given

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

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

$\Rightarrow {A}^{3} = 2^{3}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 3 & 0 & 1 \end{bmatrix}$ and so on

we can observe that

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

By comparing we get

$a= 2^{n} , b= n2^{n}$

$3\begin{bmatrix}2 & 3\\ -4 & 1\end{bmatrix} - 2 \begin{bmatrix} x& y\\ 3 & 4\end{bmatrix} = \begin{bmatrix}10 & 11\\ z & -5\end{bmatrix}$ , then

$x + y - z =$

0%
$-21$
0%
$-15$
0%
$0$
0%
$15$
0%
$21$

$O\left( A \right) =2\times 3,$

$O\left( B \right) =3\times 2$ and

$O\left( C \right) =3\times 3$ , which one of the following is not defined?

0%
$CB+{ A }^{ ' }$
0%
$BAC$
0%
$C{ \left( A+{ B }^{ ' } \right) }^{ ' }$
0%
$C\left( A+{ B }^{ ' } \right)$

Explanation

Given that,

$O\left( A \right) =2\times 3,$

$O\left( B \right) =3\times 2$ and

$O\left( C \right) =3\times 3$

$\Rightarrow O\left( { A }^{ ' } \right) =3\times 2, O\left( { B }^{ ' } \right) =2\times 3$
(a)

$CB+{ A }^{ ' }$
Now, Order of

$CB=$ (Order of

$C$ ) (Order of

$B$ )

$=$ (Order of

$C$ is

$3\times 3$ ) (Order of

$B$ is

$3\times 2$ )

$=$ Order of

$CB$ is

$3\times 2$
Since,

$O\left( { A }^{ ' } \right) =3\times 2$
Therefore, matrix

$CB+{ A }^{ ' }$ can be determined.
(b)

$O\left( BA \right) =3\times 3$ and

$O\left( C \right) =3\times 3$
Therefore, matrix

$BAC$ can be determined.
(c)

$C{ \left( A+{ B }^{ ' } \right) }^{ ' }$

$O\left( A+{ B }^{ ' } \right) =2\times 3$

$\Rightarrow O{ \left( A+{ B }^{ ' } \right) }^{ ' }=3\times 2$
and

$O\left( C \right) =3\times 3$
Therefore, matrix

$C{ \left( A+{ B }^{ ' } \right) }^{ ' }$ can be determined.
(d)

$C\left( A+{ B }^{ ' } \right)$

$O\left( A+{ B }^{ ' } \right) =2\times 3$
and

$O\left( C \right) =3\times 3$

$\therefore$ Matrix

$C\left( A+{ B }^{ ' } \right)$ cannot be determined
Hence, option D is correct.

$A$ is of order

$m \times n$ and

$B$ is of order

$p \times q,$ addition of

$A$ and

$B$ is possible only if

0%
$m = p$
0%
$n = q$
0%
$n = p$
0%
$m = p, n = q$

Explanation

Since, the order of

$A$ is

$m\times n$ and order of

$B$ is

$p\times q$

Then

$A+B$ is only possible if

$m=p$

$\&$

$n=q.$

Hence, the answer is

$m=p$

$\&$

$n=q.$

Let

$A=\begin{bmatrix} 3 & 1\\ -1 & 2\end{bmatrix}$ , then

0%
$A^2+7A-5I=0$
0%
$A^2-7A+5I=0$
0%
$A^2+5A-7I=0$
0%
$A^2-5A+7I=0$

Explanation

${ A }^{ 2 }=\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}=\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

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

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

We find (d) satisfy the condition as

${ A }^{ 2 }-5A-7I=0$

$\quad \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}-\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}+\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}=0$

Thus,

${ A }^{ 2 }-5A-7I=0$ is correct

What is the inverse of the matrix

$A=\begin{bmatrix} \cos { \theta } & \sin { \theta } & 0 \\ -\sin { \theta } & \cos { \theta } & 0 \\ 0 & 0 & 1 \end{bmatrix}$ ?

0%
$\begin{bmatrix} \cos { \theta } & -\sin { \theta } & 0 \\ \sin { \theta } & \cos { \theta } & 0 \\ 0 & 0 & 1 \end{bmatrix}$
0%
$\begin{bmatrix} \cos { \theta } & 0 & -\sin { \theta } \\ 0 & 1 & 0 \\ \sin { \theta } & 0 & \cos { \theta } \end{bmatrix}$
0%
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos { \theta } & -\sin { \theta } \\ 0 & \sin { \theta } & \cos { \theta } \end{bmatrix}$
0%
$\begin{bmatrix} \cos { \theta } & \sin { \theta } & 0 \\ -\sin { \theta } & \cos { \theta } & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Explanation

$A = \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Calculate first minors.

$M_{11} = \cos \theta , M_{13} = 0, M_{22} = \cos \theta$

$M_{12} = -\sin \theta, M_{21} = \sin \theta, M_{23} = 0$

$M_{31} = 0, M_{32} = 0, M_{33} = \cos^{2}\theta + \sin^{2}\theta = 1$

Cofactor Matrix

$= \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0\\ 0 & 0 & 1 \end{bmatrix} = C$

$det|A| = \cos^{2}\theta + \sin^{2}\theta = 1$

$adj (A) = C^{T} = \begin{bmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0 & 0 & 1\end{bmatrix}$

$A^{-1} = \dfrac {adj(A)}{(A)} = \begin{bmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0 & 1\end{bmatrix}$ .

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

$I$ is the unit matrix of order

$2$ , then

$A^2$ equals

0%
$4A-3I$
0%
$3A-AI$
0%
$A-I$
0%
$A+I$

Explanation

$A=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ { A }^{ 2 }=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\times \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ \quad =\begin{bmatrix} 4+1 & -2-2 \\ -2-2 & 1+4 \end{bmatrix}\\ \quad =\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}$

Answer will be A option because

$4A-3I=\begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\\ \quad \quad \quad \ =\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}$

Let

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

$10B=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$ . If

$B$ is the inverse of matrix

$A$ , then

$\alpha$ is

0%
$-2$
0%
$1$
0%
$2$
0%
$5$

Explanation

Since,

$B$ is the inverse of matrix

$A$ .
So,

$10{ A }^{ -1 }=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$

$\Rightarrow 10{ A }^{ -1 }\cdot A=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$

$\Rightarrow 10I = \begin{bmatrix} 4+4+2 & -4+2+2 & 4-6+2 \\ -5+0+\alpha & 5+0+\alpha & -5+0+\alpha \\ 1-4+3 & -1-2+3 & 1+6+3 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix}=\begin{bmatrix} 10 & 0 & 0 \\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0 & 0 & 10 \end{bmatrix}$

$\Rightarrow -5+\alpha =0$

$\Rightarrow \alpha =5$

$If\quad A=\begin{bmatrix} ab & { b }^{ 2 } \\ -{ a }^{ 2 } & -ab \end{bmatrix},then\quad { A }^{ 2 }\quad is\quad equal\quad to$

0%
$O$
0%
$I$
0%
$-I$
0%
$None\quad of\quad these$

The value of

$x$ , so that

$\left[ 1 \quad x \quad 1 \right] \begin{bmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$ is/are

0%
$\pm\ 2$
0%
$0$
0%
$\dfrac { -7\pm \sqrt { 35 } }{ 2 }$
0%
$\dfrac { -9\pm \sqrt { 53 } }{ 2 }$

If the matrix

$\begin{bmatrix} 0 & 2\beta & \Upsilon \\ \alpha & \beta & -\Upsilon \\ \alpha & -\beta & \Upsilon \end{bmatrix}$ is orthogonal, then

0%
$\alpha = \pm\dfrac{1}{\sqrt{2}}$
0%
$\beta = \pm\dfrac{1}{\sqrt{6}}$
0%
$\gamma = \pm\dfrac{1}{\sqrt{3}}$
0%
all of these

Explanation

$\begin{bmatrix} 0 & 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{bmatrix}$

for orthogonal matrix we have

$A.A^{T}=I$

$\begin{bmatrix} 0 & 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{bmatrix}\begin{bmatrix} 0 & \alpha & \alpha \\ 2\beta & \beta & -\beta \\ \gamma & -\gamma & \gamma \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\begin{bmatrix} 0+4{ \beta }^{ 2 }+{ \gamma }^{ 2 } & 0+2{ \beta }^{ 2 }-{ \gamma }^{ 2 } & 0-2{ \beta }^{ 2 }+{ \gamma }^{ 2 } \\ 0+2{ \beta }^{ 2 }-{ \gamma }^{ 2 } & { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 } & { \alpha }^{ 2 }-{ \beta }^{ 2 }-{ \gamma }^{ 2 } \\ 0-2{ \beta }^{ 2 }+{ \gamma }^{ 2 } & { \alpha }^{ 2 }-{ \beta }^{ 2 }-{ \gamma }^{ 2 } & { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 } \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$4\beta^{2}+\gamma^{2}=1, 2\beta^{2}-\gamma^{2}=0$ ,

$4\left(\dfrac{\gamma^{2}}{2}\right)+\gamma^{2}=1$

$\beta^{2}=\dfrac{r^{2}}{2}$

$r^{2}[3]=1$

$r=\pm \dfrac{1}{\sqrt{3}}$

$2\beta^{2}-\gamma^{2}=0, \alpha^{2}+\beta^{2}+\gamma^{2}+\gamma^{2}=1, \alpha^{2}-\beta^{2}-\gamma^{2}=0$

$\beta^{2}=\dfrac{\gamma^{2}}{2}, \alpha^{2}+\dfrac{\gamma^{2}}{2}+\dfrac{\gamma^{2}}{1}=1$

$\alpha^{2}+\dfrac{3\gamma^{2}}{2}=1$

$\beta^{2}=\dfrac{1}{6}\alpha^{2}+\dfrac{3}{2}\times \dfrac{1}{3}=1$

$\beta=\pm \dfrac{1}{\sqrt{6}}$

$\alpha=\pm \dfrac{1}{\sqrt{2}}$

$\left[ \begin{matrix} x & 4 & -1 \end{matrix} \right] \left[ \begin{matrix} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4 \end{matrix} \right] \left[ \begin{matrix} x \\ 4 \\ -1 \end{matrix} \right] =0,$ then

$x=$

0%
$-1+\sqrt { 6 }$
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$8\pm \sqrt { 5 }$
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$-2\pm \sqrt { 10 }$
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$3\pm \sqrt { 6 }$

$A=\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$ , then :

$A^{-1}$ =

0%
$A$
0%
$A^{2}$
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$A^{3}$
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$A^{4}$

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 11 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 11 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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