If A = [1, 2, 3], then the set of elements of A is:
[1, 2, 3]
[2, 0]
Only 2
None of these
Answer
[1, 2, 3]
Solution:
Since, $ \text{A}=\begin{bmatrix} 1 &\text{amp; } 2 &\text{amp; }3 \end{bmatrix}$ represents a matrix with three,
elements 1, 2, 3 $\therefore$ Elements of A = (1, 2, 3)
Choose the correct answer from the given four options.
The matrix $\begin{bmatrix}1&0&0\\0&2&0\\0&0&4\end{bmatrix}$ is a:
Identity matrix.
Symmetric matrix.
Skew-symmetric matrix.
None of these.
Answer
Symmetric matrix.
Solution:
We have $\text{A}=\begin{bmatrix}1&0&0\\0&2&0\\0&0&4\end{bmatrix}$
$\therefore\ \text{A}'=\text{A}$
So, the given matrix is a symmetric matrix.
Choose the correct answer from the given four options.
The matrix $\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$ is a:
Diagonal matrix.
Symmetric matrix.
Skew-symmetric matrix.
Scalar matrix.
Answer
Skew-symmetric matrix.
Solution:
We have $\text{B}=\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$
$\Rightarrow\ \text{B}'=\begin{bmatrix}0&5&8\\-5&0&-12\\8&12&0\end{bmatrix}$
$=-\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$
$=-\text{B}$
Since, B' = -B,
Thus, B is a skew-symmetric matrix.
The matrix $\begin{bmatrix}0&\text{amp; }1\\1&\text{amp; }0\end{bmatrix}$ is the matrix reflection in the line:
x = 1
x + y = 1
y = 1
x = y
Answer
x = y
Solution:
We know that the reflection matrix through a line $\text{y}=\text{mx}$ making an $\angle \theta$ with x - axis is given as.
$\begin{bmatrix} \cos { 2\theta }&\text{amp;}\sin { 2\theta } \\ \sin { 2\theta } &\text{amp;} -\cos { 2\theta }\end{bmatrix}$
Given transformation matrix is $\begin{bmatrix}0&\text{amp; }1\\1&\text{amp; }0\end{bmatrix}$
$\Rightarrow\cos2\theta=0\sin2\theta=1$
$\Rightarrow 2\theta ={90}^{0}$
$\Rightarrow \theta={45}^{0}$
$\Rightarrow \tan \theta=1$
Hence, the line of reflection is $\text{y}=\text{x}$
The possibility for the formation of rectangular matrices in the matrix algebra are?
rows greater than columns
rows lesser than columns
rows greater than column by 2 times
None of these
Answer
rows greater than columns
Solution:
The possibilities of formation of rectangular matrix are the following:
(1) Rows are greater then columns.
(2) Columns are greater then rows.
(3) Rows greater then column by 2 times.
The matrix $\begin{bmatrix}0&5&-7\\-5&0&11\\7&-11&0\end{bmatrix}$ is:
A skew-symmetric matrix.
A symmetric matrix.
A diagonal matrix.
An uppertriangular matrix.
Answer
A skew-symmetric matrix.
Solution:
Here,
$\text{A}=\begin{bmatrix}0&5&-7\\-5&0&11\\7&-11&0\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}0&-5&7\\5&0&-11\\-7&11&0\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=-\begin{bmatrix}0&5&-7\\-5&0&11\\7&-11&0\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=-\text{A}$
Thus, A is a skew-symmetric matrix.
The number of possible matrices of order $3\times 3$ with each entry $2$ or $0$ is:
A
$9$
B
$27$
C
$81$
✓
None of these.
Answer
Correct option: D.
None of these.
Let us consider a matrix matrix $\begin{bmatrix}\text{a}&\text{b}&\text{c}\\\text{d}&\text{e}&\text{f}\\\text{g}&\text{h}&\text{i}\end{bmatrix}$
The element a can have two values $0$ or $2$ in two ways.
Similarly all other elements can also have two values $0$ or $2$ in two ways each.
So, the total number of combinations is $2^9$.
So, total no of matrices will be $2^9$.
If $A$ is a matrix of order $m \times n$ and $B$ is a matrix such that $AB^{T }$ and $B^{T }A$ are both defined, then the order of matrix $B$ is:
A
$m \times m$
B
$n \times n$
C
$n \times m$
✓
$m \times n$
Answer
Correct option: D.
$m \times n$
Given $A$ is a matrix of order $m\times m\times n$ and $B$ is a matrix such
that $AB^T$ and $B^{T }A$ are both defined.
Since $AB^T$ is defined then number of columns of $A$ must be equal to number of rows of $B^T.$
So number of rows in $B^T$ is $n$.This gives number of columns in $B$ is $n.$
Again since $AB^{T }A$ is defined then number of columns of $B^T$ is equal to the number of rows of $A.$
So number of columns of $B^{T }$ is $m$ this gives the number of rows of $B$ is $m.$
So order of $B$ is $m \times n.$
If $\text{A}=\begin{bmatrix}1&-1\\2&-1\end{bmatrix},\text{B}=\begin{bmatrix}\text{a}&1\\\text{b}&-1\end {bmatrix}$ and $(A + B)^2 = A^2 + B^2,$ values of $a$ and $b$ are:
A
$a = 4, b = 1$
✓
$a = 1, b = 4$
C
$a = 0, b = 4$
D
$a = 2, b = 4$
Answer
Correct option: B.
$a = 1, b = 4$
Here,
$(\text{A+B})^2=\text{A}^2+\text{B}^2$
$\Rightarrow\text{A}^2+\text{AB}+\text{BA}+\text{B}^2=\text{A}^2+\text{B}^2$
$\Rightarrow\text{AB}+\text{BA}=0$
$\Rightarrow\text{AB}=-\text{BA}$
$\Rightarrow\begin{bmatrix}1&-1\\2&-1\end{bmatrix}\begin{bmatrix}\text{a}&1\\\text{b}&-1\end{bmatrix}=-\begin{bmatrix}\text{a}&1\\\text{b}&-1\end{bmatrix}\begin{bmatrix}1&-1\\2&-1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{a}-\text{b}&2\\2\text{a}-\text{b}&3\end{bmatrix}=-\begin{bmatrix}\text{a}+2&-\text{a}-1\\\text{b}-2&-\text{b}+1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{a}-\text{b}&2\\2\text{a}-\text{b}&3\end{bmatrix}=\begin{bmatrix}-\text{a}-2&\text{a}+1\\\text{b}+2&\text{b}-1\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\Rightarrow\text{a}+1=2$ and $b-1=3$
$\therefore\ \text{a}=1$ and $b=4$
The number of all possible matrices of order $3 \times 3$ with each entry $0$ or $1$ is:
A
$27$
B
$18$
C
$81$
✓
$512$
Answer
Correct option: D.
$512$
A matrix of order $3 \times 3$ has $9$ elements.
Now each elementscan be $0$ or $1$.
$\therefore 9$ places can be filled up in $2^9$ waysrequired number of matrices $= 2^9{= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}\ {= 512,}$
If $\text{A}=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}$ is expressed as the sum of a symmetric and skew-symmetric matrix, then the symmetric matrix is:
Solution:
Here,
$\text{A}=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}2&4&-5\\0&3&7\\-3&1&2\end{bmatrix}$
Now,
$\text{A}+\text{A}^\text{T}=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}+\begin{bmatrix}2&4&-5\\0&3&7\\-3&1&2\end{bmatrix}$
$\Rightarrow\text{A}+\text{A}^\text{T}=\begin{bmatrix}2+2&0+4&-3-5\\4+0&3+3&1+7\\-5-3&7+1&2+2\end{bmatrix}$
$\Rightarrow\text{A}+\text{A}^\text{T}=\begin{bmatrix}4&4&-8\\4&6&8\\-8&8&4\end{bmatrix}$
$\text{A}-\text{A}^\text{T}=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}-\begin{bmatrix}2&4&-5\\0&3&7\\-3&1&2\end{bmatrix}$
$\Rightarrow\text{A}-\text{A}^\text{T}=\begin{bmatrix}2-2&0-4&-3+5\\4-0&3-3&1-7\\-5+3&7-1&2-2\end{bmatrix}$
$\Rightarrow\text{A}-\text{A}^\text{T}=\begin{bmatrix}0&-4&2\\4&0&-6\\-2&6&0\end{bmatrix}$
Let $\text{P}=\frac{1}{2}(\text{A}+\text{A}^\text{T})=\frac{1}{2}\begin{bmatrix}4&4&-8\\4&6&8\\-8&8&4\end{bmatrix}$ $=\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}$
$\text{Q}=\frac{1}{2}(\text{A}-\text{A}^\text{T})=\frac{1}{2}\begin{bmatrix}0&-4&2\\4&0&-6\\-2&6&0\end{bmatrix}$$=\begin{bmatrix}0&-2&1\\2&0&-3\\-1&3&0\end{bmatrix}$
Now,
$\text{P}^\text{T}=\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}^\text{T}=\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}=\text{P}$
$\text{Q}^\text{T}=\begin{bmatrix}0&-2&1\\2&0&-3\\-1&3&0\end{bmatrix}=\begin{bmatrix}0&2&-1\\-2&0&3\\1&-3&0\end{bmatrix}$$=-\begin{bmatrix}0&-2&1\\2&0&-3\\-1&3&0\end{bmatrix}=-\text{Q}$
$\text{P+Q}=\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}+\begin{bmatrix}0&-2&1\\2&0&-3\\-1&3&0\end{bmatrix}$
$=\begin{bmatrix}2+0&2-2&-4+1\\2+2&3+0&4-3\\-4-1&4+3&2+0\end{bmatrix}$
$=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}=\text{A}$
Thus, we have expressed A is the sum of a symmetric and a skew-symmetric matrix.
Hence,the symmetric matrix is $\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}.$
If a matrix P has 8 elements then how many different values the order of the matrix can take?
3
4
8
6
Answer
4
Solution:
A matrix of mm rows and n columns has m × n elements.
8 can be got by all combinations of 1 × 8, 8 × 1, 2 × 4, 4 × 2
Hence, there are 4 possible matrices which have 8 elements.
The order the matrix is $\begin{bmatrix}2&\text{amp; }3&\text{amp; }4\\9&\text{amp; }8&\text{amp; }7\end{bmatrix}$ is:
4 × 3
3 × 2
2 × 3
3 × 1
Answer
2 × 3
Solution:
If A is a matrix with mm rows and n columns.Then the order of a matrix is nothing but a size of a matrix, which is given by m × n.
Since, in the given matrix, there are 2 rows and 3 columns.So, order of given matrix will be 2 × 3.
If $A$ is a matrix of order $m \times n$ and $B$ is a matrix such that $AB^T$ and $B^TA$ are both defined, then the order of matrix $B$ is:
A
$m\times n$
B
$n\times n$
C
$n\times m$
✓
$m\times n$
Answer
Correct option: D.
$m\times n$
$A$ is $m\times n$ matrix and $AB^T$ is defined then
number of columns in $A =$ number of rows in $B^T =^{n }$
$B^{T}A$ is also defined then number of columns in $B^T =$ number of rows in $A = m$
Order of $B$ is $m\times n$
If $\text{A}=\begin{bmatrix}1&2&\text{x}\\0&1&0\\0&0&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&-2&\text{y}\\0&1&0\\0&0&1\end{bmatrix}$ and $AB = I_3,$ then $x + y$ equals$:$
✓
$0$
B
$-1$
C
$2$
D
None of these.
Answer
Correct option: A.
$0$
Given: $AB = I_3$
$\Rightarrow\begin{bmatrix}1&2&\text{x}\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}1&-2&\text{y}\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&0&\text{y+x}\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore\ \text{y}+\text{x}=0$
If $\begin{bmatrix}\text{r}+4&\text{amp; 6}\\3&\text{amp; 3}\end{bmatrix}=\begin{bmatrix}{5}&\text{amp;}\text{ r}+5\\\text{r+2}&\text{amp; 4}\end{bmatrix}$ then $\text{r}=$
1
2
3
-1
Answer
1
Solution:
We know that two matrices are equal iff their corresponding elements are equal.
Thus comparing corresponding elements we get, for the first entry of.
the given matrices r + 4 = 5 and r is satisfying other equations which are involving r ⇒ r = 1
The total number of matrices formed with the help of 6 different numbers are:
6
6!
2(6!)
4(6!)
Answer
4(6!)
Solution:
No.of numbers in Matrix is 6
The possible orientations of Matrix is.
1 × 6, 2 × 3, 3 × 2, 6 × 1
The numbers in each orientation can be arranged in 6! ways.
$\implies$The total possibilities are 4(6!).
If every row of a matrix A contains p elements and its column contains q elements, then the order of A is:
p × p
q × q
p × q
q × p
Answer
q × p
Solution:
$\begin{bmatrix}\text{a}_{11} &\text{amp;}\text{ a}_{12} \\\text{a}_{21}& \text{amp;}\text{ a}_{22}\\\text{a}_{31}&\text{amp; }\text{a}_{32} \end{bmatrix}$
Hence order of $\text{A}$ is $3\times2$
Row contains pp elements
So number of columns $=\text{P}$
Each column contains $\text{q}:$ element
So number of rows $=\text{q}$
Therefore, order $=\text{q}\times\text{p}$
A matrix having mm rows and nn columns with m = n is said to be a?
rectangular matrix
square matrix
identity matrix
scalar matrix
Answer
square matrix
Solution:
A matrix having mm rows and nn columns with m = n, means number of rows are equal to number of columns.
$\therefore$ given matrix is square matrix.
If $\displaystyle \text{a}_{\text{ij}}=0\left (\text{i}\neq \text{j} \right )$ and $\displaystyle \text{a}_{\text{ij}}=1\left (\text{i}= \text{j} \right )$ then the matrix $\text{A}=\displaystyle \left [\text{a}_{\text{ij}} \right ]_{\text{n}\times\text{n}}$ is a _____ matrix:
Null
Identity
Scalar
Triangular
Answer
Identity
Solution:
The elements $\text{a}_\text{ij}$ of a matrix where i = j lie along its diagonal and
the elements $\text{a}_\text{ij}$ of a matrix where $\text{i}\neq\text{j}$ are not along the diagonal.
As the diagonal elements are 11 and the rest of the elements are 0, the matrix A is an identity matrix.
Let $A$ is a square matrix of order $n$ and $a$ being $a$ scalar then $∣aA∣ =$
A
$a∣A∣$
B
$∣a∣∣A∣$
✓
$a^{n }∣A∣$
D
none of these
Answer
Correct option: C.
$a^{n }∣A∣$
Given, $A$ is a square matrix of order $n$ and a being a scalar.
Now $aA$ is the matrix in which each elements of $A$ is multiplied by $a.$
So when we take determinant of $aA$ then form each row or column $a$ will be common.
Then $∣aA∣ = a^{n }∣A∣.$
If the order of matrices A and B are 3 × 2 and 2 × 1 respectively, then find the order of matrix (if possible) AB:
1 × 3
3 × 1
2 × 2
2 × 3
Answer
3 × 1
Solution:
Order of A : 3 × 2 Order of B : 2 × 1 Multiplication of matrices is possible if and only.
if the number of columns of first matrix is equal to the number of rows of second matrix In AB
No.of columns in A is No.of rows in B is 2 $\therefore$ AB exists.
Order of AB is (number of rows of A x number of columns of B)
$\therefore$ Order of AB is (3 × 1)
Choose the correct answer from the given four options.If matrix $A = [a_{ij}]_{2\times 2},$ where $a_{ij} = 1,$ if $\text{i}\neq\text{j}$ and $0$ if $i = j$ then $A^2$ equal to$:$
✓
$I$
B
$A$
C
$0$
D
None of these.
Answer
Correct option: A.
$I$
We have, $A = [a_{ij}]_{2\times 2},$ where $a_{ij} = 1,$ if $\text{i}\neq\text{j}$ and $0$ if $i = j$
$\therefore\ \text{A}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$
And $\text{A}^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=\text{I}$
Choose the correct answer from the given four options.If A and B are two matrices of the order 3 × m and 3 × n, respectively and m = n, then order of matrix (5A – 2B) is:
m × 3
3 × 3
m × n
3 × n
Answer
3 × n
Solution:
We are given that, the order of the matrices A and B are 3 × m and 3 × n respectively. Now, If m = n, then A and B have same orders as 3 × n each, so the order of (5A – 2B) should be same as 3 × n.
Given $\text{A}=\begin{bmatrix}1&\text{a}\\0&1\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}1&\text{a}\\0&1\end{bmatrix}\begin{bmatrix}1&\text{a}\\0&1\end{bmatrix}$
$=\begin{bmatrix}1&2\text{a}\\0&1\end{bmatrix}$
$\text{A}^3=\text{A}^2\text{A}$
$=\begin{bmatrix}1&2\text{a}\\0&1\end{bmatrix}\begin{bmatrix}1&\text{a}\\0&1\end{bmatrix}$
$=\begin{bmatrix}1&3\text{a}\\0&1\end{bmatrix}$
On genaralising we get
$\text{A}^\text{n}=\begin{bmatrix}1&\text{na}\\0&1\end{bmatrix}$
If $\text{A}= \begin{bmatrix} 1 &\text{amp; } 2 &\text{amp;} 3\end{bmatrix},$ then order is:
3 × 1
1 × 3
2 × 3
None of these
Answer
1 × 3
Solution:
An $\text{m}\times\text{n} $ matrix has m row and n columns.
The given matrix $\text{A}= \begin{bmatrix} 1 &\text{amp; } 2 &\text{amp;} 3\end{bmatrix},$ has 1 row and 3 columns.
Thus, order of A is $ 1\times3.$
If $\text{AB}=\text{A}$ and $\text{BA = B}$ then $\text{B}^2 $ is equal to:
$\text{B}$
$\text{A}$
$\text{-B}$
$\text{B}^2$
Answer
$\text{B}$
Solution:
We have, $\text{AB}=\text{A}$and $\text{BA = B}$
Since, $\text{B}^2=\text{B.B}$
$\text{B}^2=\text{(BA)}.\text{B}$
$\text{B}^2=\text{B}.\text{(AB)}$
$\text{B}^2=\text{B.A}$
$\text{B}^2=\text{B}$
Hence, this is the answer.
Choose the correct answer from the given four options.The matrix $\text{P}=\begin{bmatrix}0&0&4\\0&4&0\\4&0&0\end{bmatrix}$ is a:
Square matrix.
Diagonal matrix.
Unit matrix.
None of these.
Answer
Square matrix.
Solution:
We know that, in a square matrix number of rows are equal to the number of columns.
So, the matrix $\text{P}=\begin{bmatrix}0&0&4\\0&4&0\\4&0&0\end{bmatrix}$ is a square matrix.
If order of A + B is n × n, then the order of AB is:
n × n
n × m
m × n
not defined
Answer
n × n
Solution:
If order of $\text{A}+\text{B}$ is $\text{n}\times\text{n},$ then the order of both$\begin{bmatrix}\text{A}&\text{amp;}\text{ B}\end{bmatrix}$ is $\text{n}\times\text{n}$
Therefore, order of $\text {AB }$is $\text{n}\times\text{n}$
If $\text{A}=\begin{bmatrix}5&\text{x}\\\text{y}&0\end{bmatrix}$ and $A = A^T,$ then$:$
A
$x = 0, y = 5$
B
$x + y = 5$
✓
$x = y$
D
None of these.
Answer
Correct option: C.
$x = y$
Here,
$\text{A}=\begin{bmatrix}5&\text{x}\\\text{y}&0\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}5&\text{y}\\\text{x}&0\end{bmatrix}$
Now,
$\text{A}=\text{A}^\text{T}$
The corresponding elements of two equal matrices are equal.
$\therefore\ \begin{bmatrix}5&\text{x}\\\text{y}&0\end{bmatrix}=\begin{bmatrix}5&\text{y}\\\text{x}&0\end{bmatrix}$
$\Rightarrow\text{x}=\text{y}$
Choose the correct answer from the given four options.If $\begin{bmatrix}2\text{x}+\text{y}&4\text{x}\\5\text{x}-7&4\text{x}\end{bmatrix}=\begin{bmatrix}7&7\text{y}-13\\\text{y}&\text{x}+6\end{bmatrix},$ then the value of x + y is:
x = 3, y = 1
x = 2, y = 3
x = 2, y = 4
x = 3, y = 3
Answer
x = 2, y = 3
Solution:
We have, $\begin{bmatrix}2\text{x}+\text{y}&4\text{x}\\5\text{x}-7&4\text{x}\end{bmatrix}=\begin{bmatrix}7&7\text{y}-13\\\text{y}&\text{x}+6\end{bmatrix}$
⇒ 4x = x + 6 ⇒ x = 2
and 4x = 7y - 13
⇒ 8 = 7y - 13
⇒ y = 3
$\therefore$ x + y = 2 + 3 = 5
Choose the correct answer from the given four options.If $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\cot^{-1}(\pi\text{x})\end{bmatrix}$ and $\text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\tan^{-1}(\pi\text{x})\end{bmatrix}$ then A - B is:
$\text{I}$
$0$
$2\text{I}$
$\frac{1}{2}\text{I}$
Answer
$\frac{1}{2}\text{I}$
Solution:
We have, $\text{B}=\begin{bmatrix}-\frac{1}{\pi}\cos^{-1}\text{x}\pi&\frac{1}{\pi}\tan^{-1}\frac{\text{x}}{\pi}\\\frac{1}{\pi}\sin^{-1}\frac{\text{x}}{\pi}&-\frac{1}{\pi}\tan^{-1}\pi\text{x}\end{bmatrix}$
and $\text{A}=\begin{bmatrix}\frac{1}{\pi}\sin^{-1}\text{x}\pi&\frac{1}{\pi}\tan^{-1}\frac{\text{x}}{\pi}\\\frac{1}{\pi}\sin^{-1}\frac{\text{x}}{\pi}&\frac{1}{\pi}\cot^{-1}\pi\text{x}\end{bmatrix}$
$\therefore\ \text{A}-\text{B}=\begin{bmatrix}\frac{1}{\pi}(\sin^{-1}\text{x}\pi+\cos^{-1}\text{x}\pi)&0\\0&\frac{1}{\pi}\big(\cot^{-1}\text{x}\pi+\tan^{-1}\pi\text{x}\big)\end{bmatrix}$
$=\begin{bmatrix}\frac{1}{\pi}\Big(\frac{\pi}{2}\Big)&0\\0&\frac{1}{\pi}\Big(\frac{\pi}{2}\Big)\end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=\frac{1}{2}\text{I}$