MCQ 511 Mark
Choose the correct answer from the given four options. Total number of possible matrices of order $3 \times 3$ with each entry $2$ or $0$ is :
AnswerTotal number of possible matrices of order $3 \times 3$ with each entry $2$ or $0$ is $2^9$
i.e., $512 .$
View full question & answer→MCQ 521 Mark
The order of a matrix $\begin{bmatrix}2&\text{amp};5 &\text{amp};7 \end{bmatrix}$ is :
- A
$3 \times 3$
- B
$1 \times 1$
- C
$3 \times 1$
- ✓
$1 \times 3$
AnswerCorrect option: D. $1 \times 3$
Since, Order of a matrix is represented by $m × n,$ where mm is the number of rows and nn is the number of columns.
Given, $\begin{bmatrix}2&\text{amp};5 &\text{amp};7 \end{bmatrix}$ is a matrix in which number of row is $1$ and number of columns are $3.$
$\therefore\begin{bmatrix}2&\text{amp};5 &\text{amp};7 \end{bmatrix}$ is a matrix of order $1 × 3$
View full question & answer→MCQ 531 Mark
Which of the following is correct:
- A
Determinant is a square matrix
- B
Determinant is a number associated to a matrix
- ✓
Determinant is a number associated to a square matrix
- D
AnswerCorrect option: C. Determinant is a number associated to a square matrix
Determinant is defined only for a square matrix.and its denotes the value of that square matrix.
View full question & answer→MCQ 541 Mark
Choose the correct answer from the given four options. For any two matrices $A$ and $B,$ we have:
- A
$\text{AB}=\text{BA}$
- B
$\text{AB}\neq\text{BA}$
- C
$\text{AB}=\text{O}$
- ✓
View full question & answer→MCQ 551 Mark
Two matrices $A$ and $B$ are added if:
AnswerWhile adding two matrices we add the numbers which belong to some row and column of each matrixo two matrices can be added.
If there are equal number of rows and columns in both. Both matrices should have same order therefore.
View full question & answer→MCQ 561 Mark
Matrix $\text{A} = [\text{a}_\text{ij}]_{\text{m} \times \text{n}}$ is a square matrix if:
- A
$m < n$
- B
$m > n$
- C
$m = 1$
- ✓
$m = n$
AnswerCorrect option: D. $m = n$
Matrix $\text{A} = [\text{a}_\text{ij}]_{\text{m} \times \text{n}}$ is a square matrix Number of columns $=$ number of rows $= m$
View full question & answer→MCQ 571 Mark
If $A$ and $B$ are symmetric matrices, then $\text{ABA}$ is:
- ✓
- B
Skew$-$symmetric matrix.
- C
- D
AnswerLet $\text{A}=\begin{bmatrix}1&2\\2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}3&2\\2&3\end{bmatrix}$
$\text{AB}=\begin{bmatrix}1&2\\2&1\end{bmatrix}\begin{bmatrix}3&2\\2&3\end{bmatrix}=\begin{bmatrix}7&8\\8&7\end{bmatrix}$
$\text{ABA}=\begin{bmatrix}7&8\\8&7\end{bmatrix}\begin{bmatrix}1&2\\2&1\end{bmatrix}=\begin{bmatrix}23&22\\22&23\end{bmatrix}$
View full question & answer→MCQ 581 Mark
The matrix $\text{A}=\begin{bmatrix}1&0&0\\0&2&0\\0&0&4\end{bmatrix}$ is:
- A
- B
- C
Skew$-$symmetric matrix.
- ✓
AnswerA matrix is called Diagonal matrix if all the elements, except those in the leading diagonal, are zero.
View full question & answer→MCQ 591 Mark
Which of the given values of $X$ and $Y$ make the following pairs of matrices equal? $\begin{bmatrix}3\text{x}+7&5\\\text{y}+1&2-3\text{x}\end{bmatrix},\begin{bmatrix}0&\text{y}-2\\8&4\end{bmatrix}$
- A
$\text{x}=-\frac{1}{3},\text{y}=7$
- B
$\text{y}=7,\text{x}=-\frac{2}{3}$
- C
$\text{x}=-\frac{1}{3},\text{y}=-\frac{2}{5}$
- ✓
Answer$\begin{bmatrix}3\text{x}+7&5\\\text{y}+1&2-3\text{x}\end{bmatrix}$
$=\begin{bmatrix}0&\text{y}-2\\8&4\end{bmatrix}$
$\Rightarrow3\text{x}+7=0$
$\Rightarrow\text{x}=\frac{-7}{3}$
$5=\text{y}-2$
$\Rightarrow\text{y}=7$
$\text{y}+1=8$
$\Rightarrow\text{y}=7$
$2-3\text{x}=4$
$\Rightarrow\text{x}=\frac{-2}{3}$
We are getting two values of $x.$
So, it is not possible to find.
View full question & answer→MCQ 601 Mark
If $A$ and $B$ are matrices of order $3 \times 2$ and $C$ is of order $2 \times 3$, then which of the following matrices is not defined :
- ✓
$A^{\top}+B$
- B
$A^{\top}+B^{\top}$
- C
$A^{\top}+C$
- D
$B+C^{\top}$
AnswerCorrect option: A. $A^{\top}+B$
Given order of $A$ is $3 \times 2$
$\Rightarrow$ order of $A^{\top}$ is $2 \times 3$
Also, given order of $B$ is $3 \times 2$
$\Rightarrow$ order of $B^{\top}$ is $2 \times 3$
Order of $C _2 \times 3$
Since, $A^{\top}, B^{\top}, C$ have same order, so addition of any $2$ or all three matrices are defined.
$A^{\top}+B$ is not defined as their orders are different.
View full question & answer→MCQ 611 Mark
- A
$\begin{bmatrix} -1 & 3 \\ 2 & 4 \end{bmatrix}$
- B
$\begin{bmatrix} 0 & 3 \\ 2 & 0 \end{bmatrix}$
- ✓
$\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$
- D
AnswerCorrect option: C. $\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$
A diagonalmatrixwith all its main diagonal entries equal is ascalar matrix, that is, ascalarmultiple of the identity matrix
$\therefore \begin{bmatrix} 4 &\text{amp; 0} \\ 0 &\text{amp; } 4 \end{bmatrix}$ is a scalar matrix.
View full question & answer→MCQ 621 Mark
If the matrices has $13$ elements , then the possible dimension $($order$)$ it can have are:
- ✓
$1 \times 13$ or $13 \times 1$
- B
$1 \times 26$ or $26 \times 1$
- C
$2 \times 13$ or $13 \times 2$
- D
AnswerCorrect option: A. $1 \times 13$ or $13 \times 1$
As we know the number of elements in a matrix $= ($no.of rows$) \times$ No.of columns.
Therefore for $13$ elements the rows and columns could only be $(13 \times 1)$ or $(1 \times 13)$
View full question & answer→MCQ 631 Mark
If $\text{A}=\displaystyle \begin{vmatrix} 1 &\text{amp; } 0 \\ 1 &\text{amp; } 0 \end{vmatrix}$ And $\text{B}=\displaystyle \begin{vmatrix} 1 &\text{amp; } 0 \\ 0 &\text{amp; } 1 \end{vmatrix}$ then $\text{A+B}=$
AnswerCorrect option: C. $\displaystyle \begin{vmatrix}2&0 \\ 1 &1 \end{vmatrix}$
$\text{A+B}=\displaystyle \begin{vmatrix} 2 &\text{amp; } 0 \\ 1 &\text{amp; } 1 \end{vmatrix}$
View full question & answer→MCQ 641 Mark
A square matrix $A$ has $9$ elements. What is the possible order of $A\ ?$
- A
$1 \times 9$
- B
$9 \times 9$
- ✓
$3 \times 3$
- D
$2 \times 7$
AnswerCorrect option: C. $3 \times 3$
The factors of $9$ are $1, 3$ and $9$.
So, the possible orders of a matrix containing $9$ elements is $1 \times 9, 9 \times 1, 3 \times 3.$
In a square matrix, the number of rows is equal to the number of columns.
So, the required order is $3 \times 3.$
View full question & answer→MCQ 651 Mark
If A, B are symmetric matrices of same order, then AB - BA is a
AnswerThe correct answer is A.
A and B are symmetric matrices, therefore, we have:
A' = A and B' = B ...(i)
Consider $ (\text{AB} - \text{BA})' = (\text{AB})' - (\text{BA})' \ \ \big[(\text{A} - \text{B})' = \text{A}' - \text{B}'\big]$
$= \text{B}'\text{A}' - \text{A}'\text{B}'\ \ \big[(\text{AB}) = \text{B}'\text{A}'\big]$
$= \text{BA} -\text{AB}\ \ \big[\text{by}(1)\big]$
$=-(\text{A}\text{B}- \text{B}\text{A})$
$\therefore\ (\text{AB}- \text{BA})' = -(\text{AB} - \text{BA})$
Thus, (AB - BA) is a skew-symmetric matrix.
View full question & answer→MCQ 661 Mark
If $A$ and $B$ are matrices of the same order, then $A B^{\top}-B A^{\top}$ is a :
AnswerCorrect option: A. Skew $-$ symmetric matrix.
$\left(A B^{\top}-B A^{\top}\right)^{\top}=\left(A B^{\top}\right)^{\top}-\left(B A^{\top}\right)^{\top}$
$=B A^{\top}-A B^{\top}$
$=-\left(A B^{\top}-B A^{\top}\right)$
Therefore, $A B^{\top}-B A^{\top}$ is a skew $-$ symmetric matrix.
Hence, the correct option is $(a).$
View full question & answer→MCQ 671 Mark
Choose the correct answer from the given four options.
If $A$ and $B$ are matrices of same order, then $(AB\ ' – BA\ ')$ is a :
AnswerWe have matrices $A$ and $B$ of same order.
Let $P = (AB\ ' - BA\ ')$
Then, $P\ ' = (AB\ ' - BA\ ')'$
$= (AB\ ')' - (BA\ ')'$
$= (B\ ')'(A)\ ' - (A\ ')'B\ ' $
$= BA\ ' - AB\ ' = -(AB\ ' - BA\ ') = -P$
Hence, $(AB\ ' - BA\ ')$ is a Skew symmetric matrix.
View full question & answer→MCQ 681 Mark
If $A$ is square matrix of order 3 , then $\left|\operatorname{Adj}\left(\operatorname{Adj} A^2\right)\right|=$
- A
$|A|^2$
- B
$|A|^4$
- ✓
$|A|^8$
- D
$|A|^{16}$
AnswerCorrect option: C. $|A|^8$
c. $|A|^8$
$∣\text{adj}(\text{adj}\text{A}^2)∣=\text{Q}=\begin{vmatrix}\text{A}^2\end{vmatrix}^{(3-1)^2} =\begin{vmatrix}\text{ A}^2 \end{vmatrix} ^4 =\begin{vmatrix} \text{A}\end{vmatrix}^8$
View full question & answer→MCQ 691 Mark
If $A$ and $B$ are two matrices of order $3 \times m$ and $3 \times n$ respectively and $m = n,$ then the order of $5A - 2B$ is :
- A
$m \times 3$
- B
$3 \times 3$
- C
$m \times n$
- ✓
$3 \times n$
AnswerCorrect option: D. $3 \times n$
$A$ matrix of order $3 \times m$
$B$ matrix of order $3 \times n$
It is also given that $m = n$ Then the order of the matrix will be sameSo order $5A - 2B$ is $3 \times n.$
View full question & answer→MCQ 701 Mark
If $\text{A}=\displaystyle \begin{vmatrix} 5 &\text{amp; x} \\ \text{y} &\text{amp; 6} \end{vmatrix}\text{B}=\displaystyle \begin{vmatrix} -4 &\text{amp; y} \\-4 &\text{amp; 5} \end{vmatrix}$ and $\text{A}+\text{B}=1$ then the values of $x$ and $y$ respectively are :
- ✓
$-4, 4$
- B
$-4, -4$
- C
$4, 4$
- D
$4, -4$
AnswerCorrect option: A. $-4, 4$
$\text{A+B =1},\text{ i.e.,} \displaystyle \begin{vmatrix} 1&\text{amp; }\text{x+y} \\\text{y}-4 &\text{amp;} 1 \end{vmatrix}=\begin{vmatrix} 1&\text{amp; } 0 \\ 0 &\text{amp; } 1 \end{vmatrix}$
or $\text{x}=\text{y}=0,\text{ y}-4=0$
$\therefore\text{ x} = -4, \text{ y}=4$
View full question & answer→MCQ 711 Mark
A matrix has $18$ elements. Find the number of possible orders of the matrix :
AnswerA matrix of mm rows and nn columns has $m \times n $ elements.
$18$ can be got by all combinations of $1 \times 18,18 \times 1, 2 \times 9, 9 \times 2, 3 \times 6, 6 \times 3$
Hence, there are $6$ possible matrices which have $18$ elements.
View full question & answer→MCQ 721 Mark
If $\text{A}=\displaystyle \left[ \begin{matrix} 1 &\text{amp; 2} \\ 3&\text{amp; 4} \end{matrix} \right],$ then which of the following is not an element of $A ?$
Answer$0$ is not present in given matrix.
View full question & answer→MCQ 731 Mark
If $m[-3 \text{ amp} ; 4] + n[-3 \text{ amp}; 4] = [10 \text{ amp} ;-11],$ then $3m + 7n =$
AnswerGiven $m[-3 + 4] + n[4 - 3] = [10 - 11]$
$\Rightarrow -3m + 4n = 10$ and $4m - 3n = -11$
by solving we get $m= -2$ and $n=1$
$\therefore 3m + 7n = -6 + 7 = 1$
View full question & answer→MCQ 741 Mark
The matrix $\text{A}=\begin{bmatrix}0&0&4\\0&4&0\\4&0&0\end{bmatrix}$ is a:
AnswerGiven: $\text{A}=\begin{bmatrix}0&0&4\\0&4&0\\4&0&0\end{bmatrix}$
Since, number of rows is equal to number of columns.
Therefore, $A$ is a square matrix.
Hence, the correct option is $(a).$
View full question & answer→MCQ 751 Mark
If n = p, then order of matrix 7X - 5Z is:
AnswerHere n = p (given), the order of matrices X and Z are equal.
$\therefore$ 7X – 5Z is well defined and the order of 7X – 5Z is same as the order of X and Z.
$\therefore$ The order of 7X – 5Z is either equal to 2 × n or 2 × p
But it is given that n = p
Therefore, the option (B) is correct.
View full question & answer→MCQ 761 Mark
If $\text{A}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix},$ then $A^{\top}+A=I_2$, if :
- A
$\theta=\text{n}\pi,\text{n}\in\text{Z}$
- B
$\theta=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
- ✓
$\theta=2\text{n}\pi+\frac{\pi}{3},\text{n}\in\text{Z}$
- D
AnswerCorrect option: C. $\theta=2\text{n}\pi+\frac{\pi}{3},\text{n}\in\text{Z}$
Here,
$\text{A}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}$
Now,
$\text{A}^\text{T}+\text{A}=\text{I}_2$
$\Rightarrow\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}+\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\cos\theta&0\\0&2\cos\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow2\cos\theta=1$
$\Rightarrow\cos\theta=\frac{1}{2}$
$\Rightarrow\cos\theta=\cos\frac{\pi}{3}$
$\Rightarrow\theta=2\text{n}\pi\pm\frac{\pi}{3}$ $(\text{n}\in\text{Z})$
View full question & answer→MCQ 771 Mark
If $\text{A}=\begin{bmatrix}1&0&0\\0&1&0\\\text{a}&\text{b}&-1\end{bmatrix},$ then $A^2$ is equal to:
- A
$A$ null matrix
- ✓
$A$ unit matrix
- C
$-A$
- D
$A$
AnswerCorrect option: B. $A$ unit matrix
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0&0\\0&1&0\\\text{a}&\text{b}&-1\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\\text{a}&\text{b}&-1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+0+0&0+0+0&0+0-0\\0+0+0&0+1+0&0+0-0\\\text{a}+0-\text{a}&0+\text{b}-\text{b}&0+0+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
View full question & answer→MCQ 781 Mark
The number of different possible orders of matrices having $18$ identical elements is:
AnswerLet the order of the matrix is $ (\text{a}\times\text{b})$ There are $18$ elements in the matrix.
So, $\text{a}\times\text{b} = 18$
Possible orders can be $ (1\times18),( 18\times1),( 2\times9),( 9\times2),( 3\times6),( 6\times3)$
There are $6$ possible orders.
View full question & answer→MCQ 791 Mark
Choose the correct answer from the given four options. If $\text{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then $A^2$ is equal to:
- A
$\begin{bmatrix}0&1\\1&0\end{bmatrix}$
- B
$\begin{bmatrix}1&0\\1&0\end{bmatrix}$
- C
$\begin{bmatrix}0&1\\0&1\end{bmatrix}$
- ✓
$\begin{bmatrix}1&0\\0&1\end{bmatrix}$
AnswerCorrect option: D. $\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\because\ \text{A}^2=\text{A}.\text{A}$
$=\begin{bmatrix}0&1\\1&0\end{bmatrix}.\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
View full question & answer→MCQ 801 Mark
If $A$ and $B$ are symmetric matrices of the same order, then:
- A
$AB$ is a symmetric matrix.
- B
$A - B$ is askew$-$symmetric matrix.
- ✓
$AB + BA$ is a symmetric matrix.
- D
$AB - BA$ is a symmetric matrix.
AnswerCorrect option: C. $AB + BA$ is a symmetric matrix.
View full question & answer→MCQ 811 Mark
If the order of matrices $A$ and $B$ are $3 \times 2$ and $2 \times 1$ respectively, then find the order of matrix $($if possible$)\ AB :$
- A
$1 \times 3$
- ✓
$3 \times 1$
- C
$2 \times 2$
- D
$2 \times 3$
AnswerCorrect option: B. $3 \times 1$
Order of $A : 3 \times 2$ Order of $B : 2 \times 1$ 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 $A\ \times$ number of columns of $B)$
$\therefore$ Order of $AB$ is $(3 \times 1)$
View full question & answer→MCQ 821 Mark
The order of the matrix $\begin{bmatrix}1\\3\\4 < \text{br} > \end{bmatrix}$ is :
- A
$1 \times 3$
- ✓
$3 \times 1$
- C
$1 \times 1$
- D
$3 \times 3$
AnswerCorrect option: B. $3 \times 1$
Order of matrix with mm rows and nn columns is given as $\text{m} \times\text{n}$ Let $\text{A}=\begin{bmatrix}-1\\3\\4 \end{bmatrix}$
In the given matrix, there are $3 $ rows and $1$ column.
Hence, the order of $A$ is $3\times 1$.
View full question & answer→MCQ 831 Mark
$B=A+A^2+A^3+A^4$ If order of $A$ is $3$ then order of $B$ is :
AnswerThe order of matrix doesnt change when the operation are done on it
So The order of $B$ remains same as the order of $A$
View full question & answer→MCQ 841 Mark
If $A, B$ are square matrices of order $3, A$ is non $-$ singular and $AB = 0,$ then $B$ is a :
- ✓
- B
- C
Unit $-$ matrix.
- D
Non $-$ singular matrix.
AnswerSince $A$ is non $-$ singular matrix and the determinant of a non $-$ singular matrix is non $-$ zero $,B$ should be a null matrix.
View full question & answer→MCQ 851 Mark
If the sum of the matrices $\begin{bmatrix}\text{x}\\\text{x}\\\text{y}\end{bmatrix},\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$ and $\begin{bmatrix}\text{z}\\0\\0\end{bmatrix}$ is the matrix $\begin{bmatrix}10\\5\\5\end{bmatrix},$ then what is the value of $y ?$
Answer$\begin{bmatrix}\text{x}\\\text{x}\\\text{y}\end{bmatrix}+\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}+\begin{bmatrix}\text{z}\\0\\0\end{bmatrix}=\begin{bmatrix}10\\5\\5\end{bmatrix}$
$\therefore\text{x}+\text{y}+\text{z}=10,\text{ x}+\text{y}=5$
$\text{y}+\text{z}=5$ Replacing $\text{x}+\text{y}=5$ in $\text{x}+\text{y}=\text{z}=10$
We have, $\text{z}=5$
Also, $\text{y}+\text{z}=5$
$\therefore\text{y}=5-\text{z}=0$
View full question & answer→MCQ 861 Mark
If $\text{A}= \begin{bmatrix} 1 &\text{amp; } 2 &\text{amp;} 3\end{bmatrix},$ then order is :
- A
$3 \times 1$
- ✓
$1 \times 3$
- C
$2 \times 3$
- D
AnswerCorrect option: B. $1 \times 3$
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.$
View full question & answer→MCQ 871 Mark
If $ \text{A}+\displaystyle \begin{vmatrix} 4 &\text{amp; } 2 \\ 1 &\text{amp; } 3 \end{vmatrix}=\displaystyle \begin{vmatrix} 6 &\text{amp; } 9 \\ 1 &\text{amp; } 4\end{vmatrix} $ then $\text{A}=$
- ✓
$\displaystyle \begin{vmatrix} 2 & 7 \\ 0 & 1\end{vmatrix} $
- B
$\displaystyle \begin{vmatrix} 0 & 1 \\ 2 & 7\end{vmatrix} $
- C
$\displaystyle \begin{vmatrix} 1 & 0 \\ 2 & 7\end{vmatrix} $
- D
$\displaystyle \begin{vmatrix} 2 & 1 \\ 0 & 7\end{vmatrix} $
AnswerCorrect option: A. $\displaystyle \begin{vmatrix} 2 & 7 \\ 0 & 1\end{vmatrix} $
$ \text{A}=\displaystyle \begin{vmatrix} 6 &\text{amp; } 9 \\ 1 &\text{amp; } 4\end{vmatrix}-\displaystyle \begin{vmatrix} 4 &\text{amp; } 2 \\ 1 &\text{amp; } 3 \end{vmatrix}=\displaystyle \begin{vmatrix} 2 &\text{amp; } 7 \\ 0 &\text{amp; } 1 \end{vmatrix}$
View full question & answer→MCQ 881 Mark
If $A$ and $B$ are two matrices of same order, then $A + B$ is equal to :
- ✓
$B + A$
- B
$BA$
- C
$(A + B)^T$
- D
AnswerCorrect option: A. $B + A$
Yes, matrices are commutative.
We can see it as follows, Let element of $A$ matrix be denoted by $\text{a}_\text{ij}$ and $B$ matrix be denoted by $\text{b}_\text{ij},$
Then corresponding elements of $\text{ A + B}$ matrix will be $(\text{a}_\text{ij} +\text{b}_\text{ij}) $ and corresponding
elements of $\text{B + A}$ matrix will be $(\text{b}_\text{ij} +\text{ a}_\text{ij}) $ But since addition is commutative, corresponding elements
of$\text{ A + B}$ and $\text{B + A}$ matrices are the same, So they are equal.
View full question & answer→MCQ 891 Mark
If $ \displaystyle \begin{vmatrix} \text{x} &\text{amp;}\text{ y} \\ 1 &\text{amp; } 6 \end{vmatrix}= \displaystyle \begin{vmatrix} 1&\text{amp; }8 \\ 1 &\text{amp; } 6 \end{vmatrix}$ then $\text{x}+2\text{y}=$
View full question & answer→MCQ 901 Mark
If $A$ is a square matrix, then $' A – A\ ’$ is a:
- A
- ✓
Skew$-$symmetric matrix.
- C
- D
AnswerCorrect option: B. Skew$-$symmetric matrix.
View full question & answer→MCQ 911 Mark
If $A$ and $B$ are square matrices of the same order, then $(A + B)(A - B)$ is equal to :
- A
$A^2-B^2$
- B
$A^2-B A-A B-B^2$
- ✓
$A^2-B^2+B A-A B$
- D
$A^2-B A+B^2+A B$
AnswerCorrect option: C. $A^2-B^2+B A-A B$
$(A+B)(A-B)=A^2-A B+B A-B^2$
Hence, the correct option is $(c).$
View full question & answer→MCQ 921 Mark
If $A$ is a matrix of order $m \times n$ and $B$ is a matrix such that $AB\ '$ and $B\ 'A$ are both defined, the order of the matrix $B$ is :
- A
$m \times m$
- B
$n \times n$
- C
$n \times m$
- ✓
$m \times n$
AnswerCorrect option: D. $m \times n$
Given that order of matrix $A$ is $m\times n \ m \times n $ Now if $AB\ '$ is defined then
number of column of $A$ should be same as number of rows of $B\ ',$ which is
$n$ Also since $B\ 'A$ is defined,so number column of $B\ '$ should be same as number of rows of $A$
which is $m$ Thus order of $B\ '$ is $n \times m$.
Hence, order of matrix $B$ is $m \times n.$
Note : Product of two matrix $A$ and $B, AB$ is defined only if number of columns of $A$ is same as number of rows of $B.$
And $B\ '$ represents transpose of matrix $B.$
View full question & answer→MCQ 931 Mark
If $\displaystyle \text{a}_{\text{ij}}=0\left (\text{i}\neq \text{j} \right )$ and $\displaystyle\text{a}_{\text{ij}}=2\left (\text{i= j} \right )$ then the matrix $\text{A}=\displaystyle \left [ \text{a}_{\text{ij}} \right ]_{\text{n}\times\text{n}}$ is a $ ........ $ matrix ?
AnswerGiven $A$ is a square matrix as the number of rows and columns are same as $n$
The elements $\text{a}_\text{ij}$ where $\text{i} = \text{j} $ lie along the diagonal.
and the elements $\text{a}_\text{ij}$ where $\text{i}\neq\text{j}$ do not lie along the diagonal.
Given, diagonal elements $= 2$ and the rest of the elements $= 0$
Such a diagonal matrix where all diagonal elements are equal, is called a scalar matrix.
View full question & answer→MCQ 941 Mark
If the matrix is a square matrix and it contains $36$ elements, then the order of the matrix is:
- ✓
$4 \times 4$
- B
$8 \times 8$
- C
$6 \times 6$
- D
$ 3 \times 3$
AnswerCorrect option: A. $4 \times 4$
View full question & answer→MCQ 951 Mark
$A=\left[a_{i j}\right]_m \times n$ is a square matrix, if:
- A
$m < n$
- B
$m > n$
- ✓
$m = n$
- D
AnswerCorrect option: C. $m = n$
View full question & answer→MCQ 961 Mark
If a matrix has $13$ elements, then the possible dimensions $($orders$)$ of the matrix are:
- ✓
$1\times13$ or $13\times1$
- B
$1\times26$ or $26\times1$
- C
$2\times13$ or $13\times2$
- D
$13\times13$
AnswerCorrect option: A. $1\times13$ or $13\times1$
If order of matrix $\text{A}=\text{a}\times\text{b}$
Then number of element in $\text{A}=\text{ab}$
Given $\text{ab}=13$
So, $\text{a}=1,\text{b}=13$
or $\text{b}=1,\text{a}=13$
So, $1\times13$ or $13\times1$ are possible order of $\text{A}$
View full question & answer→MCQ 971 Mark
Choose the correct answer from the given four options. On using elementary column operations $C_2 \rightarrow C_2-2 C_1$ in the following matrix equation $\begin{bmatrix}1&-3\\2&4\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&1\\2&4\end{bmatrix},$ we have :
- A
$\begin{bmatrix}1&-5\\0&4\end{bmatrix}=\begin{bmatrix}1&-1\\-2&2\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$
- B
$\begin{bmatrix}1&-5\\0&4\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\-0&2\end{bmatrix}$
- C
$\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-3\\0&1\end{bmatrix}\begin{bmatrix}3&1\\-2&4\end{bmatrix}$
- ✓
$\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$
AnswerCorrect option: D. $\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$
Given that, $\begin{bmatrix}1&-3\\2&4\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&1\\2&4\end{bmatrix}$
On using $C_2 \rightarrow C_2-2 C_1$ $\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$
Since, on using elementary column operation on $X = AB,$
we apply these operations simultaneously on $X$ and on the second matrix $B$ of the product $AB$ on $\text{RHS}.$
View full question & answer→MCQ 981 Mark
If $A$ and $B$ are two matrices such that $AB = A$ and $BA = B,$ then $B^2$ is equal to :
AnswerHere, $A B=A \ldots(1)$
$B A=B \ldots(2)$
$\Rightarrow B A B=B B\ [$ Multiplying both sides by $ B]$
$\Rightarrow B A=B^2\ [$ From eq. $(1)]$
$\Rightarrow B=B^2\ [$ From eq. $(2)]$
View full question & answer→MCQ 991 Mark
The matrix $\begin{bmatrix}0&5&-7\\-5&0&11\\7&-11&0\end{bmatrix}$ is :
AnswerCorrect option: A. A skew $-$ symmetric matrix.
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.
View full question & answer→MCQ 1001 Mark
If every row of a matrix $A$ contains $p$ elements and its column contains $q$ elements, then the order of $A$ is :
- A
$p \times p$
- B
$q \times q$
- C
$p \times q$
- ✓
$q \times p$
AnswerCorrect option: D. $q \times p$
$\begin{bmatrix}\text{a}_{11} &\text{amp;}\text{ a}_{12} \\\text{a}_{21}& \text{amp;}\text{ a}_{22}\\\text{a}_{31}&\text{amp; }\text{a}_{32} \end{bmatrix}$
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 $=\text{q}\times\text{p}$
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