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MATHS M.C.Q (1 Marks)

MATRICES · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 161 questions.

Questions · Page 2 of 4

M.C.Q (1 Marks)

Question 511 Mark
If A and B are symmetric matrices, then ABA is:
  1. Symmetric matrix.
  2. Skew-symmetric matrix.
  3. Diagonal matrix.
  4. Scalar matrix.
Answer
  1. Symmetric matrix.
Solution:
Let $\text{A}=\begin{bmatrix}1&2\\2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}3&2\\2&3\end{bmatrix}$
$\text{AB}=\begin{bmatrix}1&2\\2&1\end{bmatrix}\begin{bmatrix}3&2\\2&3\end{bmatrix}=\begin{bmatrix}7&8\\8&7\end{bmatrix}$
$\text{ABA}=\begin{bmatrix}7&8\\8&7\end{bmatrix}\begin{bmatrix}1&2\\2&1\end{bmatrix}=\begin{bmatrix}23&22\\22&23\end{bmatrix}$
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Question 521 Mark
The matrix $\text{A}=\begin{bmatrix}1&0&0\\0&2&0\\0&0&4\end{bmatrix}$ is:
  1. Identity matrix.
  2. Symmetric matrix.
  3. Skew-symmetric matrix.
  4. Diagonal matrix.
Answer
  1. Diagonal matrix.
Solution:
A matrix is called Diagonal matrix if all the elements, except those in the leading diagonal, are zero.
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MCQ 531 Mark
Choose the correct answer from the given four options. If $A$ is a square matrix such that $A^2 = I,$ then $(A - I)^3 + (A + I)^3 - 7A$ is equal to:
  • ✓
    $A$
  • B
    $I - A$
  • C
    $I + A$
  • D
    $3A$
Answer
Correct option: A.
$A$
We have, $A^2 = I$
Now $, (A - I)^3 + (A + I)^3 - 7A = [(A - I) + (A + I)] \ [(A - I)^2 + (A + I)^2 - (A - I)(A + I)] - 7A$
$[\because a^3 + b^3 = (a + b)(a^2 + b^2 - ab)]$
$= [(2A){A^2 + I^2 - 2AI + A^2 + I^2 + 2AI - (A^2 - I^2)}] - 7A$
$= [(2A){AI + I^2 - 2AI + AI + I^2 + 2AI - AI +I^2}] - 7A \ [\because A^2 = AI]$
$= 2A[I + I^2 + I + I^2 - I + I^2] - 7A$
$= 2A[5I - I] - 7A$
$= 8AI - 7AI \ [\because A = AI]$
$= AI$
$= A$
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MCQ 541 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:
  • ✓
    $A^{T }+ B$
  • B
    $A^{T }+ B^T$
  • C
    $A^{T }+ C$
  • D
    $B + C^T$
Answer
Correct option: A.
$A^{T }+ B$
Given order of $A$ is $3 \times 2$
$\Rightarrow$ order of $A^T$ is $2 \times 3$
Also, given order of $B$ is $3 \times 2$
$\Rightarrow$ order of $B^T$ is $2 \times 3$
Order of $C 2 \times 3$
Since, $A^T, B^T, C$ have same order,
so addition of any $2$ or all three matrices are defined.
$A^{T }+ B$ is not defined as their orders are different.
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Question 551 Mark
The scalar matrix is:
  1. $\begin{bmatrix} -1 & 3 \\ 2 & 4 \end{bmatrix}$
  2. $\begin{bmatrix} 0 & 3 \\ 2 & 0 \end{bmatrix}$
  3. $\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$
  4. $\text{None of these}$
Answer
  1. $\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$
Solution:
A diagonalmatrixwith all its main diagonal entries equal is ascalar matrix, that is, ascalarmultiple of the identitymatrix
$\therefore \begin{bmatrix} 4 &\text{amp; 0} \\ 0 &\text{amp; } 4 \end{bmatrix}$ is a scalar matrix.
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MCQ 561 Mark
If a matrix $A$ is both symmetric and skew $-$ symmetric, then:
  • A
    $A$ is a diagonal matrix.
  • B
    $A $ is a zero matrix.
  • C
    $A$ is a scalar matrix.
  • ✓
    $A$ is a null matrix.
Answer
Correct option: D.
$A$ is a null matrix.
$A$ is symmetric $\Rightarrow a_{ij} = a_{ji} \rightarrow (1)$
$A$ is skew $-$ symmetric
$\Rightarrow a_{ij} = - a_{ij} \rightarrow (2)$ and
$a_{ij} = - a_{ij}$
$\Rightarrow a_{ij} = 0$ means the diagonal entries are zero.
From $(1)$ and $(2)$ we can write
$a_{ij} = a_{ij} = 0$ which means all the off diagonal entries are zero.
So $,A$ is a null matrix.
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Question 571 Mark
If the matrices has 13 elements , then the possible dimension (order) it can have are:
  1. 1 × 13 or 13 × 1
  2. 1 × 26 or 26 × 1
  3. 2 × 13 or 13 × 2
  4. None of these
Answer
  1. 1 × 13 or 13 × 1
Solution:
As we know the number of elements in a matrix = (no.of rows) × No.of columns.
Therefore for 13 elements the rows and columns could only be (13 × 1) or (1 × 13)
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Question 581 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 $\text{A+B}=$
  1. $\text{A}$
  2. $\text{B}$
  3. $\displaystyle \begin{vmatrix}2&0 \\ 1 &1 \end{vmatrix}$
  4. $\displaystyle \begin{vmatrix}0&2 \\ 2 &2 \end{vmatrix}$
Answer
  1. $\displaystyle \begin{vmatrix}2&0 \\ 1 &1 \end{vmatrix}$
Solution:
$\text{A+B}=\displaystyle \begin{vmatrix} 2 &\text{amp; } 0 \\ 1 &\text{amp; } 1 \end{vmatrix}$
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Question 591 Mark
A square matrix A has 9 elements. What is the possible order of A?
  1. 1 × 9
  2. 9 × 9
  3. 3 × 3
  4. 2 × 7
Answer
  1. 3 × 3
Solution:
The factors of 9 are 1, 3 and 9.So, the possible orders of a matrix containing 9 elements is 1 × 9,  9 × 1, 3 × 3.
In a square matrix, the number of rows is equal to the number of columns.So, the required order is 3 × 3.
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Question 601 Mark
If A and B are two matrices of order 3 × m and 3 × n respectively and m = n, then the order of 5A - 2B is:
  1. m × 3 
  2. 3 × 3
  3. m × n
  4. 3 × n
Answer
  1. 3 × n
Solution:
A matrix of order 3 × mB matrix of order 3 × n
It is also given that m = nThen the order of the matrix will be sameSo order 5A - 2B is 3 × n.
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Question 611 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:
  1. -4, 4
  2. -4, -4
  3. 4, 4
  4. 4, -4
Answer
  1. -4, 4
Solution:
$\text{A+B =1},\text{ i.e.,} \displaystyle \begin{vmatrix} 1&\text{amp; }\text{x+y} \\\text{y}-4 &\text{amp;} 1 \end{vmatrix}=\begin{vmatrix} 1&\text{amp; } 0 \\ 0 &\text{amp; } 1 \end{vmatrix}$
or $\text{x}=\text{y}=0,\text{ y}-4=0$
$\therefore\text{ x} = -4, \text{ y}=4$
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Question 621 Mark
A matrix has 18 elements.Find the number of possible orders of the matrix:
  1. 5
  2. 6
  3. 4
  4. 7
Answer
  1. 6
Solution:
A matrix of mm rows and nn columns has m × n elements.
18 can be got by all combinations of 1 × 18,18 × 1, 2 × 9, 9 × 2, 3 ×6, 6 × 3
Hence, there are 6 possible matrices which have 18 elements.
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Question 631 Mark
The matrix $\text{A}=\begin{bmatrix}0&0&4\\0&4&0\\4&0&0\end{bmatrix}$ is a:
  1. Square matrix.
  2. Diagonal matrix.
  3. Unit matrix.
  4. None of these.
Answer
  1. Square matrix.
Solution:
Given: $\text{A}=\begin{bmatrix}0&0&4\\0&4&0\\4&0&0\end{bmatrix}$
Since, number of rows is equal to number of columns.
Therefore, A is a square matrix.
Hence, the correct option is (a).
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MCQ 641 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
  • B
    $A$ unit matrix
  • C
    $-A$
  • ✓
    $A$
Answer
Correct option: D.
$A$
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0&0\\0&1&0\\\text{a}&\text{b}&-1\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\\text{a}&\text{b}&-1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+0+0&0+0+0&0+0-0\\0+0+0&0+1+0&0+0-0\\\text{a}+0-\text{a}&0+\text{b}-\text{b}&0+0+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
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Question 651 Mark
If A and B are symmetric matrices of the same order, then:
  1. AB is a symmetric matrix.
  2. A - Bis askew-symmetric matrix.
  3. AB + BA is a symmetric matrix.
  4. AB - BA is a symmetric matrix.
Answer
  1. AB + BA is a symmetric matrix.
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Question 661 Mark
The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
  1. 27
  2. 18
  3. 81
  4. 512
Answer
  1. 512
Solution:
The given matrix of the order $3\times3$ has 9 elements and each of these elements can be either 0 or 1.
Now, each of the 9 elements can be filled in two possible ways.
Therefore, by the multiplication principle, the required number of possible matrices is $2^9=512$
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MCQ 671 Mark
If $S = [S_{ij}]$ is a scalar matrix such that $S_{ij} = k$ and $A$ is a square matrix of the same order, then $AS = SA =$ ?
  • A
    $A^k$
  • B
    $k + A$
  • ✓
    $kA$
  • D
    $kS$
Answer
Correct option: C.
$kA$
Here,
$\text{S}=\big[\text{S}_{\text{ij}}\big]$
$\Rightarrow\text{S}=\begin{bmatrix}\text{k}&0\\0&\text{k}\end{bmatrix}$$\big[\because\ \text{S}_\text{ij} = \text{k}\big]$
Let $\text{A}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}$$\big[\because\ \text{A is square matrix}\big]$
Now,
$\text{AS}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}\begin{bmatrix}\text{k}&0\\0&\text{k}\end{bmatrix}=\begin{bmatrix}\text{ka}_{11}&\text{ka}_{12}\\\text{ka}_{21}&\text{ka}_{22}\end{bmatrix}$$=\text{k}\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}=\text{kA}$
$\text{SA}=\begin{bmatrix}\text{k}&0\\0&\text{k}\end{bmatrix}\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}=\begin{bmatrix}\text{ka}_{11}&\text{ka}_{12}\\\text{ka}_{21}&\text{ka}_{22}\end{bmatrix}$$=\text{k}\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}=\text{kA}$
$\therefore\ \text{AS}=\text{SA}=\text{kA}$
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MCQ 681 Mark
If matrix $\text{A}=\big[\text{a}_{\text{ij}}\big]_{2\times2'}$ where $\text{a}_\text{ij}=\begin{cases}1,&\text{if }\text{i }\neq\text{j}\\0,&\text{if }\text{i }=\text{j}\end{cases},$ then $A^2$ is equal to$:$
  • ✓
    $1$
  • B
    $A$
  • C
    $O$
  • D
    $-I$
Answer
Correct option: A.
$1$
$\text{A}=\begin{bmatrix}0 &1\\1&0\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=1$
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MCQ 691 Mark
If $A$ and $B$ are two matrices such that $AB = B$ and $BA = A, A^2 + B^2$ is equal to:
  • A
    $2AB$
  • B
    $2BA$
  • ✓
    $A + B$
  • D
    $AB$
Answer
Correct option: C.
$A + B$
Given $AB = A$ and $BA = B,$ then
$\Rightarrow \ce{BAB} = B^2$ and $\ce{ABA} = A^2$
$\Rightarrow BA = B^2$ and $AB = A^2$
$\Rightarrow B = B^2$ and $A = A^2$
$\Rightarrow A^2 + B^2$
$= A + B$
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Question 701 Mark
$\text{A}^2=\text{I}\Rightarrow$
  1. $|\text{A}|=0$
  2. $|\text{A}|=1$
  3. $|\text{A}|=-1$
  4. $|\text{A}|=\pm1$
Answer
  1. $|\text{A}|=\pm1$
Solution:
Given, $\text{A}^2=\text{I}$
Take determinant both sides,
$|\text{A}^2|=|\text{I}|\Rightarrow|\text{A}^2|=1\Rightarrow|\text{A}|=\pm1$
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Question 711 Mark
Which one of the following statements is not true:
  1. A scalar matrix is a square matrix
  2. A diagonal matrix is a square matrix
  3. A scalar matrix is a diagonal matrix
  4. A diagonal matrix is a scalar matrix
Answer
  1. A diagonal matrix is a scalar matrix
Solution:
Option A and Option C and option B - true A scalar matrix is a diagonal matrix and every diagonal matrix is a square matrixHence every scalar matrix is also square matrixOption D - not trueEvery diagonal matrix is square matrix but not vice versa
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Question 721 Mark
If $\text{A} = \displaystyle \left[ \begin{matrix} 1 &\text{amp ; 2} \\ 3&\text{amp; 4} \end{matrix} \right],$ then number of elements in A are:
  1. 4
  2. 3
  3. 2
  4. None of these
Answer
  1. 4
Solution:
Since, given matrix A is of order $2\times2 = 4\therefore$  Number of elements in $\text{A} = 4$
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Question 731 Mark
If A = [1] , then A is:
  1. Zero matrix
  2. SIngular matrix
  3. Non - singular matrix
  4. Data insufficient
Answer
  1. Non - singular matrix
Solution:
$\text{A} = \big[1\big] $ is an identity matrix with order $1\times1.|\text{A}|\neq0$
So A is nonsingular.
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MCQ 741 Mark
If $A = [a_{ij}]$ is a scalar matrix of order $n \times n$ such that $a_{ij} = k,$ for all $i,$ then trace of $A$ is equal to$:$
  • ✓
    $nk$
  • B
    $n + k$
  • C
    $\frac{\text{n}}{\text{k}}$
  • D
    none of these
Answer
Correct option: A.
$nk$
$\text{Trace}=\sum\limits^\text{n}_{\text{i}-1}\text{a}_{\text{ij}}=\text{nk}$
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Question 751 Mark
If A = [1 ​amp; 2​], B = [3​ amp; 4​] then A + B =
  1. [1 ​amp; 4​]
  2. [4​ amp; 4​]
  3. [4​ amp; 6​]
  4. None of these
Answer
  1. [4​ amp; 6​]
Solution:
Given, A = [1 ​amp; 2​], B = [3​ amp; 4​] then A + B =
[1 + 3 ​amp; 2 + 4​] A + B = [4 ​amp; 6​]
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Question 761 Mark
If $\text{A}$ is a square of order 3, then$|\text{Adj}(\text{Adj}\text{A}^2)|=$
  1. $|\text{A}|^2$
  2. $|\text{A}|^4$
  3. $|\text{A}|^8$
  4. $|\text{A}|^{16}$
Answer
  1. $|\text{A}|^8$
Solution:
KEY :  3
$|\text{Adj}(\text{Adj}\text{A}^2)|$
$\text{Q}=|\text{A}^2|^{(3-1)^2}=|\text{A}^2|^4=|\text{A}|^8$
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Question 771 Mark
The matrix $\text{A}=\begin{bmatrix}0&\text{amp};0&\text{amp};0\\0&\text{amp};4&\text{amp};0\\4&\text{amp};0&\text{amp};0\end{bmatrix}$ is a:
  1. Square matrix
  2. Diagonal matrix
  3. Unit matrix
  4. None of these
Answer
  1. Square matrix
Solution:
Given, $\text{A}=\begin{bmatrix}0&\text{amp};0&\text{amp};4\\0&\text{amp};4&\text{amp};0\\4&\text{amp};0&\text{amp};0\end{bmatrix}$
The matrix is a square as it has same no. of rows and columns,
But it is not a diagonal matrix as there are elements other than diagonal ones which are not zero.
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MCQ 781 Mark
The element in the second row and third column of the matrix $\begin{bmatrix}4&\text{amp; }5&\text{amp; }6 \\3 &\text{amp;}-4&\text{amp; }3\\2 &\text{amp; }1&\text{amp; }0 \end{bmatrix}$ is$:$
  • ✓
    $3$
  • B
    $1$
  • C
    $2$
  • D
    $-4$
Answer
Correct option: A.
$3$
The element in the second row, third column is represented by $a_{23}​ = 3.$
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Question 791 Mark
If a matrix has mm rows and nn columns then its order is:
  1. $\text{m}+\text{n}$
  2. $\text{n}\times\text{n}$
  3. $\text{m}\times\text{m}$
  4. $\text{m}\times\text{n}$
Answer
  1. $\text{m}\times\text{n}$
Solution:
A matrix has mm rows and n columns then its order is $\text{m}\times\text{n}$
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MCQ 801 Mark
If $\text{A}=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix},\text{n}\in\text{N},$ then $A^{4n}$ equals$:$
  • A
    $\begin{bmatrix}0&\text{i}\\\text{i}&0\end{bmatrix}$
  • B
    $\begin{bmatrix}0&0\\0&0\end{bmatrix}$
  • ✓
    $\begin{bmatrix}1&0\\0&1\end{bmatrix}$
  • D
    $\begin{bmatrix}0&\text{i}\\\text{i}&0\end{bmatrix}$
Answer
Correct option: C.
$\begin{bmatrix}1&0\\0&1\end{bmatrix}$
Given $\text{A}=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}=\text{i}\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\text{A}^4=\text{i}^4\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\text{A}^{4\text{n}}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
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Question 811 Mark
Two matrices A and B are added if:
  1. Both are rectangular
  2. Both have same order
  3. No of columns of A is equal to columns of B
  4. No of rows of A is equal to no of columns of B
Answer
  1. Both have same order
Solution:
While 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.
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Question 821 Mark
Matrix $\text{A} = [\text{a}_\text{ij}]_{\text{m} \times \text{n}}$ is a square matrix if:
  1. m < n
  2. m > n
  3. m = 1
  4. m = n
Answer
  1. m = n
Solution:
Matrix $\text{A} = [\text{a}_\text{ij}]_{\text{m} \times \text{n}}$ is a square matrix Number of columns = number of rows = m
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Question 831 Mark
If A and B are two matrices such that A + B and AB are both defined, then
  1. A and B can be any matrices
  2. A, B are square matrices not necessarily of the same order
  3. A, B are square matrices of the same order
  4. Number of columns of A = Number of rows of B
Answer
  1. A, B are square matrices of the same order
Solution:
Let A and B both have a matrices of order m × n
Also AB is defined, it means m = n
Hence A and B are square matrices of same order
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Question 841 Mark
The order of the matrix $\begin{bmatrix}1\\3\\4<\text{br}> \end{bmatrix}$ is:
  1. 1 × 3
  2. 3 × 1
  3. 1 × 1
  4. 3 × 3
Answer
  1. 3 × 1
Solution:
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 $3\times 1$.
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Question 851 Mark
If A, B are square matrices of order 3, A is non-singular and AB = 0, then B is a:
  1. Null matrix.
  2. Singular matrix.
  3. Unit-matrix.
  4. Non-singular matrix.
Answer
  1. Null matrix.
Solution:
Since A is non-singular matrix and the determinant of a non-singular matrix is non-zero, B should be a null matrix.
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Question 861 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 ?
  1. -5
  2. 0
  3. 5
  4. 10
Answer
  1. 0
Solution:
$\begin{bmatrix}\text{x}\\\text{x}\\\text{y}\end{bmatrix}+\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}+\begin{bmatrix}\text{z}\\0\\0\end{bmatrix}=\begin{bmatrix}10\\5\\5\end{bmatrix}\therefore\text{x}+\text{y}+\text{z}=10,\text{ x}+\text{y}=5$
$\text{y}+\text{z}=5$ Replacing $\text{x}+\text{y}=5$ in $\text{x}+\text{y}=\text{z}=10$
We have, $\text{z}=5$
Also, $\text{y}+\text{z}=5$
$\therefore\text{y}=5-\text{z}=0$​​​​
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Question 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 $\text{A}=$
  1. $\displaystyle \begin{vmatrix} 2 & 7 \\ 0 & 1\end{vmatrix} $
  2. $\displaystyle \begin{vmatrix} 0 & 1 \\ 2 & 7\end{vmatrix} $
  3. $\displaystyle \begin{vmatrix} 1 & 0 \\ 2 & 7\end{vmatrix} $
  4. $\displaystyle \begin{vmatrix} 2 & 1 \\ 0 & 7\end{vmatrix} $
 
Answer
  1.  $\displaystyle \begin{vmatrix} 2 & 7 \\ 0 & 1\end{vmatrix} $
Solution:
$ \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}$ 
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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
    $A - B$
Answer
Correct 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.
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Question 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 $\text{x}+2\text{y}=$
  1. 9
  2. 17
  3. 10
  4. 7
Answer
  1. 17
Solution:
x = 1.y = 8
$\therefore$ x + 2y = 17
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Question 901 Mark
If A is a square matrix, then A – A’ is a:
  1. Diagonal matrix.
  2. Skew-symmetric matrix.
  3. Symmetric matrix.
  4. None of these.
Answer
  1. Skew-symmetric matrix.
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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
    $A^2 - BA - AB - B^2$
  • ✓
    $A^2 - B^2 + BA - AB$
  • D
    $A^2 - BA + B^{2 }+ AB$
Answer
Correct option: C.
$A^2 - B^2 + BA - AB$
$(A + B)(A - B) $
$= A^2 - AB + BA - B^2$
Hence, the correct option is $(c).$
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Question 921 Mark
If A is a matrix of order m × n and B is a matrix such that AB′ and B′A are both defined, the order of the matrix B is:
  1. m × m
  2. n × n
  3. n × m
  4. m × n
Answer
  1. m × n
Solution:
Given that order of matrix A is m\times nm×nNow 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 × m.Hence, order of matrix B is m × 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.
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Question 931 Mark
If the matrix is a square matrix and it contains 36 elements, then the order of the matrix is:
  1. 4 × 4
  2. 8 × 8
  3. 6 × 6
  4. 3 × 3
Answer
  1. 6 × 6
Solution:
The number of elements in a matrix are equal to the product of number of rows and columns.
As the matrix is a square matrix so that if matrix has 36 elements then the order of the matrix is 6 × 6.
Hence, the answer is 6 × 6.
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Question 941 Mark
If $\text{A}=\begin{bmatrix}\text{a}&\text{b}\\\text{b}&\text{a}\end{bmatrix}$ and $\text{A}^2=\begin{bmatrix}\alpha&\beta\\\beta&\alpha\end{bmatrix},$ then:
  1. $\alpha=\text{a}^2+\text{b}^2,\beta=\text{ab}$
  2. $\alpha=\text{a}^2+\text{b}^2,\beta=2\text{ab}$
  3. $\alpha=\text{a}^2+\text{b}^2,\beta=\text{a}^2-\text{b}^2$
  4. $\alpha=2\text{ab},\beta=\text{a}^2+\text{b}^2$
Answer
  1. $\alpha=\text{a}^2+\text{b}^2,\beta=2\text{ab}$
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MCQ 951 Mark
$A = [a_{ij}]_{m \times n }$ is a square matrix, if$:$
  • A
    $m < n$
  • B
    $m > n$
  • ✓
    $m = n$
  • D
    None of these.
Answer
Correct option: C.
$m = n$
For $A = [a_{ij}]_{m \times n}$ to be square matrix.
number of row $s =$ number of columns
$\therefore m = n$
$\therefore (c)$ is correct.
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Question 961 Mark
If a matrix has 13 elements, then the possible dimensions (orders) of the matrix are:
  1. $1\times13$ or $13\times1$
  2. $1\times26$ or $26\times1$
  3. $2\times13$ or $13\times2$
  4. $13\times13$
Answer
  1. $1\times13$ or $13\times1$
Solution:
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 $\text{A}$
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MCQ 971 Mark
If $A$ and $B$ are two matrices such that $AB = A$ and $BA = B,$ then $B^2$ is equal to:
  • ✓
    $B$
  • B
    $A$
  • C
    $1$
  • D
    $0$
Answer
Correct option: A.
$B$
Here $, AB = A ...(1)$
$BA = B ...(2)$
$\Rightarrow BAB = BB \ [$Multiplying both sides by $B]$
$\Rightarrow BA = B^2 \ [$From eq. $(1)]$
$\Rightarrow B = B^2 \ [$From eq. $(2)]$
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MCQ 981 Mark
Choose the correct answer from the given four options. On using elementary row operation $R_1 \rightarrow R_1 – 3R_2$ in the following matrix equation $\begin{bmatrix}4&2\\3&3\end{bmatrix}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix},$ we have:
  • ✓
    $\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$
  • B
    $\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\begin{bmatrix}-1&-3\\1&1\end{bmatrix}$
  • C
    $\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&2\\1&-7\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$
  • D
    $\begin{bmatrix}4&2\\-5&-7\end{bmatrix}=\begin{bmatrix}1&2\\-3&-3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$
Answer
Correct option: A.
$\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$
We have, $\begin{bmatrix}4&2\\3&3\end{bmatrix}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$
Using elementary row operation $R_1 \rightarrow R_1 - 3R_2$
$\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$
Since, on using elementary row operation on $X = AB,$
we apply these operation simultaneously on $X$ and on the first matrix $A$ of the product $AB$ on $\text{RHS}.$
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MCQ 991 Mark
Choose the correct answer from the given four options.
If $A$ is matrix of order $m \times n$ and $B$ is a matrix such that $AB\ '$ and $B\ 'A$ are both defined, then order of matrix $B$ is:
  • A
    $m \times m$
  • B
    $n \times n$
  • C
    $n \times m$
  • ✓
    $m \times n$
Answer
Correct option: D.
$m \times n$
Let $A = [a_{ij}]_{m\times n}$ and $B = [b_{ij}]_{p} {\times q}$
$B\ ' = [b_{ji}]_{q\times p}$
Now, $AB\ ’$ is defined, so $n = q$
and $B\ ’A$ is also defined, so $p = m$
$\therefore$ Order of $B\ ' = [b_{ji}]_{n\times m}$
And order of $B = B = [b_{ij}]m_{\times n}$
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Question 1001 Mark
A matrix consisting of a single column of m elements is know as:
  1. Column matrix
  2. Row matrix
  3. Square matrix
  4. Null matrix
Answer
  1. Column matrix
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M.C.Q (1 Marks) - Page 2 - MATHS STD 12 Science Questions - Vidyadip