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MATHS M.C.Q (1 Marks)

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

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M.C.Q (1 Marks)

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MCQ 11 Mark
If $A$ is a square matrix of order $3$ and $|A| = 5,$ then the value of $|2A\ '|$ is$:$
  • A
    $-10$
  • B
    $10$
  • C
    $-40$
  • ✓
    $40$
Answer
Correct option: D.
$40$
$|2A\ '|$
$= 2^3 |A\ '|$
$= 8 |4|$
$= 8 \times 5$
$= 40$
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MCQ 21 Mark
If $A$ is a square matrix such that $A^2 = A,$ then $(I - A)^3 + A$ is equal to:
  • ✓
    $I$
  • B
    $0$
  • C
    $I - A$
  • D
    $I + A$
Answer
Correct option: A.
$I$
$A^2 = A$
$(I + A)^3 +A$
$\Rightarrow I^3 - A^3 - 3I^2A + 3IA^2 + A$
$\Rightarrow I - A^3 - 3A+ 3A + A \ [\therefore A^2 = A]$
$\Rightarrow I - A.A^2 + A$
$\Rightarrow I - A.A + A$
$\Rightarrow I - A + A$
$= I$
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Question 31 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 ?
  1. unit
  2. null
  3. scalar
  4. skew symmetric
Answer
  1. scalar
Solution:
Given 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.
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Question 41 Mark
If $\text{A}=\begin{bmatrix}3&\text{x}-1\\2\text{x}+3&\text{x}+2\end{bmatrix}$ is a symmetric matrix, then x =
  1. 4
  2. 3
  3. -4
  4. -3
Answer
  1. -4
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Question 51 Mark
The transpose of a row matrix is:
  1. zero matrix
  2. diagonal matrix
  3. column matrix
  4. row matrix
Answer
  1. column matrix
Solution:
Transpose of row matrix Let $ \text{A}=\begin{bmatrix}\text{x} &\text{amp; y} &\text{amp; z} \end{bmatrix}$ be a row
matrix $\text{A}^\text{T}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$Clearly $\text{A}^\text{T}$ is a column matrix $\therefore$ Transpose of row.
 
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Question 61 Mark
If $\text{A}=\displaystyle \begin{vmatrix} 2 &\text{amp;}-3 \\ 3 &\text{amp; 2} \end{vmatrix}$ and $\text{B}=\displaystyle \begin{vmatrix} 3 &\text{amp;}-2 \\ 2 &\text{amp; 3} \end{vmatrix}$ then $2\text{ A-B}=$
  1. $\displaystyle \begin{vmatrix} 1 &4 \\ 4 &1\end{vmatrix}$
  2. $\displaystyle \begin{vmatrix} 1 &4 \\ 1 &4\end{vmatrix}$
  3. $\displaystyle \begin{vmatrix} 1 &-4 \\ 4 &1\end{vmatrix}$
  4. $\displaystyle \begin{vmatrix} 4 &1\\ 1 &4\end{vmatrix}$
Answer
  1.  $\displaystyle \begin{vmatrix} 1 &-4 \\ 4 &1\end{vmatrix}$
Solution:
$2\text{ A-B}=\displaystyle \begin{vmatrix} 4 &\text{amp;}-6 \\ 6 &\text{amp; 4} \end{vmatrix}-\displaystyle \begin{vmatrix} 3 &\text{amp;}-2 \\ 2 &\text{amp; 3} \end{vmatrix}=\displaystyle \begin{vmatrix} 1 &\text{amp;}-4\\ 4 &\text{amp; 1} \end{vmatrix}$ 
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Question 71 Mark
The order of $\begin{bmatrix}\text{x}&\text{amp;}\text{ y}&\text{amp; }\text{z}\end{bmatrix}$ $\begin{bmatrix}\text{x} &\text{amp;}\text{ h}&\text{amp;}\text{ g} \\\text{h} &\text{amp;}\text{ b}&\text{amp; }\text{f}\\\text{g} &\text{amp;}\text{ f}&\text{amp; }\text{c} \end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$ is:
  1. $3\times1$
  2. $1\times1$
  3. $1\times3$
  4. $3\times3$
Answer
  1. $1\times1$
Solution:
Let $\text{ABC}=\begin{bmatrix}\text{x}&\text{amp;}\text{ y}&\text{amp; }\text{z}\end{bmatrix}\begin{bmatrix}\text{x} &\text{amp;}\text{ h}&\text{amp;}\text{ g} \\\text{h} &\text{amp;}\text{ b}&\text{amp; }\text{f}\\\text{g} &\text{amp;}\text{ f}&\text{amp; }\text{c} \end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$
Here, the order of $\text{A}$ is  $1\times3$
Order of $\text{B}$ is $3\times3$
Since, matrix multiplication satisfies associative property
$\text{i}.\text{e}. (\text{AB})\text{C} = \text{A}(\text{BC})$
 
Hence, the order of $\text{AB}$ is $1\times3$
Hence, the order of $\text{ABC}$ is $1\times1$
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Question 81 Mark
If A is a square matrix, then AA is a:
  1. Skew-symmetric matrix.
  2. Symmetric matrix.
  3. Diagonal matrix.
  4. None of these.
Answer
  1. None of these.
Solution:
Given: A is a square matrix.
Let $\text{A}=\begin{bmatrix}1&2\\1&0\end{bmatrix}$
$\Rightarrow\text{AA}=\begin{bmatrix}1&2\\1&0\end{bmatrix}\begin{bmatrix}1&2\\1&0\end{bmatrix}=\begin{bmatrix}3&2\\1&2\end{bmatrix}$
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Question 91 Mark
The restriction on n, k and p so that PY + WY will be defined are:
  1. k = 3, p = n
  2. k is arbitary, p = 2
  3. p is arbitary, k = 3
  4. k = 2, p = 3
Answer
  1. k = 3, p = n
Solution:
In this, order of P = p × k Order of  W = n × 3 Order of  Y = 3 × k
Thus, order of PY = p × k, when k = 3
And the order of WY = p × k, where p = n
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Question 101 Mark
If $\text{A} = \begin{bmatrix} 2 &\text{amp; } 3\\ 6 &\text{amp; x} \end{bmatrix}, \text{B} = \begin{bmatrix} 2 &\text{amp; 3}\\ \text{p} &\text{amp; }2 \end{bmatrix}$ and $\text{A} = \text{B}, $ then$\text{p}$ and $ \text{x} $ are:
  1. p = 6, x = 4
  2. p = 3, x = 4
  3. p = 4, x = 3
  4. p = 6, x = 2
Answer
  1. p = 6, x = 2
Solution:
Weve, two matrices will be same, if the given two matrices have same number of rows and columns and each elements of that two matrices are same.
Now equating the given two matrices we get, 6 = p and x = 2.
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Question 111 Mark
A matrix has 16 elements Which of the following can be the order of the matrix:
  1. 1 × 16
  2. 2 × 8
  3. 4 × 4
  4. All of these
Answer
  1. All of these
Solution:
A matrix of mm rows and nn columns has m \times nm×n elements.
On multiplying the rows and columns in the given options, we notice.
that all 1 × 16 = 16, 2 × 8 = 16, 4 × 4 = 16
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Question 121 Mark
If the matrix $\begin{bmatrix}1 &\text{amp;3}& \text{amp}\lambda+2 \\2& \text{amp;4}&\text{amp;8} \\3&\text{amp;5}&\text{amp;}10 \end{bmatrix}$ is singular, then $\lambda=$
  1. -2
  2. 4
  3. 2
  4. -4
Answer
  1. 2
Solution:
A matrix is singular if and only if it has a determinant of 0.
$\begin{bmatrix}1 &\text{amp;3}& \text{amp}\lambda+2 \\2& \text{amp;4}&\text{amp;8} \\3&\text{amp;5}&\text{amp;}10 \end{bmatrix}$
$(40-40)-2(20-24)+(\lambda+2)(10-12)=0$
$2\lambda=4$
$\Rightarrow\lambda=2$
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Question 131 Mark
If $\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}^\text{k}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then the least positive integral value of k is:
  1. 3
  2. 4
  3. 6
  4. 7
Answer
  1. 7
Solution:
$\text{A}=\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\text{A}\times\text{A}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos^2\frac{2\pi}{7}-\sin^2\frac{2\pi}{7}&\Big(-2\cos\frac{2\pi}{7}-\sin\frac{2\pi}{7}\Big)\\2\cos\frac{2\pi}{7}\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}-\sin^2\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos\frac{4\pi}{7}&-\sin\frac{4\pi}{7}\\\sin\frac{4\pi}{7}&\cos\frac{4\pi}{7}\end{bmatrix}$$\begin{bmatrix}\because\ \cos^2\theta-\sin^2\theta=\cos2\theta\\2\sin\theta\cos\theta=\sin2\theta\end{bmatrix}$
$\Rightarrow\text{A}^3=\text{A}^2\times\text{A}$
$\Rightarrow\text{A}^3=\begin{bmatrix}\cos\frac{4\pi}{7}&-\sin\frac{4\pi}{7}\\\sin\frac{4\pi}{7}&\cos\frac{4\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}\Big(\cos\frac{4\pi}{7}\cos\frac{2\pi}{7}-\sin\frac{4\pi}{7}\sin\frac{2\pi}{7}\Big)&\Big(-\cos\frac{4\pi}{7}\sin\frac{2\pi}{7}-\sin\frac{4\pi}{7}\cos\frac{2\pi}{7}\Big)\\\Big(\sin\frac{4\pi}{7}\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}\sin\frac{2\pi}{7}\Big)&\Big(-\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{2\pi}{7}\Big)\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}$$\begin{bmatrix}\because\ \cos\text{(A+B)}=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}\\\sin\text{(A+B)}=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\end{bmatrix}$
Now we check if the pattern is same for k = 6.
Here,
$\text{A}^6=\text{A}^3.\text{A}^3$
$\Rightarrow\text{A}^6=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^6=\begin{bmatrix}\cos\frac{12\pi}{7}&-\sin\frac{12\pi}{7}\\\sin\frac{12\pi}{7}&\cos\frac{12\pi}{7}\end{bmatrix}$
Now, we check if the pattern is same for k = 7.
Here,
$\text{A}^7=\text{A}^6\times\text{A}$
$\Rightarrow\text{A}^7=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^7=\begin{bmatrix}\cos\frac{14\pi}{7}&-\sin\frac{14\pi}{7}\\\sin\frac{14\pi}{7}&\cos\frac{14\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^7=\begin{bmatrix}\cos2\pi&-\sin2\pi\\\sin2\pi&\cos2\pi\end{bmatrix}$$\begin{bmatrix}\because\ \frac{14\pi}{7}=2\pi\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
So, the least positive integral value of k is 7.
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MCQ 141 Mark
If $A = [a_{ij}]$ is a square matrix of even order such that $a_{ij} = i^2 - j^2,$ then
  • A
    $A$ is a skew$-$symmetric matrix and $|A| = 0$
  • B
    $A$ is symmetric matrix and $|A|$ is a square
  • C
    $A$ is symmetric matrix and $|A| = 0$
  • ✓
    None of these.
Answer
Correct option: D.
None of these.
Let $\text{A}=\begin{bmatrix}0&-3\\3&0\end{bmatrix}$$\big[\because\ \text{a}_\text{ij} = \text{i}^2 -\text{j}^2\big]$ 
$|\text{A}|=0-(-9)=9\neq0$
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Question 151 Mark
Which of the given value of x and y make the following pair of matrices equal
$\begin{bmatrix}3\text{x}+7&5\\ \text{y}+1&2-3\text{x}\end {bmatrix},\ \begin{bmatrix}0& \text {y}-2\\8&4 \end{bmatrix}$
  1. $x = \frac{-1}{3}, y = 7$
  2.  Not possible to find
  3. $y = 7, x = \frac{-2}{3}$
  4. $x = \frac{-1}{3}, y = \frac{-2}{3}$
Answer
  1. Not possible to find
We are given that
$\begin{bmatrix}3\text{x}+7&5\\ \text{y}+1&2-3\text{x}\end {bmatrix},\ \begin{bmatrix}0& \text {y}-2\\8&4 \end{bmatrix}$
By defination of equality of matrices.
3x + 7 = 0, 5 = y - 2, y + 1 = 8, 2 - 3x = 4
$\therefore \text x = \frac{7}{3},\ \text y = 7,\ \text x = -\frac {2}{3}$
$\therefore$ (b) is correct answer.
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Question 161 Mark
If $ \displaystyle \begin{vmatrix}\text{a} &\text{amp; }\text{b} &\text{amp; 0}\\ 0 &\text{amp; a} &\text{amp; b}\\\text{b}&\text{amp; a}&\text{amp; 0}\end{vmatrix}=0,$ then the order is:
  1. 3 × 3
  2. 2 × 3
  3. 2 × 2
  4. None of these
Answer
  1. 3 × 3
Solution:
There are 3 rows and 3 columns.Therefore the order of the matrix is 3 × 3.
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Question 171 Mark
If $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&\cot^{-1}(\pi\text{x})\end{bmatrix},$ $\text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&-\tan^{-1}(\pi\text{x})\end{bmatrix},$ then A - B is equal to:
  1. I
  2. 0
  3. 2I
  4. $\frac{1}{2}\text{I}$
Answer
  1. $\frac{1}{2}\text{I}$
Solution:
Given $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&\cot^{-1}(\pi\text{x})\end{bmatrix}$ $\text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&-\tan^{-1}(\pi\text{x})\end{bmatrix}$
$\text{A}-\text{B}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\text{x}\pi)+\cos^{-1}(\pi\text{x})&0\\0&\cot^{-1}(\pi\text{x})+\tan^{-1}(\pi\text{x})\end{bmatrix}$
$=\frac{1}{\pi}\begin{bmatrix}\frac{\pi}{2}&0\\0&\frac{\pi}{2}\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1&0\\0&1\end{bmatrix}=\frac{1}{2}\text{I}$
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Question 181 Mark
Out of the given matrices, choose that matrix which is a scalar matrix:
  1. $\begin{bmatrix}0&0\\0&0\end{bmatrix}$
  2. $\begin{bmatrix}0&0&0\\0&0&0\end{bmatrix}$
  3. $\begin{bmatrix}0&0\\0&0\\0&0\end{bmatrix}$
  4. $\begin{bmatrix}0\\0\\0\end{bmatrix}$
Answer
  1. $\begin{bmatrix}0&0\\0&0\end{bmatrix}$
Solution:
A diagonal matrix with all diagonal elements are equal is a scalar matrix.
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Question 191 Mark
If the matrix AB is zero, then:
  1. It is not necessary that either A = 0 or, B = 0
  2. A = 0 or B = 0
  3. A = 0 and B = 0
  4. All the above statements are wrong
Answer
  1. It is not necessary that either A = 0 or, B = 0
Solution:
Let $\text{A}=\begin{bmatrix}0&2\\0&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$\therefore\ \text{AB}=\begin{bmatrix}0&2\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
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MCQ 201 Mark
Let $\text{A}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix},$ then $A^n$ is equal to$:$
  • A
    $\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}^\text{n}&0\\0&0&\text{a}\end{bmatrix}$
  • B
    $\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}$
  • ✓
    $\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}^\text{n}&0\\0&0&\text{a}^\text{n}\end{bmatrix}$
  • D
    $\begin{bmatrix}\text{na}&0&0\\0&\text{na}&0\\0&0&\text{na}\end{bmatrix}$
Answer
Correct option: C.
$\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}^\text{n}&0\\0&0&\text{a}^\text{n}\end{bmatrix}$
$\text{A}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}=\text{a}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\text{A}^\text{n}=\text{a}^\text{n}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}^\text{n}&0\\0&0&\text{a}^\text{n}\end{bmatrix}$
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Question 211 Mark
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 and y is:
  1. x = 3, y = 1
  2. x = 2, y = 3
  3. x = 2, y = 4
  4. x = 3, y = 3
Answer
  1. x = 2, y = 4
Solution:
$\begin{bmatrix}2\text{x}+\text{y}&4\text{x}\\5\text{x}-7&4\text{x}\end{bmatrix}=\begin{bmatrix}7&7\text{y}-13\\\text{y}&\text{x}+6\end{bmatrix},$
Equating the terms, we get
4x = x + 6
⇒ x = 2
And
2x + y = 7
⇒ y = 3
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MCQ 221 Mark
If $\text{A}=\begin{bmatrix}\text{n}&0&0\\0&\text{n}&0\\0&0&\text{n}\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3\end{bmatrix},$ then $AB$ is equal to$:$
  • A
    $B$
  • ✓
    $n^B$
  • C
    $B^n$
  • D
    $A + B$
Answer
Correct option: B.
$n^B$
Here,
$\text{A}=\begin{bmatrix}\text{n}&0&0\\0&\text{n}&0\\0&0&\text{n}\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3\end{bmatrix}$ 
$\therefore\ \text{AB}=\begin{bmatrix}\text{n}&0&0\\0&\text{n}&0\\0&0&\text{n}\end{bmatrix}\begin{bmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}\text{na}_1&\text{na}_2&\text{na}_3\\\text{nb}_1&\text{nb}_2&\text{nb}_3\\\text{nc}_1&\text{nc}_2&\text{nc}_3\end{bmatrix}$
$\Rightarrow\text{AB}=\text{n}\begin{bmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3\end{bmatrix}$
$\Rightarrow\text{AB}=\text{n}\text{B}$
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Question 231 Mark
If $\text{A}=\begin{bmatrix}2&-1&3\\-4&5&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&3\\4&-2\\1&5\end{bmatrix},$ then:
  1. Only AB is defined.
  2. Only BA is defined.
  3. AB and BA both are defined.
  4. AB and BA both are not defined.
Answer
  1. AB and BA both are defined.
Soluton:
Given: $\text{A} = \begin{bmatrix}2& -1 &3\\-4 & 5 & 1 \end{bmatrix} \ \text{B} = \begin{bmatrix}2& 3\\4 & -2 \\1 & 5 \end{bmatrix}$
$\text{AB} = \begin{bmatrix}2& -1 &3\\-4 & 5 & 1 \end{bmatrix} \begin{bmatrix}2& 3 \\4 & -2 \\1 & 5 \end{bmatrix}$
$ \begin{bmatrix}3 & 23\\13 & -17 \end{bmatrix}$
So, AB is defined as of columns in A is equal to number of rows in B.
$\text{BA} = \begin{bmatrix}2&3\\4 &-2\\1 & 5 \end{bmatrix} \begin{bmatrix}2& -1 &3\\-4 & 5 & 1 \end{bmatrix}$
$ = \begin{bmatrix}-8& 13 &9\\16 & -14 & 10\\-18 & 24 & 8 \end{bmatrix}$
So, BA is also defined of columns in B is equal to number of rows in A.
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Question 241 Mark
The possible dimension of a matrix consisting 27 elements is 4.Reason: The number of ways of expressing 27 as a product of two positive integers is 4.
  1. Both Assertion & Reason are individually correct & Reason is correct explanation of Assertion,
  2. Both Assertion & Reason are individually true but Reason is Not the correct explanation of Assertion.
  3. Assertion is correct but Reason is incorrect.
  4. Assertion is incorrect but Reason is correct.
Answer
  1. Assertion is correct but Reason is incorrect.
Solution:
27 = 1 × 27 and 3 × 9
Thus the number of ways of expressing 27 as a product of two numbers is only 2 Thus the Reason is false.
Also the possible dimensions of a matrix having 27 elements are  27, 27 × 1, 3 × 9, 9 × 3
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MCQ 251 Mark
If $\text{A}=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}$ is such that $A^2 = I,$ then$:$
  • A
    $1+\alpha^2+\beta\gamma=0$
  • B
    $1-\alpha^2+\beta\gamma=0$
  • ✓
    $1-\alpha^2-\beta\gamma=0$
  • D
    $1+\alpha^2-\beta\gamma=0$
Answer
Correct option: C.
$1-\alpha^2-\beta\gamma=0$
Given $\text{A}=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}$ and $A^2 = I,$ then
$\text{A}^2=\text{I}$
$\Rightarrow\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\alpha^2+\beta\gamma&0\\0&\alpha^2+\beta\gamma\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\alpha^2+\beta\gamma=1$
$\Rightarrow1-\alpha^2-\beta\gamma=0$
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Question 261 Mark
If matrix A is of order p × q and matrix B is of order r × s then  A − B will exist if:
  1. p = q
  2. p = r, q = s
  3. p = q, r = s
  4. p = s, q = r
Answer
  1. p = r, q = s
Solution:
If matrix A is of order p × q and matrix B is of order r × s then  A − B will exist if order of  A and B is same.
Therefore, p = r, q = s
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Question 271 Mark
The matrix $\text{A}=\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$ is a:
  1. Diagonal matrix.
  2. Symmetric matrix.
  3. Skew-symmetric matrix.
  4. Scalar matrix.
Answer
  1. Skew-symmetric matrix.
Solution:
Given $\text{A}=\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}0&5&-8\\-5&0&-12\\8&12&0\end{bmatrix}$
$\Rightarrow\text{A}=-\text{A}^\text{T}$
So, A is skew-symmetric matrix.
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Question 281 Mark
The possible number of different orders that a matrix can have when it has 24 elements, is:
  1. 8
  2. 16
  3. 4
  4. None of these
Answer
  1. 8
Solution:
Possible order of matrices 24 × 1, 1 × 24, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, 6 × 4
So, the number of possible matrices with 24 elements is 8.
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Question 291 Mark
If A and B are non - zero square matrices of the same order such that  AB = 0, then
  1. adj A = 0 or adj B = 0
  2. adj A = 0 and adj B = 0
  3. |A| = 0 or |B| = 0
  4. None of these
Answer
  1. |A| = 0 or |B| = 0
Soluton:
From the properties of the matrices, if A, B are non - zero square matrices of same order such that AB = 0 then the either of the matrices must be singular matrix.
A singular matrix is a matrix whose determinant is zero.
$\therefore$ |A| = 0 or |B| = 0
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Question 301 Mark
If $\displaystyle \begin{vmatrix}\text{x} &\text{amp; } 1 \\ \text{y} &\text{amp; } 2 \end{vmatrix}-\displaystyle \begin{vmatrix}\text{y} &\text{amp; } 1 \\ 8&\text{amp; } 0\end{vmatrix}=\displaystyle \begin{vmatrix}2 &\text{amp; } 0 \\ \text{-x}&\text{amp; } 2\end{vmatrix}$ then the values of x and y respectively are:
  1. 5 and 1 
  2. 5 and 3
  3. 5 and 2
  4. 3 and 4
Answer
  1. 5 and 3
Solution:
x - y = 2
y - 8 = -x
Solving we get x = 5 and y = 3
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Question 311 Mark
If A is 3×4 matrix and B is a matrix such that A'B and BA' are both defined. Then, B is of the type:
  1. 3×4
  2. 3×3
  3. 4×4
  4. 4×3
Answer
  1. 3×4
Solution:
The order of A is 3×4. So, the order of A' is 4×3.
Now, both A'B and BA' are defined. So, the number of columns in A' should be equal to the number of rows in B for A'B.
Also, the number of columns in B should be equal to number of rows in A' for BA'.
Hence, the order of matrix B is 3×4.
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Question 321 Mark
If A and B are square matrices such that AB = I and BA = I, then B is:
  1. Unit matrix
  2. Null matrix
  3. Multiplicative inverse matrix of A
  4. -A
Answer
  1. Multiplicative inverse matrix of A
Solution:
$\text{AB}=\begin{bmatrix}\text{I}&\text{amp; }\end{bmatrix}\text{BA}=\text{I}$ is the multiplicative inverse of A.
Hence, the answer is multiplicative inverse matrix of A.
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Question 331 Mark
If $\text{A}=\begin{bmatrix}0&2\\3&-4\end{bmatrix}$ and $\text{kA}=\begin{bmatrix}0&3\text{a}\\2\text{b}&24\end{bmatrix},$ then the values of k, a, b, are respectively
  1. -6, -12, -18
  2. -6, 4, 9
  3. -6, -4, -9
  4. -6, 12, 18
Answer
  1. -6, -4, -9
Solution:
$\text{A}=\begin{bmatrix}0&2\\3&-4\end{bmatrix}$
$\text{kA}=\begin{bmatrix}0&3\text{a}\\2\text{b}&24\end{bmatrix}$
$\Rightarrow\begin{bmatrix}0&2\text{k}\\3\text{k}&-4\text{k}\end{bmatrix}=\begin{bmatrix}0&3\text{a}\\2\text{b}&24\end{bmatrix}$
$\Rightarrow-4\text{k}=24$
$\Rightarrow\text{k}=-6$
$2\text{k}=3\text {a}$
$\Rightarrow\text{a}=-4$
$3\text{k}=2\text{b}$
$\Rightarrow\text{b}=-9$
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Question 341 Mark
The order of a matrix $\begin{bmatrix}2&\text{amp};5 &\text{amp};7 \end{bmatrix}$ is:
  1. 3 × 3
  2. 1 × 1
  3. 3 × 1
  4. 1 × 3
Answer
  1. 1 × 3
Solution:
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
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Question 351 Mark
Which of the following is correct:
  1. Determinant is a square matrix
  2. Determinant is a number associated to a matrix
  3. Determinant is a number associated to a square matrix
  4. None of these
Answer
  1. Determinant is a number associated to a square matrix
Solution:
Determinant is defined only for a square matrix.and its denotes the value of that square matrix.
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Question 361 Mark
Which of the given values of X and Y make the following pairs of matrices equal? $\begin{bmatrix}3\text{x}+7&5\\\text{y}+1&2-3\text{x}\end{bmatrix},\begin{bmatrix}0&\text{y}-2\\8&4\end{bmatrix}$
  1. $\text{x}=-\frac{1}{3},\text{y}=7$
  2. $\text{y}=7,\text{x}=-\frac{2}{3}$
  3. $\text{x}=-\frac{1}{3},\text{y}=-\frac{2}{5}$
  4. $\text{Not possible to find}$
Answer
  1. $\text{Not possible to find}$
Solution:
$\begin{bmatrix}3\text{x}+7&5\\\text{y}+1&2-3\text{x}\end{bmatrix}=\begin{bmatrix}0&\text{y}-2\\8&4\end{bmatrix}$
$\Rightarrow3\text{x}+7=0$
$\Rightarrow\text{x}=\frac{-7}{3}$
$5=\text{y}-2$
$\Rightarrow\text{y}=7$
$\text{y}+1=8$
$\Rightarrow\text{y}=7$
$2-3\text{x}=4$
$\Rightarrow\text{x}=\frac{-2}{3}$
We are getting two values of x.
So, it is not possible to find.
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Question 371 Mark
If m[-3 amp; 4] + n[-3 amp; 4] = [10 amp;-11], then 3m + 7n =
  1. 3
  2. 5
  3. 10
  4. 1
Answer
  1. 1
Solution:
Given m[-3 + 4] + n[4 - 3] = [10 - 11]
⇒ -3m + 4n = 10 and 4m - 3n = -11
by solving we get m= -2 and n=1
$\therefore$ 3m + 7n = -6 + 7 = 1
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MCQ 381 Mark
If $A$ is square matrix of order $3,$ then $∣​\text{adj}(\text{adj}\ \text{A}^2)​∣$
  • A
    $|A|^2$
  • B
    $|A|^4$
  • ✓
    $|A|^8$
  • D
    $|A|^{16}$
Answer
Correct option: 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 $
$=\begin{vmatrix} \text{A}\end{vmatrix}^8$
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MCQ 391 Mark
If $\text{A}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix},$ then $A^T + A = I_2,$ 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
    None of these
Answer
Correct 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,
$\text{A}^\text{T}+\text{A}=\text{I}_2$
$\Rightarrow\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}+\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\cos\theta&0\\0&2\cos\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow2\cos\theta=1$
$\Rightarrow\cos\theta=\frac{1}{2}$
$\Rightarrow\cos\theta=\cos\frac{\pi}{3}$
$\Rightarrow\theta=2\text{n}\pi\pm\frac{\pi}{3}$ $(\text{n}\in\text{Z})$
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MCQ 401 Mark
If $A$ and $B$ are matrices of the same order, then $AB^T - BA^T$ is a$:$
  • ✓
    Skew$-$symmetric matrix.
  • B
    Null matrix.
  • C
    Unit matrix.
  • D
    Symmetric matrix.
Answer
Correct option: A.
Skew$-$symmetric matrix.
$(AB^T - BA^T)^T = (AB^T)^T - (BA^T)^T$
$= BA^T - AB^T$
$= -(AB^T - BA^T)$
Therefore, $AB^T - BA^T$ is a skew$-$symmetric matrix.
Hence, the correct option is $(a).$
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Question 411 Mark
If A is 3 × 4 matrix and B is matrix such that AB and BA are both defined, then B is of the type:
  1. 3 × 4
  2. 3 × 3
  3. 4 × 4
  4. 4 × 3
Answer
  1. 3 × 4
Solution:
Given that matrix A is 3 x 4.
Let the B matrix be P x Q.
$\therefore$ A is 4 x 3.
Since AB is defined, so number of columns of A must be equal to number of rows of B, therefore, P = 3.
Also, BA is defined, so the number of columns of B must be equal to number of rows of A, then Q = 4.
Therefore, matrix B is 3 x 4.
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Question 421 Mark
Choose the correct answer from the given four options.
If A and B are matrices of same order, then (AB′ – BA′) is a:
  1. Skew symmetric matrix.
  2. Null matrix.
  3. Symmetric matrix.
  4. Unit matrix.
Answer
  1. Skew symmetric matrix.
Solution:
We 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.
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MCQ 431 Mark
$B = A + A^{2 }+ A^3 + A^{4 }$ If order of $A$ is $3$ then order of $B$ is:
  • ✓
    $3$
  • B
    $6$
  • C
    $2$
  • D
    $9$
Answer
Correct option: A.
$3$
The order of matrix doesnt change when the operation are done on it So The order of $B$ remains same as the order of $A$
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Question 441 Mark
The number of different possible orders of matrices having 18 identical elements is:
  1. 3
  2. 1
  3. 6
  4. 4
Answer
  1. 6
Solution:
Let 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.
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MCQ 451 Mark
Choose the correct answer from the given four options. On using elementary column operations $C_2 \rightarrow C_2 – 2C_1$ 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
    $\begin{bmatrix}1&-5\\0&4\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\-0&2\end{bmatrix}$
  • ✓
    $\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-3\\0&1\end{bmatrix}\begin{bmatrix}3&1\\-2&4\end{bmatrix}$
  • D
    $\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$
Answer
Correct option: C.
$\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-3\\0&1\end{bmatrix}\begin{bmatrix}3&1\\-2&4\end{bmatrix}$
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 $C_2 \rightarrow C_2-2C_1;$ $\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$
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 $\text{RHS}.$
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Question 461 Mark
The order of the matrix $ \displaystyle \left[ \begin{matrix} 1 &\text{amp; }2 \\ 3&\text{amp; } 4 \end{matrix} \right]$ is:
  1. 2 × 2
  2. 4 × 1
  3. 1 × 4
  4. None of these
Answer
  1. 2 × 2
Solution:
If a matrix has mm rows and n columns then its order is m × n Clearly in the given matrix, number of rows and columns are each 2Hence its order is 2 × 2.
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Question 471 Mark
What is the order of the product- $\begin{bmatrix}\text{x}&\text{amp;}\text{ y}&\text{amp;}\text{ z}\end{bmatrix}\begin{bmatrix}\text{a} &\text{amp;}\text{ h}&\text{amp;}\text{ g} \\\text{h} &\text{amp;}\text{ b}&\text{amp; }\text{f}\\\text{g} &\text{amp;}\text{ f}&\text{amp; }\text{c} \end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$ is:
  1. $3\times1$
  2. $1\times1$
  3. $1\times3$
  4. $3\times3$
Answer
  1. $1\times1$
Solution:
If two matrix of order $\text{x}\times\text{m}$ and $\text{m}\times\text{n}$ are multiplied then the order of resultant matrix is $\text{x}\times\text{n}$
Now in the given equation going from left, matrices of order $1\times3$ and $3\times3$ are multiplied.So, the order of resultant matrix is  $1\times3 $ And now this is multiplied by matrix of order $3\times1.$
This will give resultant matrix of order $1\times1$
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Question 481 Mark
The transpose of a square matrix is a?
  1. rectangular matrix
  2. diagonal matrix
  3. square matrix
  4. scaler matrix
Answer
  1. square matrix
Solution:
The transpose of square matrix is a new square matrix whose rows are.
the columns of original. this makes the columns the new square matrix row of the original. Answer is square matrix.
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Question 491 Mark
If $\text{A} =\displaystyle \begin{bmatrix} -1 &\text{amp; } 0 &\text{amp; }0 \\ 0 &\text{amp; }\text{x} &\text{amp; } 0 \\ 0 &\text{amp; } 0 &\text{amp; }\text{m} \end{bmatrix}$is a scalar matrix then $\text{x}+\text{m}=$
  1. 0
  2. -1
  3. -2
  4. -3
Answer
  1. -2
Solution:
A scalar matrix has all the elements of the diagonals same.For example:$ \begin{bmatrix} 3 &\text{amp; } 0\\ 0&\text{amp; } 3 \end{bmatrix}$
In our case A is given to be a scalar matrix hence all the diagonal elements must be same.
So, $\text{x} = \text{m} = -1$
And $\text{x}+\text{m} = -1 -1 = -2$
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Question 501 Mark
If $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{J}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix},$ then B equals:
  1. $\text{I}\cos\theta+\text{J}\sin\theta$
  2. $\text{I}\sin\theta+\text{J}\cos\theta$
  3. $\text{I}\cos\theta-\text{J}\sin\theta$
  4. $-\text{I}\cos\theta+\text{J}\sin\theta$
Answer
  1. $\text{I}\cos\theta+\text{J}\sin\theta$
Solution:
Here,
$\text{I}\cos\theta+\text{J}\sin\theta$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}\cos\theta+\begin{bmatrix}0&1\\-1&0\end{bmatrix}\sin\theta$
$=\begin{bmatrix}\cos\theta&0\\0&\cos\theta\end{bmatrix}+\begin{bmatrix}0&\sin\theta\\-\sin\theta&0\end{bmatrix}$
$=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}=\text{B}$
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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip