Matrices
36 Board Maths previous year questions on Matrices — options free on every question; 4 include the answer & explanation free, the rest unlock with PYQ Pass.
If a matrix has 36 elements, the number of possible orders it can have, is :
9
Factor pairs of 36:
\((1,36),(2,18),(3,12),(4,9),(6,6),(9,4),(12,3),(18,2),(36,1)\)
Number of possible orders = 9.
Find the matrix \({\mathrm{A}}^{2}\), where \(\mathrm{A}=\left[{a}_{ij}\right]\) is a \(2\times 2\) matrix whose elements are given by \({a}_{ij}=maximum(\mathrm{i},\mathrm{j})-minimum(\mathrm{i},\mathrm{j})\) :
\(\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]\)
Given: \({a}_{ij}=\max (i,j)-\min (i,j)\)
So,
\({a}_{11}=\max (1,1)-\min (1,1)=1-1=0,\\ {a}_{12}=\max (1,2)-\min (1,2)=2-1=1,\\ {a}_{21}=\max (2,1)-\min (2,1)=2-1=1,\\ {a}_{22}=\max (2,2)-\min (2,2)=2-2=0\)
The matrix \(A\) is: \(A=\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]\)
Hence,
\({A}^{2}=A\times A\\ =\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]\times \left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]\\ =\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]\)
If \(\left[\begin{array}{cc}x+y & 2 \\ 5 & x y\end{array}\right]=\left[\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right]\), then the value of \(\left(\frac{24}{x}+\frac{24}{y}\right)\) is :
18
Given: \(\left[\begin{array}{cc}x+y & 2 \\ 5 & x y\end{array}\right]=\left[\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right]\)
Corresponding elements must be equal,
\(x+y=6\), \(xy=8\)
Now, \(\frac{24}{x}+\frac{24}{y}=24\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=24\left(\frac{x+y}{xy}\right)\\ =24\times \frac{6}{8}\\ =18\)
If \(\left[\begin{array}{cc}x+y & 2 \\ 5 & x y\end{array}\right]=\left[\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right]\), then the value of \(\left(\frac{24}{x}+\frac{24}{y}\right)\) is :
18
From equality of matrices:
\(x+y=6\)
\(xy=8\)
\(\frac{24}{x}+\frac{24}{y}=24\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\)
\(\frac{x+y}{xy}=\frac{6}{8}=\frac{3}{4}\)
\(\Rightarrow 24\left(\frac{1}{x}+\frac{1}{y}\right)=24\cdot \frac{3}{4}=18\)
If the sum of all the elements of a \(3 \times 3\) scalar matrix is \(9\) , then the product of all its elements is :
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Let A and B be two square matrices of order 3 such that \(|\mathrm{A}|=3\) and \(|\mathrm{B}|=2\).
Then \(\left|{\mathrm{A}}^{\mathrm{T}}\mathrm{A}(adj(2\mathrm{A}){)}^{-1}(adj(4\mathrm{B}))(adj(\mathrm{AB}){)}^{-1}{\mathrm{AA}}^{\mathrm{T}}\right|\)
is equal to :
[JEE Main 2024, 05 Apr (Shift 1)]
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If \(F(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]\) and \([F(x)]^2=F(k x)\), then the value of \(k\) is :
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If \(F(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]\) and \([F(x)]^2=F(k x)\), then the value of \(k\) is :
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If a matrix has \(36\) elements, the number of possible orders it can have, is :
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If the sum of all the elements of a \(3 \times 3\) scalar matrix is 9 , then the product of all its elements is :
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If \(\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]\) be a \(3 \times 3\) matrix, where \(\mathrm{a}_{\mathrm{ij}}=\mathrm{i}-3 \mathrm{j}\), then which of the following is false?
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Which of the following can be both a symmetric and skew-symmetric matrix ?
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If \(\left[\begin{matrix}x & 2 & 0\end{matrix}\right]\left[\begin{matrix}5 \\ -1 \\ x\end{matrix}\right]=\left[\begin{matrix}3 & 1\end{matrix}\right]\left[\begin{matrix}-2 \\ x\end{matrix}\right]\) then value of \(x\) is :
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If \(\left[\begin{matrix}x & 2 & 0\end{matrix}\right]\left[\begin{matrix}5 \\ -1 \\ x\end{matrix}\right]=\left[\begin{matrix}3 & 1\end{matrix}\right]\left[\begin{matrix}-2 \\ x\end{matrix}\right]\) then value of \(x\) is :
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If \(A\) and \(B\) are square matrices of order \(m\) such that \({A}^{2}-{B}^{2}=(A-B)(A+B)\), then which of the following is always correct?
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Which of the following can be both a symmetric and skew-symmetric matrix ?
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Four friends Abhay, Bina, Chhaya and Devesh were asked to simplify \(4 \mathrm{AB}+3(\mathrm{AB}+\mathrm{BA})-4 \mathrm{BA}\), where \(A\) and \(B\) are both matrices of order \(2 \times 2\). It is known that \(A \neq B \neq I\) and \(A^{-1} \neq B\).
Their answers are given as :
Abhay : \(6 \mathrm{AB}\)
Bina : \(7 \mathrm{AB}-\mathrm{BA}\)
Chhaya: \(8 \mathrm{AB}\)
Devesh : \(7 \mathrm{BA}-\mathrm{AB}\)
Who answered it correctly ?
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Four friends Abhay, Bina, Chhaya and Devesh were asked to simplify \(4 \mathrm{AB}+3(\mathrm{AB}+\mathrm{BA})-4 \mathrm{BA}\), where A and B are both matrices of order \(2 \times 2\). It is known that \(A \neq B \neq I\) and \(A^{-1} \neq B\).
Their answers are given as :
Abhay :\(6 \mathrm{AB}\)
Bina : \(7 \mathrm{AB}-\mathrm{BA}\)
Chhaya: \(8 \mathrm{AB}\)
Devesh : \(7 \mathrm{BA}-\mathrm{AB}\)
Who answered it correctly ?
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Find the matrix \({\mathrm{A}}^{2}\), where \(\mathrm{A}=\left[{\mathrm{a}}_{\mathrm{ij}}\right]\) is a \(2\times 2\) matrix whose elements are given by \({\mathrm{a}}_{\mathrm{ij}}=maximum(\mathrm{i},\mathrm{j})-minimum(\mathrm{i},\mathrm{j})\) :
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If \(A\) and \(B\) are square matrices of order \(m\) such that \({A}^{2}-{B}^{2}=(A-B)(A+B)\), then which of the following is always correct?
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If \(\mathrm{A}=\left[\begin{matrix}\mathrm{a} & \mathrm{c} & -1 \\ \mathrm{b} & 0 & 5 \\ 1 & -5 & 0\end{matrix}\right]\) is a skew-symmetric matrix, then the value of \(2\mathrm{a}-(\mathrm{b}+\mathrm{c})\) is :
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If \(A=\left[\begin{matrix}1 & 12 & 4y \\ 6x & 5 & 2x \\ 8x & 4 & 6\end{matrix}\right]\) is a symmetric matrix, then \((2x+y)\) is
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If A and B are two non-zero square matrices of same order such that \((A+B)^2=A^2+B^2\), then :
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If \(\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]\) be a \(3 \times 3\) matrix, where \(\mathrm{a}_{\mathrm{ij}}=\mathrm{i}-3 \mathrm{j}\), then which of the following is false?
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If \(A=\left[\begin{matrix}1 & 12 & 4y \\ 6x & 5 & 2x \\ 8x & 4 & 6\end{matrix}\right]\) is a symmetric matrix, then \((2x+y)\) is
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If \(\mathrm{A}=\left[\begin{matrix}\mathrm{a} & \mathrm{c} & -1 \\ \mathrm{b} & 0 & 5 \\ 1 & -5 & 0\end{matrix}\right]\) is a skew-symmetric matrix, then the value of \(2\mathrm{a}-(\mathrm{b}+\mathrm{c})\) is :
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If \(A\) and \(B\) are two non-zero square matrices of same order such that \((A+B)^2=A^2+B^2\), then :
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If \(2A+B+X=0\), where \(A=\left[\begin{matrix}-1 & 2 \\ 3 & 4\end{matrix}\right]\) and \(B=\left[\begin{matrix}3 & -2 \\ 1 & 5\end{matrix}\right]\) then \(X=\)
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If A and B are symmetric matrices of same Border, then AB-BA is a :
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(iii) The number of all possible matrices of order \(2\times 2\) with entry 0 or 1 is -
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Assertion (A) : A = diag [ \(\left.\begin{array}{lll}3 & 5 & 2\end{array}\right]\) is a scalar matrix of order \(3 \times 3\).
Reason (R) : If a diagonal matrix has all non-zero elements equal, it is known as a scalar matrix.
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If \(A=\left[\begin{matrix}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{matrix}\right]\) and \(A+{A}^{'}=I\), then the value of \(\alpha\) is
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Assertion (A) : For any symmetric matrix \(\mathrm{A}, \mathrm{B}^{\prime} \mathrm{AB}\) is a skew-symmetric matrix.
Reason \((R)\) : A square matrix P is skew-symmetric if \(\mathrm{P}^{\prime}=-\mathrm{P}\).
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Assertion (A) : For any symmetric matrix \(\mathrm{A}, \mathrm{B}^{\prime} \mathrm{AB}\) is a skew-symmetric matrix.
Reason \((R)\) : A square matrix \(P\) is skew-symmetric if \(\mathrm{P}^{\prime}=-\mathrm{P}\).
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Assertion (A) : \(A=\operatorname{diag}[3,5,2]\) is a scalar matrix of order \(3 \times 3\).
Reason (R) : If a diagonal matrix has all non-zero elements equal, it is known as a scalar matrix.
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If the order of a matrix is \(m\times n\),
then the number of elements in it are -
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