CUET UG Mathematics Applied Matrices Previous Year Questions (PYQs)
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If $A$ and B are symmetric matrices of the same order, then AB-BA is a:
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Solution
If $[A]_{3\times 2}[B]_{x\times y}=[C]_{3\times 1}$, then:
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Solution
If $A$ is a square matrix and $I$ is an identity matrix such that $A^{2}=A$, then $A(I-2A)^{3}+2A^{3}$ is equal to:
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Solution
If $A=\begin{bmatrix}2 & -3 \\ 3 & 5\end{bmatrix}$, then which of the following statements are correct?
A. A is a square matrix
B. $A^{-1}$ exists
C. A is a symmetric matrix
D. $|A|=19$
E. A is a null matrix
Choose the correct answer from the options given below.
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Solution
Given
$A=\begin{bmatrix}2 & -3 \ 3 & 5\end{bmatrix}$
1.Order of matrix = $2\times 2$
So, A is a square matrix → True
2.Determinant:
$|A|=(2)(5)-(-3)(3)$
$=10+9=19$
Since $|A|\neq 0$, therefore $A^{-1}$ exists → True
3.Transpose:
$A^T=\begin{bmatrix}2 & 3 \ -3 & 5\end{bmatrix}$
Since $A^T\neq A$, so A is not symmetric → False
4. $|A|=19$ → True
5. A is clearly not a null matrix → False
Correct statements: A, B, D
Let A and B be two non-zero square matrices and AB and BA both are defined. It means
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Solution
Let A be of order $m \times m$ and B be of order $n \times n$.
For AB to be defined: number of columns of A = number of rows of B
⇒ $m = n$
For BA to be defined: number of columns of B = number of rows of A
⇒ $n = m$
Hence $m = n$, so both matrices must have same order.
The number of all possible matrices of order $2 \times 2$ with each entry $0$ or $1$ is:
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Solution
A $2 \times 2$ matrix has $4$ entries.
Each entry can be either $0$ or $1$ (2 choices).
Total possible matrices
$=2^4=16$
If $A=\begin{bmatrix}x & y & z \\ 2 & u & v \\ -1 & 6 & w\end{bmatrix}$ is skew symmetric matrix, then value of $x^2+y^2+z^2+u^2+v^2+w^2$ is:
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Solution
For skew symmetric matrix,
$A^T=-A$
So diagonal elements must be zero:
$x=0,; u=0,; w=0$
Now compare off-diagonal elements:
From $(1,2)$ and $(2,1)$:
$y=-2$
From $(1,3)$ and $(3,1)$:
$z=1$
From $(2,3)$ and $(3,2)$:
$v=-6$
Now compute:
$x^2+y^2+z^2+u^2+v^2+w^2$
$=0^2+(-2)^2+1^2+0^2+(-6)^2+0^2$
$=4+1+36$
$=41$
If
$A=\begin{bmatrix}
\cos\alpha & \sin\alpha \\
-\sin\alpha & \cos\alpha
\end{bmatrix}$
then:
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Solution
Transpose of $A$ is:
$A'=\begin{bmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{bmatrix}$
Now,
$A'A=\begin{bmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha\end{bmatrix}$
$A'A=\begin{bmatrix}\cos^2\alpha+\sin^2\alpha & 0 \\ 0 & \sin^2\alpha+\cos^2\alpha\end{bmatrix}$
Since $\sin^2\alpha+\cos^2\alpha=1$
$A'A=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}=I$
CUET UG Mathematics Applied
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CUET UG Mathematics Applied
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
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