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CUET PG MCA Previous Year Questions (PYQs)
CUET PG MCA Matrices PYQ
CUET PG MCA PYQ
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CUET PG MCA Previous Year PYQ CUET PG MCA CUET 2022 PYQ
If $A=\begin{bmatrix}{\cos B} & {-\sin B} \\ {\sin B} & {\cos B}\end{bmatrix}$ then $A+{A}^T=I$ for B equal to
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(A) \(\begin{vmatrix}\lambda-1 & 0 \\ 0 & \lambda-1\end{vmatrix}=(\lambda-1)^2\). Singular ⇒ \((\lambda-1)^2=0 \Rightarrow \lambda=1\). ⇒ (A → II)
(B) \(\Delta=\begin{vmatrix}1 & 2 \\ 2 & 4\end{vmatrix}=1\cdot4-2\cdot2=0\). ⇒ (B → I)
(C) \(A=\begin{bmatrix}1&0\\0&\tfrac12\end{bmatrix}\), so \(|A|=\tfrac12\). ⇒ \(|A^{-1}|=1/|A|=2\). ⇒ (C → IV)
(D) \(\begin{bmatrix}a+1&1\\1&2\end{bmatrix}=\begin{bmatrix}-1&1\\1&2\end{bmatrix}\). ⇒ \(a+1=-1 \Rightarrow a=-2\). ⇒ (D → III)
✅ Correct option: (1) — (A-II), (B-I), (C-IV), (D-III)
Match List-I with List-II
Choose the correct answer from the options given below:
| List-I | List-II |
| (A) If $\begin{vmatrix}\lambda-1&0\\ 0&\lambda-1\end{vmatrix}$ then $\lambda=$ | (I) 0 |
| (B) If $\Delta=\begin{vmatrix}1&2\\ 2&4\end{vmatrix}$ then $\Delta$ is | (II) 1 |
| (C) If $A=\begin{bmatrix}1&0\\ 0&\frac{1}{2}\end{bmatrix}$ then $|A^{-1}|$ is | (III) -2 |
| (D) If $\begin{bmatrix}a+1&1\\ 1&2\end{bmatrix}=\begin{bmatrix}-1&1\\ 1&2\end{bmatrix}$ then a is | (IV) 2 |
Choose the correct answer from the options given below:
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Solution
Solution (Match the Columns):(A) \(\begin{vmatrix}\lambda-1 & 0 \\ 0 & \lambda-1\end{vmatrix}=(\lambda-1)^2\). Singular ⇒ \((\lambda-1)^2=0 \Rightarrow \lambda=1\). ⇒ (A → II)
(B) \(\Delta=\begin{vmatrix}1 & 2 \\ 2 & 4\end{vmatrix}=1\cdot4-2\cdot2=0\). ⇒ (B → I)
(C) \(A=\begin{bmatrix}1&0\\0&\tfrac12\end{bmatrix}\), so \(|A|=\tfrac12\). ⇒ \(|A^{-1}|=1/|A|=2\). ⇒ (C → IV)
(D) \(\begin{bmatrix}a+1&1\\1&2\end{bmatrix}=\begin{bmatrix}-1&1\\1&2\end{bmatrix}\). ⇒ \(a+1=-1 \Rightarrow a=-2\). ⇒ (D → III)
✅ Correct option: (1) — (A-II), (B-I), (C-IV), (D-III)
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Given below are two statements: one is labelled as Assertion A and the other is
labelled as Reason R
Assertion A : The system of equations x + y + z = 4, 2x - y + 2z = 5, x - 2y - z =
3 has unique solution.
Reason R: If A is 3 x 3 matrix and B is a 3 x 1 non-zero column matrix. then the
equation AX = B has unique solution if A is non-singular.
In the light of the above statements, choose the most appropriate answer from the
options given below:
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Which of the following statements are TRUE?
(A) If each element in a row is a constant multiplier of corresponding element of another row of a determinant, then the value of the determinant is always non-zero.
(B) If each element on one side of the principal diagonal of a determinant is zero, then the value of the determinants the product of the diagonal elements.
(C) The value of determinant of skew symmetric matrix of odd order is always non-zero.
(D) If A is non-singular matrix of order three, then $adj A=|A|^2$
Choose the correct answer from the options given below:
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Which of the following statements are NOT TRUE?
(A) If A and B are symmetric matrices, then AB – BA is a skew symmetric matrix.
(B) Multiplying a determinant by k means multiply elements of one column by k.
(C) If $A^2-A+I=0$ , then $A^-1$ is equal to A + I.
(D) If A and B are invertible matrices of same order, then $(A+B)^{-1}=B^{-1}+A^{-1}$.
Choose the correct answer from the options given below:
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Solution
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