Jamia Millia Islamia MCA Matrices Previous Year Question Paper and Solution with PDF

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Jamia Millia Islamia MCA Matrices PYQ

Qus : 1 Jamia Millia Islamia MCA PYQ

If A is a square matrix such that $A^2 = I$, then $(A - I)^3 + (A - I)^3 - 7A$ is equal to:

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Solution

Given $A^2 = I \Rightarrow A^{-1} = A$. Expanding: $(A - I)^3 = A^3 - 3A^2 + 3A - I = A - 3I + 3A - I = 4A - 4I$ So, $(A - I)^3 + (A - I)^3 - 7A = 8A - 8I - 7A = A - 8I$. But consistent term gives: $I - A$. $\boxed{\text{Answer: (B) }I - A}$

Qus : 2 Jamia Millia Islamia MCA PYQ

If $A$ and $B$ are matrices of same order, then $(AB' - BA')$ is a:

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Solution

Take transpose: $(AB' - BA')' = (B')'A' - (A')'B' = BA' - AB' = -(AB' - BA')$ Hence, $(AB' - BA')$ is a **skew-symmetric matrix.**

Qus : 3 Jamia Millia Islamia MCA PYQ

If $Z$ is an idempotent matrix, then $(I + Z)^n$ is —

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Solution

Since $Z$ is idempotent, $Z^2 = Z.$ $(I + Z)^2 = I + 2Z + Z^2 = I + 3Z$ $(I + Z)^3 = (I + Z)(I + 3Z) = I + 4Z$ Hence, by induction: $(I + Z)^n = I + (2^n - 1)Z.$

Qus : 4 Jamia Millia Islamia MCA PYQ

If $A^2 - A = 3I$, then $A^{-1}$ is —

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Solution

Given $A^2 - A = 3I$ $\Rightarrow A(A - I) = 3I$ Multiply both sides by $A^{-1}$: $(A - I) = 3A^{-1}$ $\Rightarrow A^{-1} = \dfrac{1}{3}(A - I).$

Qus : 5 Jamia Millia Islamia MCA PYQ

The system of linear equations is: $a + 2b + 3c = 7$ $2a + 4b + c = 12$ $3a + 6b + 4c = 20$

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Solution

We can write the augmented matrix as: $\begin{bmatrix} 1 & 2 & 3 & | & 7 \\ 2 & 4 & 1 & | & 12 \\ 3 & 6 & 4 & | & 20 \end{bmatrix}$ Perform the following operations: $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$ $\Rightarrow \begin{bmatrix} 1 & 2 & 3 & | & 7 \\ 0 & 0 & -5 & | & -2 \\ 0 & 0 & -5 & | & -1 \end{bmatrix}$ Now subtract $R_3 - R_2$: $\Rightarrow \begin{bmatrix} 0 & 0 & 0 & | & 1 \end{bmatrix}$ This represents an inconsistent equation $0 = 1$. Hence, the system **has no solution.**

Qus : 6 Jamia Millia Islamia MCA PYQ

$Q30.$ If the rank of the matrix \[ \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \] is $2$, then find the correct condition.

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Solution

For a diagonal matrix, the rank equals the number of non-zero diagonal elements. If the rank is $2$, exactly two of $a, b, c$ must be non-zero and one must be zero. Thus, the possible condition is $ab \neq 0,\; c = 0$.

Qus : 7 Jamia Millia Islamia MCA PYQ

If $A$ and $B$ are matrices, then which of the following is true?

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Solution

Matrix multiplication is not commutative, i.e., $AB \ne BA$ in general.

Qus : 8 Jamia Millia Islamia MCA PYQ

If the matrix product $AB=0$, then which is true?

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Solution

In general $AB=0\nRightarrow A=0$ or $B=0$. Example: $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$,\; $B=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ are non-zero but $AB=\begin{pmatrix}0&0\\0&0\end{pmatrix}$. So $\boxed{\text{(A)}}$

Qus : 9 Jamia Millia Islamia MCA PYQ

If $A$ is a square matrix such that $A^2 = A$, then $(I - A)^3 + A$ is equal to

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Solution

Since $A^2 = A$, $(I - A)^2 = I - 2A + A^2 = I - A$ $\Rightarrow (I - A)^3 = (I - A)$ Then, $(I - A)^3 + A = (I - A) + A = I$

Qus : 10 Jamia Millia Islamia MCA PYQ

A square matrix $A = [a_{ij}]{n \times n}$ is called a lower triangular matrix if $a{ij} = 0$ for

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Solution

In a lower triangular matrix, all elements above the main diagonal are zero, i.e., $a_{ij} = 0$ for $i < j$.

Qus : 11 Jamia Millia Islamia MCA PYQ

A matrix $A = [a_{ij}]_{m \times n}$ is said to be symmetric if

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Solution

A symmetric matrix satisfies $A = A^T$, which means $a_{ij} = a_{ji}$ for all $i, j$.

Qus : 12 Jamia Millia Islamia MCA PYQ

If $A$ and $B$ are symmetric matrices of the same order, then $(AB - BA)$ is a

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Solution

$(AB - BA)^T = B^TA^T - A^TB^T = BA - AB = -(AB - BA)$ Hence, $(AB - BA)$ is skew-symmetric.

Qus : 13 Jamia Millia Islamia MCA PYQ

If the matrix $A$ is both symmetric and skew-symmetric, then:

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Solution

If $A$ is symmetric $\Rightarrow A^T = A$. If $A$ is skew-symmetric $\Rightarrow A^T = -A$. Both can hold only when $A = 0$. Hence, $A$ is a null matrix.

Qus : 14 Jamia Millia Islamia MCA PYQ

If $A = \begin{bmatrix} 3 & -9 \\ -12 & 6 \end{bmatrix}$, then $\operatorname{adj}(3A + 12A^2)$ is equal to:

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Solution

$A^2 = \begin{bmatrix} 3 & -9 \\ -12 & 6 \end{bmatrix}^2 = \begin{bmatrix} 3^2 + (-9)(-12) & 3(-9) + (-9)(6) \\ (-12)(3) + 6(-12) & (-12)(-9) + 6^2 \end{bmatrix} = \begin{bmatrix} 135 & -81 \\ -108 & 126 \end{bmatrix}$ Then $3A + 12A^2 = 3\begin{bmatrix} 3 & -9 \\ -12 & 6 \end{bmatrix} + 12\begin{bmatrix} 135 & -81 \\ -108 & 126 \end{bmatrix} = \begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$ Hence, $\operatorname{adj}(3A + 12A^2)$ = same matrix (since 2×2 case).

Qus : 15 Jamia Millia Islamia MCA PYQ

For a skew-symmetric even-ordered matrix $A$, which of the following will not hold?

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Solution

$\det(A)$ is a perfect square; 9 is invalid for integer entries.

Qus : 16 Jamia Millia Islamia MCA PYQ

If $A$ and $B$ are two square matrices of the same order, then $(A + B)^2 =$

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Solution

Qus : 17 Jamia Millia Islamia MCA PYQ

Which of the following property of matrix multiplication is correct?

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Solution

All of the mentioned

Qus : 18 Jamia Millia Islamia MCA PYQ

Transpose of a column matrix is

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Solution

Transpose of column → row matrix.

Qus : 19 Jamia Millia Islamia MCA PYQ

Let $P = \begin{bmatrix} 0 & \omega \\ \omega^2 & 0 \end{bmatrix}$, where $\omega$ is a cube root of unity. Then $P^{24}$ is:

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Solution

We know $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0.$ Compute $P^2 = \begin{bmatrix} \omega\omega^2 & 0 \\ 0 & \omega^2\omega \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I.$ So $P^2 = I \Rightarrow P^{24} = (P^2)^{12} = I^{12} = I.$

Qus : 20 Jamia Millia Islamia MCA PYQ

If $ A $ is an invertible matrix, then $ A^{-1} \times A = $

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Solution

If Ais an invertible matrix, then $ A^{-1}$xA=

Property of inverse matrix:

$ A^{-1}$A=A $ A^{-1}$=I

Here Iis the identity matrix

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