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AMU MCA Matrices PYQ

AMU MCA PYQ
The following system of equations:
$2x_1 + x_2 - x_3 = 2$
$3x_1 + 2x_2 + x_3 = 3$
has:





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2025 PYQ

Solution

Rank of coefficient matrix equals rank of augmented matrix but less than number of variables, hence there are 2 degenerate and 1 non-degenerate solutions.
AMU MCA PYQ
Let $T:\mathbb{R}^4 \rightarrow \mathbb{R}^3$ be a linear transformation defined by $T(x_1,x_2,x_3,x_4)=C(x_1-x_2,;x_2-x_3,;x_3-x_4)$ Then which of the following is true?(i) $\dim(\ker T)=1$ if $C \ne 0$ (ii) $\dim(\ker T)=0$ if $C=0$ (iii) $\dim(\ker T)=1$ if $T$ is onto





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Solution

For $C \ne 0$, kernel has dimension 1. If $T$ is onto, rank is 3, hence nullity is 1.
AMU MCA PYQ
Let $T : P_2(x) \to P_2(x)$ be a linear transformation on vector space $P_2(x)$ (polynomials of degree $\le 2$ over $\mathbb{R}$) such that $T(f(x)) = \dfrac{d}{dx}(f(x))$. Then the matrix of $T$ w.r.t. basis ${1, x, x^2}$ is:





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Solution

AMU MCA PYQ
If $\lambda$ is an eigenvalue of a non-singular matrix $A$, then the characteristic root of adj $A$ is





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Solution

Eigenvalue of adj $A$ is $\dfrac{|A|}{\lambda}$
AMU MCA PYQ
If $A$ and $B$ are two odd order skew-symmetric matrices such that $AB = BA$, then $AB$ is





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Solution

$A^T = -A$, $B^T = -B$ $(AB)^T = B^T A^T = (-B)(-A) = AB$ So $AB$ is symmetric.
AMU MCA PYQ
Which of the following is not true?





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Solution

Definition (b) is incorrect.
AMU MCA PYQ
Let $T$ be a linear operator on $\mathbb{R}^3$ defined by $T(x,y,z) = (2x,; x-y,; 5x + 4y + z)$ Then $T^{-1}$ is





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

Given:

$T(x,y,z) = (2x,\; x-y,\; 5x+4y+z)$

Write system:

$u = 2x$

$v = x - y$

$w = 5x + 4y + z$

From first:

$x = \frac{u}{2}$

From second:

$y = x - v = \frac{u}{2} - v$

From third:

$z = w - 5x - 4y$

$z = w - \frac{5u}{2} - 4\left(\frac{u}{2} - v\right)$

$z = w - \frac{5u}{2} - 2u + 4v$

$z = w - \frac{9u}{2} + 4v$

Thus inverse:

$T^{-1}(x,y,z) = \left(\frac{x}{2}, \frac{x - 2y}{2}, \frac{-9x + 8y + 2z}{2}\right)$

AMU MCA PYQ
If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ satisfies the matrix equation $A^2 - kA + 2I = 0$, then the value of $k$ is:






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Solution

Trace of $A = 3 + (-2) = 1$

Determinant of $A = (3)(-2) - (-8) = -6 + 8 = 2$

Characteristic equation of $A$ is:

$\lambda^2 - (\text{trace})\lambda + \det = 0$

$\lambda^2 - \lambda + 2 = 0$

By Cayley–Hamilton theorem:

$A^2 - A + 2I = 0$

Comparing with $A^2 - kA + 2I = 0$

$k = 1$

AMU MCA PYQ
Let $T : \mathbb{R}^2 \to \mathbb{R}^3$ defined by $T(x,y) = (-x-y, 3x+8y, 9x-11y)$. Then the rank and nullity of $T$ are respectively






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Solution

Matrix form:

$\begin{bmatrix} -1 & -1 \\ 3 & 8 \\ 9 & -11 \end{bmatrix}$

Columns are independent ⇒ Rank = 2

Nullity = 2 - 2 = 0


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