AMU MCA Matrices Previous Year Questions (PYQs) – Page 1 of 2
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If $ A=\left(\matrix{0&1&2\cr 1&2&3\cr 3&\alpha&1}\right) $,
$ A^{-1}=\left(\matrix{\frac{1}{2}&-\frac{1}{2}&\frac{1}{2}\cr -4&3&\beta\cr \frac{5}{2}&-\frac{3}{2}&\frac{1}{2}}\right) $, then
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Solution
If $A=\left[\matrix{k&l\cr m&n}\right]$ and $kn\ne lm$, then the value of
$A^2-(k+n)A+(kn-lm)I$ equals
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Solution
If $A$ is an orthogonal matrix, then
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Solution
For orthogonal matrix:
$ A^TA=I $
Taking determinant:
$ |A^T||A|=|I|=1 $
$ \Rightarrow |A|^2=1 $
$ \Rightarrow |A|=\pm1 $
The following system of equations:
$2x_1 + x_2 - x_3 = 2$
$3x_1 + 2x_2 + x_3 = 3$
has:
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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.
If $A=\begin{bmatrix}2 & 3 \\ 5 & -2\end{bmatrix}$ be such that $A^{-1}=\lambda A$. Then, the value of $\lambda$ is
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Solution
$|A| = (2)(-2) - (3)(5) = -4 - 15 = -19$
$\text{adj }A = \begin{bmatrix}-2 & -3 \\ -5 & 2\end{bmatrix}$
$A^{-1} = \frac{1}{|A|}\text{adj }A = \frac{-1}{19}\begin{bmatrix}-2 & -3 \\ -5 & 2\end{bmatrix}$
$= \begin{bmatrix}\frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & \frac{-2}{19}\end{bmatrix}$
Given $A^{-1} = \lambda A$
$\Rightarrow \begin{bmatrix}\frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & \frac{-2}{19}\end{bmatrix} = \lambda \begin{bmatrix}2 & 3 \\ 5 & -2\end{bmatrix}$
Comparing, $\lambda = \frac{1}{19}$
$\boxed{\frac{1}{19}}$
If $A=\begin{bmatrix}\alpha & 0 \\ 1 & 1\end{bmatrix}$ and $B=\begin{bmatrix}1 & 0 \\ 5 & 1\end{bmatrix}$ and $A^2=B$, then the value of $\alpha$ is
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Solution
$A^2 = \begin{bmatrix}\alpha & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}\alpha & 0 \\ 1 & 1\end{bmatrix} = \begin{bmatrix}\alpha^2 & 0 \\ \alpha + 1 & 1\end{bmatrix}$
Given $A^2 = B$
$\Rightarrow \begin{bmatrix}\alpha^2 & 0 \\ \alpha + 1 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 5 & 1\end{bmatrix}$
Comparing:
$\alpha^2 = 1 \quad \text{and} \quad \alpha + 1 = 5$
$\alpha = \pm1 \quad \text{and} \quad \alpha = 4$
Both cannot occur simultaneously
$\Rightarrow$ No value possible
$\boxed{\text{not possible}}$
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.
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
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
$\dfrac{|A|}{\lambda}$
If $\left[\matrix{\alpha & \beta \cr \gamma & -\alpha}\right]$ is to be square root of the two rowed unit matrix, then $\alpha, \beta$ and $\gamma$ should satisfy
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Solution
Let $A = \left[\matrix{\alpha & \beta \cr \gamma & -\alpha}\right]$
$A^2 = I$
Multiply:
$A^2 = \left[\matrix{\alpha^2 + \beta\gamma & 0 \cr 0 & \alpha^2 + \beta\gamma}\right]$
Equating with identity:
$\alpha^2 + \beta\gamma = 1$
$\Rightarrow 1 - \alpha^2 - \beta\gamma = 0$
Final Answer: $\boxed{1 - \alpha^2 - \beta\gamma = 0}$
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