NIMCET 2027 Matrices Previous Year Questions (PYQs) – Page 1 of 3
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Showing 24 questions
If A and B are square matrices such that $B=-A^{-1} BA$, then $(A + B)^2$ is
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Solution
If $A$ is a $3\times 3$ matrix with $\det(A)=3$, then $\det(\operatorname{adj}A)$ is:
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Solution
For an $n \times n$ matrix:
$\det(\operatorname{adj}A) = (\det A)^{,n-1}$
Here:
$n = 3$, $\det A = 3$
So,
$\det(\operatorname{adj}A) = 3^{3-1} = 3^2 = 9$
If $a + b + c \neq 0$, the system of equations:
$(b+c)(y+z) - ax = b - c$
$(c+a)(z+x) - by = c - a$
$(a+b)(x+y) - cz = a - b$
has:
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Solution
This is a symmetric linear system.
Solving gives the unique solution:
$x = y = z = 0$
Since $a+b+c \neq 0$, determinant $\neq 0$ → unique solution.
Car travels half distance with $v_1$, half with $v_2$.
Average speed is
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Solution
For equal distances:
Average velocity $=\dfrac{2v_1v_2}{v_1+v_2}$
If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, then
$I + A + A^2 + A^3 + \cdots \infty$ equals:
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Solution
Since $I + A + A^2 + \cdots = (I - A)^{-1}$,
$I - A = \begin{bmatrix} 0 & -2 \\ -3 & -3 \end{bmatrix}$
$(I - A)^{-1}
= \begin{bmatrix} \tfrac{1}{2} & -\tfrac{1}{3} \\ -\tfrac{1}{2} & 0 \end{bmatrix}$.
If $A=\begin{bmatrix} 1 &0 &0 \\ 0& 1 &1 \\ 0&-2 & 4 \end{bmatrix}$ and $6A^{–1} = A^{2} + cA + dI$, where $A^{–1}$ is A- inverse, I is the identify matrix, then (c, d)
is
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Solution
$ \text{If } B = -A^{-1}BA,\ \text{then } (A+B)^{2} = $
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Solution
From $ B = -A^{-1}BA $, multiply both sides by $A$:
$ BA = -BA $
So $ BA = 0 $.
Then
$ (A+B)^{2} = A^{2} + AB + BA + B^{2} = A^{2} + AB + 0 + B^{2} $
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