Jamia Millia Islamia MCA Determinants Previous Year Questions (PYQs) – Page 1 of 2
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The determinant of a singular matrix is:
1
2
3
4
Solution
Definition:
A matrix is singular if
$ |A| = 0 $
If $x, y, z$ are all different from zero and
$\begin{vmatrix}
1 + x & 1 & 1 \\
1 & 1 + y & 1 \\
1 & 1 & 1 + z
\end{vmatrix} = 0$,
then the value of $x^{-1} + y^{-1} + z^{-1}$ is:
1
2
3
4
Solution
Expand the determinant:
$(1+x)(1+y)(1+z) - (1+x) - (1+y) - (1+z) + 2 = 0$
Simplify: $xyz(x^{-1} + y^{-1} + z^{-1}) + ... = 0$
Final result gives $\boxed{x^{-1} + y^{-1} + z^{-1} = -1}$
$\boxed{\text{Answer: (D) -1}}$
If $A = \left[\begin{array}{cc}
x & 2 \\
2 & x
\end{array}\right]$ and $|A^2| = 0$, then $x$ is equal to
1
2
3
4
Solution
Since $|A^2| = |A|^2 = 0$, we have $|A| = 0$.
$\therefore |A| = x^2 - 4 = 0 \Rightarrow x = \pm 2.$
The area of the triangle with vertices $A(a, b+c)$, $B(b, c+a)$ and $C(c, a+b)$ is equal to
1
2
3
4
Solution
Area $=\dfrac{1}{2}\left|\begin{array}{ccc}
a & b+c & 1 \\
b & c+a & 1 \\
c & a+b & 1
\end{array}\right|$
If $
\begin{vmatrix}
x & 3 & 6 \\
3 & 6 & x \\
6 & x & 3
\end{vmatrix}
=
\begin{vmatrix}
2 & x & 7 \\
x & 7 & 2 \\
7 & 2 & x
\end{vmatrix}
=
\begin{vmatrix}
4 & 5 & x \\
5 & x & 4 \\
x & 4 & 5
\end{vmatrix}
= 0
$, then $x$ is equal to:
1
2
3
4
Solution
For determinant $\begin{vmatrix}x & 3 & 6 \ 3 & 6 & x \ 6 & x & 3\end{vmatrix} = 0$
Expanding, we get:
$x(6×3 - x×x) - 3(3×3 - x×6) + 6(3×x - 6×6) = 0$
$\Rightarrow x(18 - x^2) - 3(9 - 6x) + 6(3x - 36) = 0$
$\Rightarrow -x^3 + 18x - 27 + 18x + 18x - 216 = 0$
$\Rightarrow -x^3 + 54x - 243 = 0$
$\Rightarrow x^3 - 54x + 243 = 0$
By trial, $x=9$ satisfies it.
Hence $\boxed{x = 9}$
If $
\begin{vmatrix}
a & p & x \\
b & q & y \\
c & r & z
\end{vmatrix}
= 16
$, then the value of
$
\begin{vmatrix}
p+q & a+x & a+p \\
q+y & b+y & b+q \\
x+z & c+z & c+r
\end{vmatrix}
$ is:
1
2
3
4
Solution
By properties of determinants,
each column in the second determinant is the **sum of two columns** of the first.
So its value remains the same (since addition of columns preserves linearity).
Hence $\boxed{16}$.
If $
\begin{vmatrix}
x & 3 & 6 \\
3 & 6 & x \\
6 & x & 3
\end{vmatrix}
=
\begin{vmatrix}
2 & x & 7 \\
x & 7 & 2 \\
7 & 2 & x
\end{vmatrix}
=
\begin{vmatrix}
4 & 5 & x \\
5 & x & 4 \\
x & 4 & 5
\end{vmatrix}
= 0
$, then $x$ is equal to:
1
2
3
4
Solution
Expanding the first determinant,
$
x(6\times3 - x\times x) - 3(3\times3 - x\times6) + 6(3\times x - 6\times6)
$
$
= x(18 - x^2) - 3(9 - 6x) + 6(3x - 36)
$
$
= -x^3 + 54x - 243 = 0
$
$\Rightarrow x^3 - 54x + 243 = 0$
Trial gives $x = 9$.
$\boxed{x = 9}$ → (D) None of these.
If $A,B,C$ are angles of a triangle, then the value of
$
\begin{vmatrix}
\sin 2A & \sin C & \sin B \\
\sin C & \sin 2B & \sin A \\
\sin B & \sin A & \sin 2C
\end{vmatrix}
$
is:
1
2
3
4
Solution
** Since $A+B+C=\pi$, we have
$\cos A=-\cos(B+C)=\sin B\sin C-\cos B\cos C$ and similarly for $B,C$.
Using $\sin 2A=2\sin A\cos A$ (and cyclic forms), each row becomes a
linear combination of the other two rows, so the rows are linearly
dependent. Hence the determinant is $0$.
$\boxed{0}$
Let $A$ be a non-singular matrix of order $2 \times 2$. Then
$|A^{-1}| =$
1
2
3
4
Solution
For any invertible matrix $A$,
$\det(A^{-1}) = \dfrac{1}{\det(A)}$
If $a,b,c$ are in A.P., then the value of determinant
$\begin{vmatrix}
x+2 & x+3 & x+2a \\
x+3 & x+4 & x+2b \\
x+4 & x+5 & x+2c
\end{vmatrix}$ is:
1
2
3
4
Solution
Since $a,b,c$ are in A.P. ⇒ $2b = a + c$.
Expanding determinant by properties of A.P., it becomes zero.
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