## Matrices and Determinants Questions and Answers Part-12

1. For a fixed positive integer n, let
$D=\begin{bmatrix}\left(n-1\right)! & \left(n+1\right)! & \left(n+3\right)!/n\left(n+1\right)\\\left(n+1\right)! & \left(n+3\right)! & \left(n+5\right)!/\left(n+2\right)\left(n+3\right) \\\left(n+3\right)! & \left(n+5\right)! & \left(n+7\right)!/\left(n+4\right)\left(n+5\right)\end{bmatrix}$
then $\frac{D}{\left(n-1\right)!\left(n+1\right)!\left(n+3\right)!}$     is equal to
a) -8
b) -16
c) -32
d) -64

Explanation: Taking (n – 1)! common from R1, (n + 1)! from R2 and (n + 3)! from R3, we get

2. If $\triangle=\begin{bmatrix}\sqrt{6} & 2i & 3+\sqrt{6} \\\sqrt{12} & \sqrt{3}+\sqrt{8}i& 3\sqrt{2}+\sqrt{6}i \\ \sqrt{18} & \sqrt{2}+\sqrt{12}i & \sqrt{27}+2i\end{bmatrix}$
then $\triangle$ is
a) a natural number
b) a negative integer
c) an irrational number
d) an imaginary number

Explanation:

3. If x, y, z are different from zero and $\triangle=\begin{bmatrix}a & b-y & c-z \\a-x & b & c-z \\a-x & b-y & c\end{bmatrix}=0$
then the value of the expression $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$     is
a) 0
b) -1
c) 1
d) 2

Explanation:

4. Let $f\left(x\right)=\begin{bmatrix}\cos x & x & 1 \\2\sin x & x^{2} & 2x \\\tan x & x &2\end{bmatrix}$
$\lim_{x \rightarrow 0}\frac{f\left(x\right)}{x^{2}}$   is equal to
a) -1
b) 0
c) 2
d) 3

Explanation: : For x $\neq$ 0, we have

5. If $p \neq a, q\neq b, r\neq c$     and the system of equations
px + ay + az = 0
bx + qy + bz = 0
cx + cy + rz = 0
has a non-trivial solution, then the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$
a) -1
b) 0
c) 1
d) 2

Explanation: As the given system of equations has a non-trivial solution

6. If $p\neq 0$  , solution set of the equation
$\triangle=\begin{bmatrix}1 & 1 & x \\p+1 & p+1 & p+x \\3 & x+1 & x+2\end{bmatrix}=0$
is
a) {1, 2}
b) {2, 3}
c) {1, p, 2}
d) {1, 2, – p}

Explanation: Applying C2 $\rightarrow$ C2 – C1, we get

7. Let a, b, $c\epsilon R$   be such that $a+b+c\neq 0$   . If the system of equations
ax + by + cz = 0
bx + cy + az = 0
cx + ay + bz = 0
has a non-trivial solution, then
a) a + c – b = 0
b) b + c – a = 0
c) a + b – c = 0
d) a = b = c

Explanation: As the system of equations has a non-trivial solution,

8. If the system of linear equations x + y + z = 6, x + 2y + 3z = 14 and $2x+5y+\lambda z=\mu\left(\lambda,\mu\epsilon R\right)$
has a unique solution, then
a) $\lambda\neq 8$
b) $\lambda=8,\mu\neq 36$
c) $\lambda=8,\mu=36$
d) none of these

Explanation: The given system of linear equations has a unique solution if

9. Let $A\left(z_{1}\right)$   , $B\left(z_{2}\right)$   and $C\left(z_{3}\right)$  be the vertices of a triangle. Let a = BC, b = CA and c = AB, then
$b\begin{bmatrix}z & \bar{z} & 1 \\z_{1} & \bar{z_{1}} & 1 \\z_{2} & \bar{z_{2}} & 1\end{bmatrix}+c\begin{bmatrix}z & \bar{z} & 1 \\z_{1} & \bar{z_{1}} & 1 \\z_{3} & \bar{z_{3}} & 1\end{bmatrix}=0$
represents
a) median of $\triangle ABC$   through A
b) angle bisector of $\angle A$
c) altitude of $\triangle ABC$   through A
d) perpendicular bisector of side BC

$S=\begin{bmatrix}1 & a & a^{2} \\1 & b & b^{2} \\1 & c & c^{2}\end{bmatrix} S_{1}=\begin{bmatrix}1 & a & a^{3} \\1 & b & b^{3} \\1 & c & c^{3}\end{bmatrix} S_{2}=\begin{bmatrix}1 & a^{2} & a^{3} \\1 & b^{2} & b^{3} \\1 & c^{2} & c^{3}\end{bmatrix} S_{3}=\begin{bmatrix}a & a^{2} & a^{3} \\b & b^{2} & b^{3} \\c & c^{2} & c^{3}\end{bmatrix}$
a) $\frac{s_{1}}{s}=a+b+c$
b) $\frac{s_{2}}{s}=ab+bc+ca$
c) $\frac{s_{3}}{s}=abc$