M.C.Q (1 Marks)

MCQ 2511 Mark

For what value of $x$, matrix $A=\left[\begin{array}{ll}6-x & 4 \\ 3-x & 1\end{array}\right]$ is a singular matrix?

A
1
✓
2
C
-1
D
-2

Answer

Correct option: B.

(b) : Matrix $A$ is singular, when $|A|=0$
$\Rightarrow\left|\begin{array}{ll}6-x & 4 \\ 3-x & 1\end{array}\right|=0$
$\Rightarrow 6-x-12+4 x=0 \Rightarrow 3 x=6 \Rightarrow x=2$

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MCQ 2521 Mark

The value of $\Delta=\left|\begin{array}{ccc}0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0\end{array}\right|$ is

✓
$0$
B
$1$
C
$2$
D
$3$

Answer

Correct option: A.

$0$

Expanding along $R_1$, we get
$ \Delta=0-\sin \alpha(0-\sin \beta \cos \alpha)-\cos \alpha(\sin \alpha \sin \beta-0)$
$=\sin \alpha \sin \beta \cos \alpha-\cos \alpha \sin \alpha \sin \beta=0$

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MCQ 2531 Mark

A matrix $A$ of order $3 \times 3$ has determinant What is the value of $|3 A|$ ?

✓
135
B
45
C
25
D
5

Answer

Correct option: A.

135

(a) : Given, $|A|=5$, order of $A$ is $3 \times 3$.
$\therefore \quad|3 A|=3^3|A|=27 \times 5=135$.

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MCQ 2541 Mark

Evaluate the following determinant : $\left|\begin{array}{cc}x & -7 \\ x & 5 x+1\end{array}\right|$

A
$3 x^2+4$
✓
$x(5 x+8)$
C
$3 x+4 x^2$
D
$x(3 x+4)$

Answer

Correct option: B.

$x(5 x+8)$

(b) : We have, $\left|\begin{array}{cc}x & -7 \\ x & 5 x+1\end{array}\right|=x(5 x+1)+7(x)$
$
=5 x^2+x+7 x=5 x^2+8 x=x(5 x+8)
$

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MCQ 2551 Mark

What positive value of $x$ makes the following pair of determinants equal? $\left|\begin{array}{cc}2 x & 3 \\5 & x\end{array}\right|, \left|\begin{array}{cc}16 & 3 \\5 & 2\end{array}\right|$

A
$1$
B
$2$
C
$3$
✓
$4$

Answer

Correct option: D.

$4$

We have, $\left|\begin{array}{rr}2 x & 3 \\ 5 & x\end{array}\right|=\left|\begin{array}{rr}16 & 3 \\ 5 &2\end{array}\right|$
$\Rightarrow 2 x^2-15=32-15$
$ \Rightarrow 2 x^2=32 $
$\Rightarrow x^2=16$
$\Rightarrow x=4\ [\because x>0]$

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MCQ 2561 Mark

If $\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|=8$, then find the value of $x$.

A
$2$
✓
$-2$
C
$1$
D
$-1$

Answer

Correct option: B.

$-2$

Expanding the given determinant, we get
$x\left(-x^2-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta)=8$
$\Rightarrow-x^3-x+x=8$
$\Rightarrow x^3+8=0$
$\Rightarrow(x+2)\left(x^2-2 x+4\right)=0$
$\Rightarrow x+2=0$
$\Rightarrow x=-2\left[\because x^2-2 x+4>0 \forall x\right]$
$\left[\because x^2-2 x+4>0 \forall x\right]$

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MATHS M.C.Q (1 Marks)