Determinants Test

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Determinants Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

Find the determinants of minors and cofactors of the determinant $$\begin{vmatrix}2 & 3 & 4\\ 7 & 2 & -5\\ 8 & -1 & 3\end{vmatrix}$$

A
$$\begin{vmatrix}1 & 61 & -23\\ 13 & -26 & -26\\ -23 & -38 & -17\end{vmatrix}$$ and $$\begin{vmatrix}1 & 61 & -23\\ -13 & -26 & 26\\ -23 & 38 & -17\end{vmatrix}$$

B
$$\begin{vmatrix}1 & 61 & -23\\ 13 & -26 & -26\\ -23 & -38 & -17\end{vmatrix}$$ and $$\begin{vmatrix}1 & -61 & 23\\ -13 & -26 & 26\\ -23 & 38 & -17\end{vmatrix}$$

C
$$\begin{vmatrix}1 & 61 & -23\\ 13 & -26 & -26\\ -23 & -38 & -17\end{vmatrix}$$ and $$\begin{vmatrix}1 & 61 & 23\\ -13 & -26 & 26\\ -23 & 38 & -17\end{vmatrix}$$

D
None of these.

Solution

Here $$M_{11} = \begin{vmatrix}2 & -5 \\ -1 & 3\end{vmatrix}$$ (Delete 1st row and first column)
$$= 6-5$$
$$M_{11} = 1$$
$$\therefore C_{11} = 1$$ $$(\because (-1)^{1 + 1} = 1)$$
$$M_2 = \begin{vmatrix}7 & -5\\ 8 & 3 \end{vmatrix}$$ (Delete 1st row and 2nd column)
$$= 21 - (-40)$$
$$M_2 = 61$$
$$\therefore C_{12} = - 61$$, $$(\because (-1)^{1+2} = - 1)$$
$$M_{13} = \begin{vmatrix}7 & 2 \\ 8 & -1\end{vmatrix}$$ (Delete 1st row and 3rd column)
$$= - 7 - 16$$
$$M_{13} = - 23$$
$$\therefore C_{13} = - 23, (\because (-1)^{1 + 3} = 1)$$
$$M_{21} = \begin{vmatrix}3 & 4 \\ -1 & 3\end{vmatrix}$$ (Delete 2nd row and 1st column)
$$= 9 - (-4)$$
$$M_{21} = 13$$
$$\therefore C_{21} = - 13, (\because (-1)^{2 + 1} = - 1)$$
$$M_{22} = \begin{vmatrix}2 & 4 \\ 8 & 3\end{vmatrix}$$ (Delete 2nd row and 2nd column)
$$= 6 - 32$$
$$M_{22} = - 26$$
$$\therefore C_{22} = - 26, (\because (-1)^{2 + 2} = 1)$$
$$M_{23} = \begin{vmatrix}2 & 3 \\ 8 & -1\end{vmatrix}$$ (Delete 2nd row and 3rd column)
$$= -2 - 24$$
$$M_{23} = - 26$$
$$\therefore C_{23} = 26, (\because (-1)^{2 + 3} = - 1)$$
$$M_{31} = \begin{vmatrix}3 & 4 \\ 2 & -5\end{vmatrix}$$ (Delete 3rd row and 1st column)
$$= -15 -8$$
$$M_{31} = - 23$$
$$\therefore C_{31} = - 23, (\because (-1)^{3 + 1} = 1)$$
$$M_{32} = \begin{vmatrix}2 & 4 \\ 7 & -5\end{vmatrix}$$ (Delete 3rd row and 2nd column)
$$= -10 - 28$$
$$M_{32} = - 38$$
$$\therefore C_{32} = 38, (\because (-1)^{3 + 2} = - 1)$$
$$M_{33} = \begin{vmatrix}2 & 3 \\ 7 & 2\end{vmatrix}$$ (Delete 3rd row and 3rd column)
$$= 4 - 21$$
$$M_{33} = - 17$$
$$\therefore C_{33} = - 17, (\because (-1)^{3 + 3} = 1)$$
Hence Determinants of Minors and Cofactors are
$$\begin{vmatrix}1

& 61 & -23\\ 13 & -26 & -26\\ -23 & -38 &

-17\end{vmatrix}$$ and $$\begin{vmatrix}1 & -61 & -23\\ -13

& -26 & 26\\ -23 & 38 & -17\end{vmatrix}$$
Question 2
1 / -0

Find the adjoint of the matrix $$A = \begin{bmatrix}1 & 2 & 3 \\ 1 & 3 & 5 \\ 1 & 5 & 12\end{bmatrix}$$.
If $$\mbox{Adjoint A: } \begin{bmatrix}a & -9 & 1 \\ b & 9 & -2 \\ 2 & c & 1\end{bmatrix} \\$$, find the value of $$abc$$.

A
231

B
213

C
321

D
312

Solution
Question 3
1 / -0

If $$A^2 = I$$, then the value of $$det(A - I)$$ is (where $$A$$ has order $$3$$)

A
$$1$$

B
$$-1$$

C
$$0$$

D
cannot say anything

Solution

$$A^2=I$$
$$\Rightarrow A^2-I=O$$
$$\Rightarrow (A-I)(A+I)=O$$
$$\Rightarrow |A-I||A+I|=0$$
$$\therefore$$ Either $$|A-I|=0$$ or $$|A+I|=0$$
Hence, cannot say anything about $$|A-I|$$.
Hence, option D.
Question 4
1 / -0

If $$A=\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}$$and $$f(x) = \displaystyle \frac{1 + x}{1- x}$$, then $$f(|A|)$$ is

A
$$\dfrac{-1}{2}$$

B
$$\dfrac{1}{2}$$

C
$$\dfrac{-1}{3}$$

D
none of these

Solution

Here $$|A| =1\times 1-2\times 2 = -3$$

$$\therefore f(|A|)=\cfrac{1+(-3)}{1+3}=-\cfrac{1}{2}$$
Question 5
1 / -0

$$\Delta =\begin{vmatrix} a & { a }^{ 2 } & 0 \\ 1 & 2a+b & (a+b) \\ 0 & 1 & 2a+3b \end{vmatrix}$$ is divisible by

A
$$a+b$$

B
$$a+2b$$

C
$$2a+3b$$

D
$${ a }^{ 2 }$$

Solution

Expanding by $$R_{3}$$,
$$=-1(a(a+b))+(2a+3b)(2a^{2}+ab-a^{2})$$
$$=-1(a(a+b))+(2a+3b)a(a+b)$$
$$=a(a+b)(2a+3b-1)$$
Question 6
1 / -0

If matrix A is given by $$A = \begin{bmatrix}6 & 11\\ 2 & 4\end{bmatrix}$$, then the determinant of $$A^{2005} - 6A^{2004}$$ is

A
$$2^{2006}$$

B
$$(-11) 2^{2005}$$

C
$$-2^{2005}$$

D
$$(-9)2^{2004}$$

Solution

Given, $$A = \begin{bmatrix}6 & 11\\ 2 & 4\end{bmatrix}$$
$$|A|=2$$

Now, $$|A^{2005}-6A^{2004}|=|A|^{2004}|A-6I|$$
$$=2^{2004}\begin{vmatrix} 0 & 11 \\ 2 & -2 \end{vmatrix}$$
$$=2^{2004}(-22)$$
$$=2^{2005}(-11)$$
Question 7
1 / -0

$$\displaystyle \left | \begin{matrix}0 &p-q &p-r \\ q-p &0 &q-r \\ r-p &r-q &0 \end{matrix} \right |$$ is equal to

A
$$\displaystyle p+q+r$$

B
$$0$$

C
$$\displaystyle p-q-r$$

D
$$\displaystyle -p+q+r$$

Solution

$$\left| \begin{matrix} 0 & p-q & p-r \\ q-p & 0 & q-r \\ r-p & r-q & 0 \end{matrix} \right| \\$$

$$ =0\left( 0-\left( r-q \right) \left( q-r \right) \right) -\left( p-q \right) \left( 0-\left( r-p \right) \left( q-r \right) \right) +\left( p-r \right) \left( \left( q-p \right) \left( r-q \right) -0 \right) \\$$

$$ =\left( p-q \right) \left( r-p \right) \left( q-r \right) +\left( p-r \right) \left( q-p \right) \left( r-q \right) =0$$
Question 8
1 / -0

If $$\displaystyle \left | \begin{matrix}6i &-3i &1 \\ 4 &3i &-1 \\ 20 &3 &i \end{matrix} \right |=x+iy$$ then

A
$$\displaystyle x=3, y=1$$

B
$$\displaystyle x=1, y=3$$

C
$$\displaystyle x=0, y=3$$

D
$$\displaystyle x=0, y=0$$

Solution

$$\Delta =\left| \begin{matrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{matrix} \right| \\$$

$$ =6i\left( -3+3 \right) +3i\left( 4i+20 \right) +1\left( 12-60i \right) \\$$

$$ =0-12+60i+12-60i$$

$$=0\\$$

$$ \Rightarrow x=0,y=0$$

Hence, the option 'D' is correct.
Question 9
1 / -0

The matrix $$\begin{bmatrix}1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 1 & 1 \end{bmatrix}$$ is

A
non-singular

B
singular

C
skew-symmetric

D
symmetric

Solution

Given $$A=\begin{bmatrix}1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 1 & 1 \end{bmatrix}$$

$$|A|=\begin{vmatrix}1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 1 & 1 \end{vmatrix}$$

$$\Rightarrow |A|=1-1=0$$

Hence, $$A$$ is singular.
Question 10
1 / -0

The value of the determinant $$\displaystyle \left | \begin{matrix}^{5}C_{0} &^{5}C_{3} &14 \\ ^{5}C_{1} &^{5}C_{4} &1 \\ ^{5}C_{2} &^{5}C_{5} &1 \end{matrix} \right |$$ is

A
$$0$$

B
$$\displaystyle -576$$

C
$$80$$

D
none of these

Solution

$$\left| \begin{matrix} ^{ 5 }C_{ 0 } & ^{ 5 }C_{ 3 } & 14 \\ ^{ 5 }C_{ 1 } & ^{ 5 }C_{ 4 } & 1 \\ ^{ 5 }C_{ 2 } & ^{ 5 }C_{ 5 } & 1 \end{matrix} \right| =\left| \begin{matrix} 1 & 10 & 14 \\ 5 & 5 & 1 \\ 10 & 1 & 1 \end{matrix} \right|$$

$$ =1\left( 5-1 \right) -10\left( 5-10 \right) +14\left( 5-50 \right)$$

$$ =4+50-630=-576$$

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Determinants Test - 34

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Determinants Test - 34

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Question 1

$$\begin{vmatrix}1 & 61 & -23\\ 13 & -26 & -26\\ -23 & -38 & -17\end{vmatrix}$$ and $$\begin{vmatrix}1 & 61 & -23\\ -13 & -26 & 26\\ -23 & 38 & -17\end{vmatrix}$$

$$\begin{vmatrix}1 & 61 & -23\\ 13 & -26 & -26\\ -23 & -38 & -17\end{vmatrix}$$ and $$\begin{vmatrix}1 & -61 & 23\\ -13 & -26 & 26\\ -23 & 38 & -17\end{vmatrix}$$

$$\begin{vmatrix}1 & 61 & -23\\ 13 & -26 & -26\\ -23 & -38 & -17\end{vmatrix}$$ and $$\begin{vmatrix}1 & 61 & 23\\ -13 & -26 & 26\\ -23 & 38 & -17\end{vmatrix}$$

None of these.

Solution

Question 2

231

213

321

312

Solution

Question 3

$$1$$

$$-1$$

$$0$$

cannot say anything

Solution

Question 4

$$\dfrac{-1}{2}$$

$$\dfrac{1}{2}$$

$$\dfrac{-1}{3}$$

none of these

Solution

Question 5

$$a+b$$

$$a+2b$$

$$2a+3b$$

$${ a }^{ 2 }$$

Solution

Question 6

$$2^{2006}$$

$$(-11) 2^{2005}$$

$$-2^{2005}$$

$$(-9)2^{2004}$$

Solution

Question 7

$$\displaystyle p+q+r$$

$$0$$

$$\displaystyle p-q-r$$

$$\displaystyle -p+q+r$$

Solution

Question 8

$$\displaystyle x=3, y=1$$

$$\displaystyle x=1, y=3$$

$$\displaystyle x=0, y=3$$

$$\displaystyle x=0, y=0$$

Solution

Question 9

non-singular

singular

skew-symmetric

symmetric

Solution

Question 10

$$0$$

$$\displaystyle -576$$

$$80$$

none of these

Solution

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