Determinants Test

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Determinants Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

$$A = \begin{bmatrix}1&2\\3&4\end{bmatrix}, B = \begin{bmatrix}2&1\\3&4\end{bmatrix}$$ then $$\left|(B^TA^T)^{-1}\right|$$ is equal to

A
$$10$$

B
$$\dfrac{1}{10}$$

C
$$1$$

D
$$- 1$$

Solution

$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ $$B = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$$

$$B^{T}A^{T} = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 2 + 6 & 6 + 12 \\ 1 + 8 & 3 + 16 \end{bmatrix} = \begin{bmatrix} 8 & 18 \\ 9 & 19 \end{bmatrix}$$

$$det (B^{T}A^{T}) = 8 \times 19 - 9 \times 18 = 152 - 162 = -10$$
$$(B^TA^T)^{-1} = \dfrac1{det(B^TA^T)} adj(B^TA^T) = \dfrac{1}{-10}\begin{bmatrix}19&-18\\-9&8\end{bmatrix}$$

$$det( (B^{T}A^{T})^{-1}) =\dfrac{19\times8 - 18\times9}{-10} = 1$$
Question 2
1 / -0

If $$\begin{vmatrix}a&a^3 &a^4-1\\b&b^3&b^4-1\\c&c^3&c^4-1\end{vmatrix}=0$$ and $$a,b,c$$ are all distinct then $$abc (ab+bc+ca)$$ is equal to

A
$$a+b+c$$

B
$$abc$$

C
$$0$$

D
none of these

Solution

Given, $$\begin{vmatrix}a&a^3 &a^4-1\\b&b^3&b^4-1\\c&c^3&c^4-1\end{vmatrix}=0$$
Doing Row operation $$R_1\to R_1-R_2,R_3\to R_3-R_2$$
$$\begin{vmatrix}a-b&a^3-b^3 &a^4-b^4\\b&b^3&b^4-1\\c-b&c^3-b^3&c^4-b^4\end{vmatrix}=0$$
$$\begin{vmatrix}a-b&(a-b)(a^2+b^2+ab) &(a^2+b^2)(a+b)(a-b)\\b&b^3&b^4-1\\c-b&(c-b)(b^2+c^2+cb)&(c^2+b^2)(c+b)(c-b)\end{vmatrix}=0$$
Taking out $$(a-b)$$ and $$(c-b)$$ from the determinant and Doing Row operations $$R_1 \to R_1-R_3$$
$$(a-b)(c-b)\begin{vmatrix}0&a^2-c^2+b(a-c) &(a^3-c^3)+b^2(a-c)+b(a^2-c^2)\\b&b^3&b^4-1\\1&(b^2+c^2+cb)&(c^2+b^2)(c+b)\end{vmatrix}=0$$
$$(a-b)(c-b)(a-c)\begin{vmatrix}0&a+c+b &(a^2+c^2+ac)+b^2+b(a+c)\\b&b^3&b^4-1\\1&(b^2+c^2+cb)&(c^2+b^2)(c+b)\end{vmatrix}=0$$
Since $$a,b,c$$ are distinct.
$$\begin{vmatrix}0&a+c+b &(a^2+c^2+ac)+b^2+b(a+c)\\b&b^3&b^4-1\\1&(b^2+c^2+cb)&(c^2+b^2)(c+b)\end{vmatrix}=0$$
Doing Row operation $$R_2 \to R_2-bR_3$$
$$\begin{vmatrix}0&a+c+b &(a^2+c^2+ac)+b^2+b(a+c)\\0&-bc^2-cb^2&-bc^3-c^2b^2-b^3c-1\\1&(b^2+c^2+cb)&(c^2+b^2)(c+b)\end{vmatrix}=0$$
Doing Row operation $$R_2 \to R_2+cbR_1$$
$$\begin{vmatrix}0&a+c+b &(a^2+c^2+ac)+b^2+b(a+c)\\0&abc&abc(a+b+c)-1\\1&(b^2+c^2+cb)&(c^2+b^2)(c+b)\end{vmatrix}=0$$
Expanding Determinant, w eget
$$abc(a+b+c)^2-(a+b+c)-abc(a^2+b^2+c^2)-abc(ab+bc+ac)=0$$
$$\Rightarrow abc(ab+bc+ac)=abc((a+b+c)^2-a^2-b^2-c^2)-(a+b+c)$$
$$\Rightarrow abc(ab+bc+ca)=abc(2(ab+bc+ac))-(a+b+c)$$
$$\Rightarrow abc(ab+bc+ac)=a+b+c$$
Hence, option A is correct.
Question 3
1 / -0

If three points $$(k, 2k), (2k, 3k), (3, 1)$$ are collinear, then $$k$$ is equal to:

A
$$-2$$

B
$$1$$

C
$$\displaystyle\frac{1}{2}$$

D
$$-\displaystyle\frac{1}{2}$$

Solution

If given points are collinear, then
$$\begin{vmatrix} k&2k&1\\2k&3k&1\\3&1&1\end{vmatrix}=0$$
$$\Rightarrow \begin{vmatrix} k&2k&1\\k&k&0\\3&1&1\end{vmatrix}=0, R_2\rightarrow R_2-R_1$$
$$\Rightarrow 1(k-3k)+1(k^2-2k^2)=0$$
$$\Rightarrow k^2+2k=0$$
$$\Rightarrow k=0,-2$$
Question 4
1 / -0

$$\left|\begin{array}{lllll}
0 & & \mathrm{c}\mathrm{o}\mathrm{s}\alpha & \mathrm{c}\mathrm{o}\mathrm{s} & \beta\\
\mathrm{c}\mathrm{o}\mathrm{s} & \alpha & 0 & \mathrm{c}\mathrm{o}\mathrm{s} & \gamma\\
\mathrm{c}\mathrm{o}\mathrm{s} & \beta & \mathrm{c}\mathrm{o}\mathrm{s}\gamma & 0 &
\end{array}\right|=$$

A
$$\cos\alpha+\cos\beta+\cos\gamma$$

B
$$\cos\alpha\cos\beta\cos\gamma$$

C
$$ 2\cos\alpha\cos\beta\cos\gamma$$

D
$$2 \displaystyle \sum \cos\alpha\cos\beta$$

Solution

Let $$A=\begin{vmatrix}
0 & cos \alpha & \cos \beta\\
\cos \alpha & 0 & cos \gamma\\
cos \beta & \cos \gamma & 0
\end{vmatrix}$$

Expanding the determinant,
$$A=-cos \alpha[-cos \beta cos \gamma]+cos \beta[cos \alpha \cos \gamma]$$
$$=2 \cos \alpha \cos \beta \cos \gamma$$
Question 5
1 / -0

If A $$ =\begin{bmatrix}
0 & c &-b \\
-c& 0& a\\
b & -a & 0
\end{bmatrix}$$ then $$\left ( a^{2}+b^{2}-c^{2} \right )\left | A \right |=$$

A
$$abc$$

B
$$a + b + c$$

C
$$\left ( a^{3}+b^{3}+c^{3} \right )$$

D
$$0$$

Solution

$$|A|=0\times(a^2)-c(-ab)-b(ac)$$
$$=0+abc-abc$$
$$=0$$
Question 6
1 / -0

$$\begin{vmatrix}
x^{2}+3 &x-1 &x+3 \\
x+3 & -2x &x-4 \\
x-3& x+4 & 3x
\end{vmatrix}$$ $$=px^{4}+qx^3+rx^{2}+sx+t,$$ then $$t = $$

A
$$72$$

B
$$0$$

C
$$24$$

D
$$-48$$

Solution

On expanding the determinant we get,

$$(x^2+3)(-6x^2-x^2+16)-(x-1)(3x^2+9x-x^2+7x-12)+(x+3)(x^2+7x+12-2x^2+6x)$$

Let us consider only the constant terms,

since $$t$$ is constant term

$$\therefore (16)(3)-(-1)(-12)+3(12) =72$$

Hence, $$t = 72 $$
Question 7
1 / -0

Maximum value of a second order determinant whose every element is either 0,1 or 2 only is:

A
0

B
1

C
2

D
4

Solution

So, $$A=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
$$ Given a,b,c & D can only be 0,1,2
det A= ad-bc
So for max. value of A,
a=2 and d=2 and $$b,c \epsilon {0,0}$$
So, Max value of det $$A=\begin{bmatrix}
2 & 0\\
0 & 2
\end{bmatrix}=4$$
Question 8
1 / -0

If abc $$\neq $$0 and if $$\begin{vmatrix}
a & b & c\\
b & c & a\\
c & a & b
\end{vmatrix}$$ = 0 then $$\dfrac{a^{3}+b^{3}+c^{3}}{abc}$$

A
3

B
-3

C
2

D
-2

Solution

$$\begin{vmatrix}
a & b & c\\
b & c & a\\
c & a & b
\end{vmatrix}$$
$$=a(bc-a^2)-b(b^2-ac)+c(ab-c^2)$$
$$=3abc-a^3-b^3-c^3$$
Given,
$$\begin{vmatrix}
a & b & c\\
b & c & a\\
c & a & b
\end{vmatrix} = 0$$

So, $$\dfrac{a^3+b^3+c^3}{abc}=3$$
Question 9
1 / -0

If P =$$\begin{bmatrix}
1 & 4\\
2 & 6
\end{bmatrix}$$ ,then adj (P)

A
$$\begin{bmatrix}1 & 4\\ 2 & 6\end{bmatrix}$$

B
$$\begin{bmatrix}6& -4\\ -2 & 1\end{bmatrix}$$

C
$$\begin{bmatrix}6& -2\\ -4 & 1\end{bmatrix}$$

D
$$\begin{bmatrix}2& 1\\ 6 & 4\end{bmatrix}$$

Solution

Cofactor of the martix is
$$\begin{bmatrix} M11 & -M12 \\ -M21 & M22 \end{bmatrix}\\$$
Where$$ M11 = 6$$
$$M12 = 2$$
$$M21 = 4$$
$$M22 = 1$$
Substituting all value we get
Cofactor of the matrix as
$$\begin{bmatrix} 6 & -2 \\ -4 & 1 \end{bmatrix}\\$$
The adjoint of the matrix is transpose of cofactor of the given matrix
$$\therefore $$ Adjoint of the matrix is
$$\begin{bmatrix} 6 & -4 \\ -2 & 1 \end{bmatrix}$$
Hence the answer is option B.
Question 10
1 / -0

If $$A=\left\{\begin{array}{lll}
a & b & c\\
b & c & a\\
c & a & b
\end{array}\right\}$$ then cofactor of $$\mathrm{a}_{21}$$ is:

A
$$\mathrm{b}^{2}-$$ ac

B
$$ac-b^2$$

C
$$\mathrm{a}^{2}-\mathrm{b}\mathrm{c}$$

D
$$\mathrm{b}\mathrm{c}-\mathrm{a}^{2}$$

Solution

By property of co-factor of matrix(2),
cofactor of $$a_{ij}=M_{ij}\times (-1)^{i+j}$$
$$\therefore$$ minor of $$a_{21}=\begin{vmatrix}
a_{12} & a_{13}\\
a_{32} & a_{33}
\end{vmatrix}=\begin{vmatrix}
b & c\\
a & b
\end{vmatrix}$$
$$M_{21}=b^2-ac$$
So, co factor of $$a_{21}=M_{21}\times(-1)^{1+2}=-M_{21}$$
$$=ac-b^2$$

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Determinants Test - 18

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Determinants Test - 18

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SHARING IS CARING

Question 1

$$10$$

$$\dfrac{1}{10}$$

$$1$$

$$- 1$$

Solution

Question 2

$$a+b+c$$

$$abc$$

$$0$$

none of these

Solution

Question 3

$$-2$$

$$1$$

$$\displaystyle\frac{1}{2}$$

$$-\displaystyle\frac{1}{2}$$

Solution

Question 4

$$\cos\alpha+\cos\beta+\cos\gamma$$

$$\cos\alpha\cos\beta\cos\gamma$$

$$ 2\cos\alpha\cos\beta\cos\gamma$$

$$2 \displaystyle \sum \cos\alpha\cos\beta$$

Solution

Question 5

$$abc$$

$$a + b + c$$

$$\left ( a^{3}+b^{3}+c^{3} \right )$$

$$0$$

Solution

Question 6

$$72$$

$$0$$

$$24$$

$$-48$$

Solution

Question 7

0

1

2

4

Solution

Question 8

3

-3

2

-2

Solution

Question 9

$$\begin{bmatrix}1 & 4\\ 2 & 6\end{bmatrix}$$

$$\begin{bmatrix}6& -4\\ -2 & 1\end{bmatrix}$$

$$\begin{bmatrix}6& -2\\ -4 & 1\end{bmatrix}$$

$$\begin{bmatrix}2& 1\\ 6 & 4\end{bmatrix}$$

Solution

Question 10

$$\mathrm{b}^{2}-$$ ac

$$ac-b^2$$

$$\mathrm{a}^{2}-\mathrm{b}\mathrm{c}$$

$$\mathrm{b}\mathrm{c}-\mathrm{a}^{2}$$

Solution

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