Determinants Test

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Determinants Test - 21

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Question 1
1 / -0

The points which are not collinear are:

A
$$(0, 1),\ (8, 3)\ and\ (6, 7)$$

B
$$(4, 3),\ (5, 1)\ and\ (1, 9)$$

C
$$(2, 5),\ (-1, 2)\ and\ (4, 7)$$

D
$$(-3, 2)\, (1, -2)\ and\ (9, -10)$$

Solution

$$ Let\quad us\quad take\quad three\quad points\quad A,\quad B\quad \& \quad C.\\ Three\quad points\quad A,\quad B\quad \& \quad C\quad will\quad be\quad collinear\\ If\quad AB+BC=AC.\\ Three\quad points\quad A,\quad B\quad \& \quad C\quad will\quad not\quad be\quad collinear\\ If\quad AB+BC=AC.\\ Let\quad us\quad investigate\quad thr\quad given\quad options.\\ Option\quad A\longrightarrow The\quad points\quad are\\ A(0,1)\quad B(8,3)\quad \& \quad C(6,7).\\ \therefore \quad AB=\sqrt { { \left( 0-8 \right) }^{ 2 }+{ \left( 1-3 \right) }^{ 2 } } units=\sqrt { 68 } =8.25units,\\ \quad \quad \quad BC=\sqrt { { \left( 8-6 \right) }^{ 2 }+{ \left( 3-7 \right) }^{ 2 } } units=\sqrt { 28 } =5.29units,\\ and\quad AC=\sqrt { { \left( 0-6 \right) }^{ 2 }+{ \left( 1-7 \right) }^{ 2 } } units=\sqrt { 72 } =8.49units.\\ \therefore \quad AB+BC=\left( 8.25+5.29 \right) units.=13.54units.\\ So\quad AB+BC\neq AC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad not\quad collinear.\\ Option\quad B\longrightarrow The\quad points\quad are\\ A(4,3)\quad B(5,1)\quad \& \quad C(1,9).\\ \therefore \quad AB=\sqrt { { \left( 4-5 \right) }^{ 2 }+{ \left( 3-1 \right) }^{ 2 } } units=\sqrt { 5 } =2.24units,\\ \quad \quad \quad BC=\sqrt { { \left( 5-1 \right) }^{ 2 }+{ \left( 1-9 \right) }^{ 2 } } units=\sqrt { 80 } =8.94units,\\ and\quad AC=\sqrt { { \left( 4-1 \right) }^{ 2 }+{ \left( 3-9 \right) }^{ 2 } } units=\sqrt { 45 } =6.71units.\\ \therefore \quad AB+AC=\left( 2.24+6.71 \right) units=8.95units.\\ So\quad AB+AC=BC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad \quad collinear.\\ Option\quad C\longrightarrow The\quad points\quad are\\ A(2,5)\quad B(-1,2)\quad \& \quad C(4,7).\\ \therefore \quad AB=\sqrt { { \left( 2+1 \right) }^{ 2 }+{ \left( 5-2 \right) }^{ 2 } } units=\sqrt { 18 } units=4.24units,\\ \quad \quad \quad BC=\sqrt { { \left( -1-4 \right) }^{ 2 }+{ \left( 2-7 \right) }^{ 2 } } units=\sqrt { 50 } units=7.07units,\\ and\quad AC=\sqrt { { \left( 2-4 \right) }^{ 2 }+{ \left( 5-7 \right) }^{ 2 } } units=\sqrt { 8 } units=2.83units.\\ \therefore \quad AB+AC=\left( 4.24+2.83 \right) units=7.07units.\\ So\quad AB+AC=BC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad \quad collinear.\\ Option\quad C\longrightarrow The\quad points\quad are\\ A(-3,2)\quad B(1,-2)\quad \& \quad C(9,-10).\\ \therefore \quad AB=\sqrt { { \left( -3-1 \right) }^{ 2 }+{ \left( 2+2 \right) }^{ 2 } } units=\sqrt { 32 } units=5.66units,\\ \quad \quad \quad BC=\sqrt { { \left( 1-9 \right) }^{ 2 }+{ \left( -2+10 \right) }^{ 2 } } units=\sqrt { 128 } units=11.31units,\\ and\quad AC=\sqrt { { \left( -3-9 \right) }^{ 2 }+{ \left( 2+10 \right) }^{ 2 } } units=\sqrt { 288 } units=16.97units.\\ \therefore \quad AB+BC=\left( 5.66+11.31 \right) units=16.97units.\\ So\quad AB+BC=AC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad \quad collinear.\\ Ans-\quad Option\quad A.\\ \\ \\ $$
Question 2
1 / -0

If $$A=\begin{bmatrix} a & b \\ b & a \end{bmatrix}$$, then $$\left| A+{ A }^{ T } \right| $$ equals

A
$$4({a}^{2}-{b}^{2})$$

B
$$2({a}^{2}-{b}^{2}$$

C
$${a}^{2}-{b}^{2})$$

D
$$4ab$$

Solution

$$A+{A}^{T}=\begin{bmatrix} a & b \\ b & a \end{bmatrix}\begin{bmatrix} a & b \\ b & a \end{bmatrix}=\begin{bmatrix} 2a & 2b \\ 2b & 2a \end{bmatrix}$$

$$\left| A+{ A }^{ T } \right| =4{ a }^{ 2 }-4{ b }^{ 2 }\quad $$

Ans: A
Question 3
1 / -0

If $$\Delta_1=\begin{vmatrix} 1 & 0\\ a & b\end{vmatrix}$$ and $$\Delta_2=\begin{vmatrix} 1 & 0\\ c & d\end{vmatrix}$$ then $$\Delta_2 \Delta_1$$ is equal to

A
$$ac$$

B
$$bd$$

C
$$(b-a)(d-c)$$

D
none of these

Solution

$$\Delta_1=\begin{vmatrix} 1 & 0\\ a & b\end{vmatrix}=1.b-0.a=b$$
$$\Delta_2=\begin{vmatrix} 1 & 0\\ c & d\end{vmatrix}=1.d-0.c=d$$
then $$\Delta_2 \Delta_1=bd$$

Ans: B
Question 4
1 / -0

If $$A$$ and $$B$$ are two matrices of same order $$3\times 3$$, where
$$A=\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 5 & 6 & 8 \end{bmatrix}$$ and $$B=\begin{bmatrix} 3 & 2 & 5 \\ 2 & 3 & 8 \\ 7 & 2 & 9 \end{bmatrix}$$
The value of $$\text{Adj} (\text{Adj}\, A)$$ is equal to

A
$$-A$$

B
$$A$$

C
$$8A$$

D
$$16A$$

Solution

$$\left| A \right| =1\left( 3\times8-4\times6 \right) -2\left( 2\times8-5\times4 \right) +3\left( 2\times6-5\times3 \right) =-1$$
Use following results
$$\text{Adj} (\text{Adj}\, A)={ \left| A \right| }^{ n-2 }A \Rightarrow \text{Adj}(\text{Adj}\,A)={ \left| A \right| }^{ n-2 }A={ \left( -1 \right) }^{ 1 }A=-A$$

Ans: $$A$$
Question 5
1 / -0

If $$\omega$$ is a cube root of unity and $$\Delta=\begin{vmatrix}1 & 2\omega \\ \omega & \omega^2\end{vmatrix}$$, then $$\Delta^2$$ is equal to

A
$$-\omega$$

B
$$\omega$$

C
$$1$$

D
$$\omega^2$$

Solution

$$\Delta=\omega^2-2\omega^2=-\omega^2$$
$$\Delta^2=\omega^4=\omega$$

Ans: B
Question 6
1 / -0

If $$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 3x & 6 \end{vmatrix}$$, then $$x$$ is equal to

A
$$6$$

B
$$\pm 6$$

C
$$-6$$

D
$$0$$

Solution

$$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 3x & 6 \end{vmatrix}$$
$$\Rightarrow x^2-36=36-6x$$
$$\Rightarrow x^2+6x-72=0$$
$$\Rightarrow \displaystyle x=\frac{-6\pm\sqrt{6^2+4\times 72}}{2}$$
$$\quad \displaystyle =\frac{-6\pm \sqrt{36(1+8)}}{2}$$
$$\quad =-3\pm 9=6\ or\ -12$$
Question 7
1 / -0

Which of the following is correct?

A
Determinant is a square matrix

B
Determinant is a number associated to a matrix

C
Determinant is a number associated to a square matrix

D
None of these

Solution

Determinant is defined only for a square matrix.
and its denotes the value of that square matrix.
Question 8
1 / -0

If $$|A| \displaystyle \neq 0$$, then $$A$$ is

A
zero matrix

B
singular matrix

C
non - singular matrix

D
diagonal matrix

Solution

If $$|A| \neq 0$$ then by definition, $$A$$ is non-singular matrix
Question 9
1 / -0

What is the determinant of the matrix $$\left [\begin{matrix} 3& 6\\ -1 & 2\end {matrix} \right]$$?

A
$$0$$

B
$$12$$

C
$$|0|$$

D
$$|6|$$

Solution

Given,
$$\begin{vmatrix} 3 & 6 \\ -1 & 2 \end{vmatrix}$$
Let determinent be $$\left| d \right| $$
Value of $$\left| d \right| $$ will be
$$\left| d \right| $$$$=$$$$3\times 2-\left( 6\times -1 \right) $$
$$=6+6=12$$
Question 10
1 / -0

If $$A = \begin{bmatrix} 1& \log_{b}a\\ \log_{a}b & 1\end{bmatrix}$$ then $$|A|$$ is equal to

A
$$0$$

B
$$\log_{a}b$$

C
$$-1$$

D
$$\log_{b}a$$

Solution

On solving the given matrix,
$$|A| = 1 - \log_{a}b . \log_{b}a = 1 - 1 = 0$$

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Determinants Test - 21

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Determinants Test - 21

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Question 1

$$(0, 1),\ (8, 3)\ and\ (6, 7)$$

$$(4, 3),\ (5, 1)\ and\ (1, 9)$$

$$(2, 5),\ (-1, 2)\ and\ (4, 7)$$

$$(-3, 2)\, (1, -2)\ and\ (9, -10)$$

Solution

Question 2

$$4({a}^{2}-{b}^{2})$$

$$2({a}^{2}-{b}^{2}$$

$${a}^{2}-{b}^{2})$$

$$4ab$$

Solution

Question 3

$$ac$$

$$bd$$

$$(b-a)(d-c)$$

none of these

Solution

Question 4

$$-A$$

$$A$$

$$8A$$

$$16A$$

Solution

Question 5

$$-\omega$$

$$\omega$$

$$1$$

$$\omega^2$$

Solution

Question 6

$$6$$

$$\pm 6$$

$$-6$$

$$0$$

Solution

Question 7

Determinant is a square matrix

Determinant is a number associated to a matrix

Determinant is a number associated to a square matrix

None of these

Solution

Question 8

zero matrix

singular matrix

non - singular matrix

diagonal matrix

Solution

Question 9

$$0$$

$$12$$

$$|0|$$

$$|6|$$

Solution

Question 10

$$0$$

$$\log_{a}b$$

$$-1$$

$$\log_{b}a$$

Solution

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