Application of Determinants MCQ Quiz - Objective Question with Answer for Application of Determinants - Download Free PDF

Concept 

The area of a triangle given its three vertices is 

$A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$   

Explanation:

$A=\frac{1}{2}[|1(5-3)+2(3-2)+4(2-5)|]$  

$A=\frac{1}{2}|1\times 2+2\times 1+4\times (-3)|$

$A=\frac{1}{2}|2+2-12|$  

$A=\frac{1}{2}|-8|$  

$A=\frac{1}{2}\times 8=4$   

Hence Option(2) is the correct answer.

Concept 

The area of a triangle given its three vertices is 

$A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$   

Explanation:

$A=\frac{1}{2}[|1(5-3)+2(3-2)+4(2-5)|]$  

$A=\frac{1}{2}|1\times 2+2\times 1+4\times (-3)|$

$A=\frac{1}{2}|2+2-12|$  

$A=\frac{1}{2}|-8|$  

$A=\frac{1}{2}\times 8=4$   

Hence Option(2) is the correct answer.

Calculation:

Given:

System of linear equations:

$2x+y-z=0$

$4x-py+4z=4$

$x-y+z=q$

$p,q\in I$ and $p,q\in [1,10]$

The coefficient matrix A is:

$A=\left[\begin{array}{ccc}2& 1& -1\\ 1& -\frac{p}{4}& 1\\ 1& -1& 1\end{array}\right]$

The augmented matrix [A|B] is:

$[A|B]=\left[\begin{array}{cccc}2& 1& -1& 0\\ 1& -\frac{p}{4}& 1& 1\\ 1& -1& 1& q\end{array}\right]$

Calculate the determinant of A, |A|:

$|A|=2(-\frac{p}{4}+1)-1(1-1)-1(-1+\frac{p}{4})$

$|A|=-\frac{p}{2}+2+0+1-\frac{p}{4}$

$|A|=3-\frac{3p}{4}$

⇒ For unique solution, $|A|\ne 0$:

$3-\frac{3p}{4}\ne 0$

$3\ne \frac{3p}{4}$

$p\ne 4$

For no solution or infinite solutions, $|A|=0$, so $p=4$.

⇒ If $p=4$, the system becomes:

$2x+y-z=0$

$x-y+z=1$

$x-y+z=q$

⇒ From the second and third equations, for a solution to exist, $1=q$.

⇒ If $p=4$ and $q=1$, infinite solutions exist.

⇒ If $p=4$ and $q\ne 1$, no solution exists.

(I) Unique solution: $p\ne 4$. $p$ can take 9 values (1 to 10 except 4). $q$ can take 10 values. Total pairs: 9 × 10 = 90.

(II) No solution: $p=4$ and $q\ne 1$. 9 values for $q$. 1 pair for $p$. Total pairs: 1 × 9 = 9.

(III) Infinite solutions: $p=4$ and $q=1$. 1 pair only.

(IV) At least one solution: Total pairs - No solution pairs = 100 - 9 = 91.

∴ (I) - (S), (II) - (Q), (III) - (P), (IV) - (R)

Given:

Points are (5, 2, 4), (6, -1, 2) & (8, -7, k)

Concept used:

Area of the triangle formed by using these points = 0

$\frac{1}{2}$ ×= 0

Calculation:

⇒ 5(- k + 14) - 2(6k - 16) + 4(- 42 + 8) = 0

⇒ - 5k + 70 - 12k + 32 + 4(- 34) = 0

⇒ -17k + 102 - 136 = 0

⇒ -17k - 34 = 0

⇒ -17k = 34

⇒ k = -2

Hence the correct answer is "-2".

$\mathrm{\Delta }=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 4\\ 1& 4& 10\end{array}\right|=1(20-16)-1(10-4)+1(4-2)$

= 4 – 6 + 2 = 0

For infinite solutions

Δ_x = Δ_y = Δ_z = 0

m² – 3x + 2 = 0

m = 1, 2 

α = 1, β = 2

${\displaystyle \therefore \sum _{\mathrm{n}=1}^{10}({\mathrm{n}}^{\alpha }+{\mathrm{n}}^{\beta })=\sum _{\mathrm{n}=1}^{10}{\mathrm{n}}^1+\sum _{\mathrm{n}=1}^{10}{\mathrm{n}}^2}$

$=\frac{10(11)}{2}+\frac{10(11)(21)}{6}$

= 55 + 385

= 440

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

$\Rightarrow \left[\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}{d}_1\\ d_2\\ d_3\end{array}\right]$

⇒ AX = B

⇒ X = A^-1 B = $\frac{\mathrm{a}\mathrm{d}\mathrm{j}\left(\mathrm{A}\right)}{det(\mathrm{A})}B$

⇒ If det (A) ≠ 0, the system is consistent having unique solution.

⇒ If det (A) = 0 and (adj A). B = 0, system is consistent, with infinitely many solutions.

⇒ If det (A) = 0 and (adj A). B ≠ 0, system is inconsistent (no solution)

Calculation:

Given: 

kx + y + z = 1, x + ky + z = k and x + y + kz = k²

$\Rightarrow \mathrm{A}=\left[\begin{array}{ccc}\mathrm{k}& 1& 1\\ 1& \mathrm{k}& 1\\ 1& 1& \mathrm{k}\end{array}\right],\mathrm{B}=\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{C}=\left[\begin{array}{c}1\\ \mathrm{k}\\ {\mathrm{k}}^2\end{array}\right]$

⇒ For the given equations to have no solution, |A| = 0

$\Rightarrow \left|\begin{array}{ccc}\mathrm{k}& 1& 1\\ 1& \mathrm{k}& 1\\ 1& 1& \mathrm{k}\end{array}\right|=0$

⇒ k(k² – 1) - 1(k – 1) + 1(1 – k) = 0

⇒ k³ – k – k + 1 + 1 – k = 0

⇒ k³ -3k +2 = 0

⇒ (k – 1) (k – 1) (k + 2) = 0

⇒ k = 1, -2

If we put k = 1 in the above given equations, then all the equations will become the same.

Hence, the given equations have no solution if k = - 2. 

Concept:

Area of a triangle with vertices (x1, y1) , (x2, y2), (x3, y3) is given by

​Area = ${\displaystyle \frac{1}{2}}\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ x_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|$

Calculations:

Given that, vertices of triangle are (-3, 0), (3, 0) and (0, k)

By using the above formula,

​⇒ Area = ${\displaystyle \frac{1}{2}}\left|\begin{array}{ccc}-3& 0& 1\\ 3& 0& 1\\ 0& k& 1\end{array}\right|$

⇒ Area = ${\displaystyle \frac{1}{2}}$[-3(0 - k) - 0 + 1(3k)]

⇒ Area = 3k

According to the question, area of triangle is 9 square unit,

⇒ 3k = 9

⇒ k = 3

∴ Required value of k is 3 unit.

Concept:

Area of equilateral triangle = $\frac{\sqrt{3}}{4}$×a2

Calculation:

Given: The co-ordinates of its vertices are (x1, y1); (x2, y2): (x3, y3) then the square of the determinant $\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

⇒ (△ ABC ) = ${\displaystyle \frac{1}{2}}$ $\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$ = ( $\frac{\sqrt{3}}{4}$) ×a²

On squaring both side, 

⇒ ${\displaystyle \frac{1}{4}}$  ${\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|}^2$ = ${\displaystyle \frac{3a^4}{16}}$

⇒ ${\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|}^2$ = ${\displaystyle \frac{3a^4}{4}}$

Concept:

Properties of determinants:​

For a n×n matrix A, det(kA) = kn det(A).

Calculation:

Given:

|A| = 5

k = 5

From the properties of the determinants, we know that |KA| = Kn |A|, where n is the order of the determinant.

Here, n = 2, therefore, the answer is K2 |A|.

|5A| = 52|A|

|5A| = 52 × 5 = 125

Concept:

If $({\mathrm{x}}_1,{\mathrm{y}}_1),({\mathrm{x}}_2,{\mathrm{y}}_2)$ and $({\mathrm{x}}_3,{\mathrm{y}}_3)$ are the vertices of a triangle then, 

Area = ${\displaystyle \frac{1}{2}}\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|$

Properties of Determinants

$\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& kx\\ {\mathrm{x}}_2& {\mathrm{y}}_2& ky\\ {\mathrm{x}}_3& {\mathrm{y}}_3& kz\end{array}\right|=\mathrm{k}\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& x\\ {\mathrm{x}}_2& {\mathrm{y}}_2& y\\ {\mathrm{x}}_3& {\mathrm{y}}_3& z\end{array}\right|$

Calculations:

If $({\mathrm{x}}_1,{\mathrm{y}}_1),({\mathrm{x}}_2,{\mathrm{y}}_2)$ and $({\mathrm{x}}_3,{\mathrm{y}}_3)$ are the vertices of a triangle then,

Area = ${\displaystyle \frac{1}{2}}\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|$

Given: the area is ‘k’ square units

⇒ k = ${\displaystyle \frac{1}{2}}\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|$

⇒ $\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|=2\mathrm{k}$

Now,  ${\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 4\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 4\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 4\end{array}\right|}^2$

= $\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 4\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 4\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 4\end{array}\right|\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 4\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 4\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 4\end{array}\right|$

= $4\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|.4\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|$

= 16.(2k)(2k)

= 64k²

Concept:

If $({\mathrm{x}}_1,{\mathrm{y}}_1),({\mathrm{x}}_2,{\mathrm{y}}_2)$ and $({\mathrm{x}}_3,{\mathrm{y}}_3)$ are the vertices of a triangle then, 

Area = ${\displaystyle \frac{1}{2}}\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|$

Note: Area is always a positive quantity, therefore we always take the absolute value of the determinant for the area. 

Calculation:

Given vertices are A (1, 1) ,B ( 6, 0) and C ( 3, 2).

We know that area of triangle ABC is given by, 

Δ = ${\displaystyle \frac{1}{2}}\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|$

⇒ Δ = $\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ 6& 0& 1\\ 3& 2& 1\end{array}\right|$ 

⇒ Δ = $\frac{1}{2}[1(0-2)-1(6-3)+1(12-0)]$ 

⇒ Δ =  $\frac{7}{2}$  

The correct option is 1.

Concept:

The area of a triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃), is given by the expression

$\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& x_2& 1\\ y_1& y_2& 1\\ z_1& z_2& 1\end{array}\right|$  

Calculation:

Given, Area of triangle, A = 4 sq. unit and vertices (K, 0), (4, 0), (0, 2) 

since the area is always Positive But the determinant can be both positive and negative.  

∴ Δ = ± 4 . 

⇒ ± 4 = $=\frac{1}{2}\left|\begin{array}{ccc}k& 0& 1\\ 4& 0& 1\\ 0& 2& 1\end{array}\right|$  

⇒ ± 4  $=\frac{1}{2}[\mathrm{k}(0-2)-0(4-0)+1(8-0)]$ 

⇒ ± 8 = -2k + 8 

So, 8 = -2k + 8 or  -8 = -2k +8 

⇒ k = 0 or  8  .

The correct option is 2.

Concept:

If the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area (A) of ΔABC is given as; 

$A=\frac{1}{2}\cdot \left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Calculation: 

Here, we have to find the value of a for  which the points (5, -2), (8, -3) and (a, -12) are collinear

Let,

x1 = 5, y1 = -2,

x2 = 8, y2 = -3,

x3 = a, y3 = -12.

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a Δ ABC then area of Δ ABC =   $A=\frac{1}{2}\cdot \left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

$\Rightarrow A=\frac{1}{2}\cdot \left|\begin{array}{ccc}5& -2& 1\\ 8& -3& 1\\ a& -12& 1\end{array}\right|$

2A = 5 (-3 + 12) + 2(8 - a) + 1(-96 + 3a)

2A = 45 + 16 - 2a - 96 + 3a

2A = a - 35  

⇒ A = (a - 35)/2

∵ The given points are collinear.

As we know that, if the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

⇒ (a - 35)/2 = 0

⇒ a = 35

Hence, option D is the correct answer.

Alternate Method 

Concept:

Three or more points are collinear if the slope of any two pairs of points is the same.

Calculation:

Let, A = (5, -2), B = (8, -3), C = (a, -12)

Now, the slope of AB = Slope of BC = Slope of AC      (∵ points are collinear)

 $$\begin{gathered} \frac{-3-(-2)}{8-5}=\frac{-12-(-3)}{\mathrm{a}-8} \\ \Rightarrow \frac{-1}{3}=\frac{-9}{\mathrm{a}-8} \end{gathered}$$

⇒ a - 8= 27

⇒ a = 27 + 8 = 35

Hence, option (4) is correct.

Explanation:

Given:

Δ = $\left|\begin{array}{lll}{\text{a}}_{11}& {\text{a}}_{12}& {\text{a}}_{13}\\ {\text{a}}_{21}& {\text{a}}_{22}& {\text{a}}_{23}\\ {\text{a}}_{31}& {\text{a}}_{32}& {\text{a}}_{33}\end{array}\right|$

 a13 = yz, a23 = zx, a33 = xy and M₁₃ = (z − y), M₂₃ = (z − x), M₃₃ = (y − x)

Expanding the determinant along Column 3,

⇒ Δ = a13 (a₂₁a₃₂ - a₂₂a₃₁) - a23 (a₁₁a₃₂ - a₁₂a₃₁) + a33 (a11a2 - a12a1)

⇒ Δ = a₁₃M₁₃ - a₂₃M₂₃ + a₃₃M₃₃

where M_ij denotes the minor of a_ij element.

⇒ Δ = yz(z - y) - zx(z - x) + xy(y - x)

⇒ Δ = yz(z - y) - z²x + zx² + xy² - x²y

⇒ Δ = yz(z - y) + (xy2 - z2x) + (zx2 - x2y)

⇒ Δ = yz(z - y) + x(y2 - z2) + x²(z - y)

⇒ Δ = (z - y)[yz - x(y + z) + x2]

⇒ Δ = (z - y)[yz - xy - xz + x2]

⇒ Δ = (z - y)[y(z - x) - x(z - x)]

⇒ Δ = (z - y)[(z - x)(y - x)]

Concept:

Consider three linear eqaution in two variable:

a₁x + b₁y + c_{1 = 0}

a₂x + b₂y + c₂ = 0

a₃x + b₃y + c₃ = 0

Condition for the consistency of three simultaneous linear equations in 2 variables:

​​​​$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0$

Formula for Quadratic equation:

ax² + bx + c = 0

x = $\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{a}\mathrm{c}}}{2\mathrm{a}}$

Calculation:

2k2x + 3y - 1 = 0      ....(1)

7x - 2y + 3 = 0      ....(2)

6kx + y + 1 = 0      ....(3)

For consistency of given simultaneous equation,

$\left|\begin{array}{ccc}2k^2& 3& -1\\ 7& -2& 3\\ 6k& 1& 1\end{array}\right|=0$

⇒ 2k²(-2- 3) - 3(7 - 18k) - 1(7 + 12k) = 0

⇒ -10k² - 21 + 54k - 7 - 12k = 0

⇒ -10k² + 42k - 28 =  0

⇒ 5k2 - 21k + 14 =  0

By using the formula,

$x=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{a}\mathrm{c}}}{2\mathrm{a}}$

$\Rightarrow k=\frac{-(-21)\pm \sqrt{(-21{)}^2-4(5)(14)}}{2\times 5}$

$\therefore k=\frac{21\pm \sqrt{161}}{10}$