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:

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

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

\(A = \frac{1}{2} | 2 + 2 - 12| \)  

\(A = \frac{1}{2} |-8| \)  

\(A = \frac{1}{2} × 8 = 4 \)   

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:

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

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

\(A = \frac{1}{2} | 2 + 2 - 12| \)  

\(A = \frac{1}{2} |-8| \)  

\(A = \frac{1}{2} × 8 = 4 \)   

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:

\(A = \begin{bmatrix} 2 & 1 & -1 \\ 1 & -\frac{p}{4} & 1 \\ 1 & -1 & 1 \end{bmatrix}\)

The augmented matrix [A|B] is:

\([A|B] = \begin{bmatrix} 2 & 1 & -1 & 0 \\ 1 & -\frac{p}{4} & 1 & 1 \\ 1 & -1 & 1 & q \end{bmatrix}\)

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

\(1\over2\) ×= 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".

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

= 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

\(\rm \displaystyle \therefore \sum_{n=1}^{10}\left(n^{\alpha}+n^{\beta}\right)=\sum_{n=1}^{10} n^{1}+\sum_{n=1}^{10} n^{2} \)

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

= 55 + 385

= 440

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

\(\; \Rightarrow \left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} {{d_1}}\\ {{d_2}}\\ {{d_3}} \end{array}} \right]\)

⇒ AX = B

⇒ X = A^-1 B = \(\frac{{{\rm{adj\;}}\left( {\rm{A}} \right)}}{{\det {\rm{\;}}({\rm{A}})}}\;B\)

⇒ 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

\( \Rightarrow {\rm{A}} = {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{k}}&1&1\\ 1&{\rm{k}}&1\\ 1&1&{\rm{k}} \end{array}} \right],{\rm{B}} = {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right]{\rm{and\;C}} = \left[ {\begin{array}{*{20}{c}} 1\\ {\rm{k}}\\ {{{\rm{k}}^2}} \end{array}} \right]\)

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

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {\rm{k}}&1&1\\ 1&{\rm{k}}&1\\ 1&1&{\rm{k}} \end{array}} \right| = 0\)

⇒ 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 = \(\rm \dfrac 12 \rm \begin{vmatrix} \rm x_1&\rm y_1 &1 \\ x_2& \rm y_2&1 \\ \rm\rm x_3 &\rm y_3&1 \end{vmatrix}\)

Calculations:

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

By using the above formula,

​⇒ Area = \(\rm \dfrac 12 \rm \begin{vmatrix} \rm -3&\rm 0 &1 \\ 3& 0&1 \\ 0 &k&1 \end{vmatrix}\)

⇒ Area = \(\dfrac 12\)[-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 \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}\)

⇒ (△ ABC ) = \(\dfrac{1}{2}\) \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}\) = ( \(\frac{\sqrt{3}}{4}\)) ×a²

On squaring both side, 

⇒ \(\dfrac{1}{4}\)  \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}^2\) = \(\dfrac{3a^4}{16}\)

⇒ \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}^2\) = \(\dfrac{3a^4}{4}\)

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 \(\rm (x_1, y_1), (x_2, y_2)\) and \(\rm (x_3, y_3)\) are the vertices of a triangle then, 

Area = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

Properties of Determinants

\(\rm \begin{vmatrix} \rm x_1 & \rm y_1 & kx \\\ \rm x_2 & \rm y_2 & ky \\\ \rm x_3 & \rm y_3 & kz \end{vmatrix} = k \begin{vmatrix} \rm x_1 & \rm y_1 & x \\\ \rm x_2 & \rm y_2 & y \\\ \rm x_3 & \rm y_3 & z \end{vmatrix}\)

Calculations:

If \(\rm (x_1, y_1), (x_2, y_2)\) and \(\rm (x_3, y_3)\) are the vertices of a triangle then,

Area = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

Given: the area is ‘k’ square units

⇒ k = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

⇒ \(\rm \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix} = 2k\)

Now,  \(\begin{vmatrix} \rm x_1 & \rm y_1 & 4 \\\ \rm x_2 & \rm y_2 & 4 \\\ \rm x_3 & \rm y_3 & 4 \end{vmatrix}^2\)

= \(\begin{vmatrix} \rm x_1 & \rm y_1 & 4 \\\ \rm x_2 & \rm y_2 & 4 \\\ \rm x_3 & \rm y_3 & 4 \end{vmatrix}\begin{vmatrix} \rm x_1 & \rm y_1 & 4 \\\ \rm x_2 & \rm y_2 & 4 \\\ \rm x_3 & \rm y_3 & 4 \end{vmatrix}\)

= \(4\begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}.4\begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

= 16.(2k)(2k)

= 64k²

Concept:

If \(\rm (x_1, y_1), (x_2, y_2)\) and \(\rm (x_3, y_3)\) are the vertices of a triangle then, 

Area = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

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 = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

⇒ Δ = \(\rm\frac{1}{2}\begin{vmatrix} 1& 1 & 1\\ 6& 0 & 1\\ 3& 2 & 1 \end{vmatrix}\) 

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

⇒ 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

\(\rm Area = \frac{1}{2}\begin{vmatrix} x_{1} &x_{2} & 1\\ y_{1}& y_{2} &1 \\ z_{1}&z_{2} & 1 \end{vmatrix}\)  

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 = \(\rm = \frac{1}{2}\begin{vmatrix} k &0 & 1\\ 4& 0 &1 \\ 0&2 & 1 \end{vmatrix}\)  

⇒ ± 4  \(\rm = \frac{1}{2} \left [ k(0-2) - 0 (4-0)+1(8-0)\right ]\) 

⇒ ± 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; 

\(A=\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

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 =   \(A= \frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

\(⇒ A = \frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{5}}&{{-2}}&1\\ {{8}}&{{-3}}&1\\ {{a}}&{{-12}}&1 \end{array}} \right|\)

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)

 \(\rm \frac{-3-(-2)}{8-5}=\frac{-12-(-3)}{a-8}\\ ⇒ \frac{-1}{3}=\frac{-9}{a-8}\)

⇒ 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:

​​​​\( \left| {\begin{array}{*{20}{c}} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3 \end{array}} \right|=0\)

Formula for Quadratic equation:

ax² + bx + c = 0

x = \(\rm \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)

Calculation:

2k2x + 3y - 1 = 0      ....(1)

7x - 2y + 3 = 0      ....(2)

6kx + y + 1 = 0      ....(3)

For consistency of given simultaneous equation,

\( \left| {\begin{array}{*{20}{c}} 2k^2&3&-1\\ 7&-2&3\\ 6k&1&1 \end{array}} \right|=0\)

⇒ 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,

\(x=\rm \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)

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

\(\therefore k=\rm \frac{21 \pm \sqrt{161}}{10}\)