Application of Determinants MCQ Quiz in मल्याळम - Objective Question with Answer for Application of Determinants - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Property of Determinant 

• \(|A^{-1}|=\frac{1}{|A|}\)

Calculation:

A = \(\begin{bmatrix} 8 & 7 & 0\\ 6& 5& 4\\ 3& 2& 1 \end{bmatrix}\)

⇒ \(|A|=\begin{vmatrix} 8 & 7 &0 \\ 6& 5 &4 \\ 3 &2 & 1 \end{vmatrix}\)

⇒ |A| = 8(5 - 8) - 7(6 - 12) + 0(12 - 15)

⇒ |A| = -24 + 42 + 0 = 18

Therefore,  \(|A^{-1}|=\frac{1}{|A|} = \frac{1}{18}\) 

Hence, option B is correct.

Concept:

If the area of the triangle formed by the vertices is zero, then three points are collinear

Formula:

Area of triangle = \(|\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1\\x_3 & y_3 & 1 \end{vmatrix}|\)

Calculation:

Given:

The points A(a , b), B(b , a) and C(2a - 2b , 3b - a) 

Area = \(|\frac{1}{2}\begin{vmatrix} a & b & 1 \\ b & a & 1\\2a-2b & 3b-a & 1 \end{vmatrix}|\)

= 1/2[a(a × 1 - (3b - a) × 1) -b (b × 1 - (2a - 2b) × 1) +1(b × (3b - a) - (2a - 2b) × a)]

⇒ Area = 0 

Hence points are collinear.

Concept:

Some properties of inverse

• A-1A = I where I is the identity matrix
• A-1 I = A^-1 where I is the identity matrix

Calculation:

∴ A satisfies the equation x2 - 4x + 3 = 0

⇒ A2 - 4A + 3I = 0

Multiply the above equation by A^-1, we get 

⇒ A^-1A2 - 4A^-1A + 3A^-1I = 0

⇒ A - 4I + 3A^-1 = 0

⇒ 3A-1 = 4I - A

⇒ A-1 = (4I - A) / 3

Hence, option 3 is correct.

Concept:

The value of the determinant \(\begin{vmatrix}a&b&c\\\ \rm d&e&f\\\ g&h&i\end{vmatrix}\) can be obtained by expanding along any row or column.

Expanding along the third column, the value of the determinant is 

c(dh - eg) - f(ah - bg) + i(ae - bd)

Calculation:

Given, \(\begin{vmatrix}1&1&0\\\ \rm x^2+2x+2&1&0\\\ 2&1&1\end{vmatrix}=0\)

Expanding LHS along column 3,

⇒ 0(x2 + 2x + 2- 2) - 0(1 - 2) + 1(1 - (x2 + 2x + 2)) = 0

⇒ 0 + 0 - x² - 2x - 1 = 0

⇒ x2 + 2x + 1 = 0

⇒ (x + 1)² = 0 

∴  x = -1

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

\(\Rightarrow \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_1}}&{{{\rm{b}}_1}}&{{{\rm{c}}_1}}\\ {{{\rm{a}}_2}}&{{{\rm{b}}_2}}&{{{\rm{c}}_2}}\\ {{{\rm{a}}_3}}&{{{\rm{b}}_3}}&{{{\rm{c}}_3}} \end{array}} \right]{\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {{{\rm{d}}_1}}\\ {{{\rm{d}}_2}}\\ {{{\rm{d}}_3}} \end{array}} \right]\)

⇒ AX = B

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

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

Calculation:

Given system of linear equation are kx + y + z = 1, x + ky + z = 1 and x + y + kz = 1

Let A = \(\left[ {\begin{array}{*{20}{c}} {\rm{k}}&1&1\\ 1&{\rm{k}}&1\\ 1&1&{\rm{k}} \end{array}} \right]\)

det (A) = |A| = k (k² – 1) – 1(k -1) + 1 (1 – k)

⇒ |A| = k³ – k – k + 1 + 1 – k = k³ – 3k + 2

For unique solution,

det (A) ≠ 0

⇒ k³ – 3k + 2 ≠ 0

⇒ (k – 1) (k² + k - 2) ≠ 0

⇒ (k – 1) (k – 1) (k + 2) ≠ 0

∴ k ≠ 1 and k ≠ -2

Concept:

Property of determinant of a matrix:

• Let A be a matrix of order n × n then det(kA) = kn det(A)

• If A and B are two square matrices then |AB| = |A||B|

Calculation:

Given: A and B are square matrices of order 2 such that |A| = 2, |B| = 4 

Here, we have to find the value of |2 AB|

As we know that, if A and B are two determinants of order n, then |AB| = |A||B|

⇒ |2 AB| = |2A||B|

As we know that, if A is a matrix of order n, then |kA| = kn |A|, where k ∈ R.

Here n = 2 So, |2A| = 22 ⋅ |A| = 4|A|

⇒ |2 AB| = |2A||B|

= 4|A||B|

= 4 × 2 × 4

= 32

Concept:

Formula:

Area of triangle = \(|\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1\\x_3 & y_3 & 1 \end{vmatrix}|\)

Calculation:

Given:

The area of the triangle formed by vertices (-2 , 0), (-2 , 8) and (t , 4) is 12 square units.

Area = \(|\frac{1}{2}\begin{vmatrix} -2 & 0& 1 \\ -2 & 8 & 1\\t & 4 & 1 \end{vmatrix}|\)

⇒ |-2(8 × 1 - 4 × 1) -0(-2 × 1 - t × 1) + 1(-2 × 4 - t × 8)| = 12 × 2

⇒ |-16 - 8t| = 24

⇒ |- 2 - t| = 3

⇒ - 2 - t = 3 or  -2 - t = -3

⇒ t = -5, 1

So, the correct answer is option 2.

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

\({\rm{\;}} ⇒ \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_1}}&{{{\rm{b}}_1}}&{{{\rm{c}}_1}}\\ {{{\rm{a}}_2}}&{{{\rm{b}}_2}}&{{{\rm{c}}_2}}\\ {{{\rm{a}}_3}}&{{{\rm{b}}_3}}&{{{\rm{c}}_3}} \end{array}} \right]{\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {{{\rm{d}}_1}}\\ {{{\rm{d}}_2}}\\ {{{\rm{d}}_3}} \end{array}} \right]\)

⇒ AX = B

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

⇒ If det (A) ≠ 0, 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 system of equation x + y + z = 8, x – y + 2z = 6 and 3x – y + 5z = k

\(⇒ \left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&{ - 1}&2\\ 3&{ - 1}&5 \end{array}} \right]{\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 8\\ 6\\ {\rm{k}} \end{array}} \right]\)

⇒ AX = B

Determinant of A = |A| = 1 (-5 + 2) – 1 (5 – 6) + 1 (-1 + 3) = -3 + 1 + 2 = 0

So we can say that equations have either an infinite solution or no solution.

A unique solution is not possible.

 ∴ Statement 3 is wrong.

We have adj A = \(\left[ {\begin{array}{*{20}{c}} -3&-6&3\\ 1&{2}&-1\\ 2&{ 4}&-2 \end{array}} \right]\)

If k = 15,

B = \(\left[ {\begin{array}{*{20}{c}} 8\\ 6\\ {\rm{15}} \end{array}} \right]\)

Now (adj A). B will be 

\(\left[ {\begin{array}{*{20}{c}} -3&-6&3\\ 1&{2}&-1\\ 2&{ 4}&-2 \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}} 8\\ 6\\ {\rm{15}} \end{array}} \right] \ne 0 \)

⇒ no solution

If K = 20,

B = \(\left[ {\begin{array}{*{20}{c}} 8\\ 6\\ {\rm{20}} \end{array}} \right]\)

Now (adj A). B will be 

\(\left[ {\begin{array}{*{20}{c}} -3&-6&3\\ 1&{2}&-1\\ 2&{ 4}&-2 \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}} 8\\ 6\\ {\rm{20}} \end{array}} \right] = 0 \)

⇒ infinitely many solutions

Hence Option 1 is correct.

The given system of linear equations:

x + y + z = 5;

x + 2y + 2z = 6;

and x + 3y + λz = μ have infinite solution.

∴ Δ = 0, Δx = Δy = Δz = 0

Now, forming determinant from the given equations,

\({\rm{\Delta }} = \left| {\begin{array}{*{20}{c}} 1&1&1\\ 1&2&2\\ 1&3&{\rm{\lambda }} \end{array}} \right| = 0\)

⇒ 1(2λ – 6)-1(λ – 2)+1(3 – 2) = 0

⇒ 2λ – 6 – λ + 2 + 3 – 2 = 0

⇒ λ – 8 + 5 = 0

⇒ λ – 3 = 0

∴ λ = 3

Now, the determinant of y is:

\({\rm{\Delta y}} = \left| {\begin{array}{*{20}{c}} 1&5&1\\ 1&6&2\\ 1&{\rm{\mu }}&3 \end{array}} \right| = 0\)

R₂ → R₂ – R₁

R₃ → R₃ – R₁

\(\Rightarrow \left| {\begin{array}{*{20}{c}} 1&5&1\\ {1 - 1}&{6 - 5}&{2 - 1}\\ {1 - 1}&{{\rm{\mu }} - 5}&{3 - 1} \end{array}} \right| = 0\)

\(\Rightarrow \left| {\begin{array}{*{20}{c}} 1&5&1\\ 0&1&1\\ 0&{{\rm{\mu }} - 5}&2 \end{array}} \right| = 0\)

⇒ 1(2 – (μ – 5)) – 5(0 – 0) + 1(0 – 0) = 0

⇒ 1(2 – (μ – 5)) = 0

⇒ 2 – μ + 5 = 0

⇒ 7 – μ = 0

∴ μ = 7

Now,

Given:

 vertices are A(4, 8),B(-6, 2) and C(5, 4)

Concept:

Area of triangle whose vertices are \(\rm (x_1,y_1),(x_2,y_2), and (x_3,y_3)\) is

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

Calculation:

vertices are A(4, 8),B(-6, 2) and C(5, 4)

Then the area is

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

\(\rm A=\frac{1}{2}|4(2-4)-6(4-8)+5(8-2)| \)

\(\rm A=\frac{1}{2}|-8+24+30|\)

\(\rm A=\frac{1}{2}\cdot46\)

\(\rm A=23 \ unit^2\)

Hence the option (4) is correct.