Solution of Linear Equations MCQ Quiz in मल्याळम - Objective Question with Answer for Solution of Linear Equations - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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)

If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a11 × {(a22 × a33) - (a23 × a32)} - a12 × {(a21 × a33) - (a23 × a31)} + a13 × {(a21 × a32) - (a22 × a31)}

Calculation:

Given: The system of equations

x + 2y + 2z = 1

2x + 2y + 3z = 3 and

x - y + 3z = 5

So, A = \(\left[ {\begin{array}{*{20}{c}} 1&2& 2\\ 2&2&3\\ 1&{ - 1}&3 \end{array}} \right]\)

det (A) = |A| = 1 × {( 2 × 3) - ( 3 × -1)} - 2 × {(2 × 3) - (3  × 1)} + 2 × {(2 × - 1) - (2 × 1)}

⇒ |A| = 1(6 + 3) -2(6 - 3) + 2(-2 - 2)

⇒ |A| = 9 - 6 - 8 = -5 

∴ |A| ≠ 0

CONCEPT:

• Since the constant terms for the given equations are non-zero so the values Δ­₁, Δ₂, and Δ₃ may be zero or may not be 0.
• For the unique solution, the value of the determinant of coefficients that means Δ should be non zero.

Δ  ≠  0

Where Δ is the determinant formed by the coefficient of the given homogeneous equation, 

CALCULATION

Given:

(1 + α)x + βy + z = 0 

αx + (1 + β)y + z = 0

αx + βy + 2z = 0

\(Δ =\left| {\begin{array}{*{20}{c}} {1 + α }&β &1\\ α &{1 + α }&1\\ α &β &2 \end{array}} \right|\)

⇒ Δ  = (α + β + 2)\(\left| {\begin{array}{*{20}{c}} 1&β &1\\ 1&{1 + α }&1\\ 1&β &2 \end{array}} \right|\) [C1 → C1 + C2 + C3]

⇒ Δ = (α + β + 2)\(\left| {\begin{array}{*{20}{c}} 1&β &1\\ 0&1&0\\ 0&0&1 \end{array}} \right|\) [R2 → R2 - R1 and R3 → R3 - R1]

⇒ Δ = (α + β + 2)

• For unique solution,

⇒ α + β + 2 ≠ 0 

⇒ α + β ≠ -2.

• This condition is satisfied only for the interval (2, 4)
• So, the correct answer is option 3.                                 

Concept:

Consider the system of equations,

a₁x + b₁y + c₁z = 0, 

a2x + b2y + c2z = 0

a3x + b3y + c3z = 0 which is linear homogeneous equations.

Every homogeneous system has either trivial solution or non trivial solution.

If Rank of Matrix = Total number of unknowns, the system possess trivial solution and 

If Rank of Matrix < Total number of unknowns, the system possess non trivial solution

Calculations:

Given,  system of equations 

3x + 2y + z = 0,

x + 4y + z = 0,

2x + y + 4z = 0 which is linear homogeneous equations.

In matrix form , we can write

\(\begin{bmatrix} 3 &2 &1 \\ 1&4 &1 \\ 2 &1 & 4 \end{bmatrix}\)\(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\) = \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)

AX = B 

Where A = \(\begin{bmatrix} 3 &2 &1 \\ 1&4 &1 \\ 2 &1 & 4 \end{bmatrix}\), X = \(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\)and B = \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)

Every homogeneous system has either trivial solution or non trivial solution.

If Rank of Matrix = Total number of unknowns, the system possess trivial solution and 

If Rank of Matrix < Total number of unknowns, the system possess non trivial solution

Here, Rank of Matrix = 3

Total number of unknowns = 3

i.e. Rank of Matrix = Total number of unknowns 

System has trivial solution.

i.e  x = 0, y = 0, z = 0

If 3x + 2y + z = 0, x + 4y + z = 0, 2x + y + 4z = 0 is a system of equations, then: it has only the trivial solution x = 0, y = 0, z = 0.

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, 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)

If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) - (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) - (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) - (a₂₂ × a₃₁)}

Calculation:

Given: The system of equations

2x + y - 3z = 5

3x - 2y + 2z = 5 and

5x - 3y - z = 16

So, A = \(\left[ {\begin{array}{*{20}{c}} 2&1&{ - 3}\\ 3&{ - 2}&2\\ 5&{ - 3}&{ - 1} \end{array}} \right]\)

det (A) = |A| = 2 × {( - 2 × - 1) - ( - 3 × 2)} - 1 × {(3 × - 1) - (2 × 5)} + ( - 3) × {(3 × - 3) - (5 × - 2)}

⇒ |A| = 2 × (2 + 6) - 1 × ( - 3 - 10) - 3 × ( - 9 + 10)

⇒ |A| = 16 + 13 - 3 = 26

∴ |A| ≠ 0

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 has 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

For no solution or infinitely many solutions

\(⇒ \left| {\begin{array}{*{20}{c}} α &1&1\\ 1&α &1\\ 1&1&α \end{array}} \right|\) = 0 

⇒ α = 1, α = -2

But for α = 1, clearly, there are infinitely many solutions as column matrix B will become zero matrices.

⇒ ​(adj A). B = O

For no solution,

When we put α = - 2 then matrix B will become a non zero column matrix with the same entries as -3 

⇒  (adj A). B ≠ O

So, α = - 2 for no solution.

Explanation:

For a matrix system AX = B of linear equations -

• It has a unique solution X = A^-1B if A is a non-singular matrix or det(A) ≠ 0.
• If det(A) = 0 and adj(A)B = 0, then there are infinite many solutions.
• If det(A) = 0 and adj(A)B ≠ 0, then there are no solutions.

Also if B = 0, then the system is homogeneous.

• This homogeneous system has only trivial solution (x = y = z = 0) if the det(A) ≠ 0
• Otherwise, for det(A) = 0; this system is consistent with infinite many solutions.

Given:

x + 2y - 3z = 2

(k + 3)z = 3

(2k + 1)y + z = 2

Concept:

The given set of equations is consistent

Therefore \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\)

Calculation:

\(\dfrac{0}{2k+1} = \dfrac{k+3}{1}\)

⇒ \(0 = 2k^2 + 6k + k + 3\)

⇒ \(0 = 2k^2 + 7k + 3\)

⇒ \(2k^2 + 7k + 3 = 0\)

\(\Delta = b^2 - 4ac\)

⇒ 49 - (4 × 2 × 3)

⇒ 49 - 24

⇒ 25 > 0

K \(=\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-7 \pm \sqrt{25}}{2 \times 2}\)

⇒ K \(=\dfrac{-7 \pm 5}{4} = \dfrac{-7-5}{4} , \dfrac{-7+5}{4}\)

⇒ K \(=\dfrac{-12}{4}, \dfrac{-2}{4}\)

∴ K = -3, -1/2

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: 2kx - 2y + 3z = 0, x + ky + 2z = 0, 2x + kz = 0

For the homogeneous system of equations,

⇒ (adj A). B = 0 for all value of k and the column matrix B is zero matrix.

So, for non trivial solution,

|A|  = 0 

⇒ 2k(k²) + 2(k - 4) + 3(-2k) = 0

⇒ k³ - 2k - 4 =0 

⇒  (k - 2)(k² + 2k + 2) = 0 

∴ k = 2 (Real value).

Calculation:

Given, x + y + z = 4,

2x + 5y + 5z = 17,

x + 2y + mz = n

∴ D = \(\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 5 & 5 \\ 1 & 2 & \mathrm{~m} \end{array}\right|\) = 0 

⇒ 1(5m - 10) - 1(2m - 5) + 1(4 - 5) = 2 

⇒ m = 2

Also, D_z = \(\left|\begin{array}{ccc} 1 & 1 & 4 \\ 2 & 5 & 17 \\ 1 & 2 & n \end{array}\right|\) = 0 

⇒ 1(5n - 34) - 1(2n - 17) + 4(4 - 5) = 0

⇒ n = 7

∴ m2 + n2 – mn = 22 + 42 – 14 = 4 + 49 – 14 = 39

∴ m and n satisfy the equation m2 + n2 – mn = 39.

The correct answer is Option 4.

Concept:

If the system of equations has non-zero solutions, then the determinant is zero. 

Calculation:

The system will have a non-zero solution, if

Δ = \( \begin{vmatrix} a^3 & (a+1)^3 & (a+2)^3 \\[0.3em] a & a+1 & a+2 \\[0.3em] 1 & 1 & 1 \end{vmatrix}=0\)

C₂ → C2 - C1

C3 → C₃ - C1

⇒ Δ = \( \begin{vmatrix} a^3 & 3a^2+3a+1 & 6a^2+12a+8 \\[0.3em] a & 1 & 2 \\[0.3em] 1 & 0 & 0 \end{vmatrix}=0\)

[(a + 1)³ - a³ = a³ + 1 + 3(a + 1) - a³ = 3a² + 3a + 1 and

(a + 2)3 - a3 = a3 + 8 + 3(a + 2) - a3 = 6a2 + 12a + 8]

Expanding along R₃,

⇒ (6a² + 6a + 2) - (6a² + 12a + 8) = 0

⇒ -6a - 6 = 0

⇒ a = -1

∴ The value of a is -1.