Cramer's Rule MCQ Quiz in मल्याळम - Objective Question with Answer for Cramer's Rule - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

The system of equations A X = 0 is said to be homogenous system of equations, then

If |A| ≠ 0, then its solution X = 0, is called trivial solution.

If |A| = 0. Then A X = 0 has a non-trivial solution which means the system will have infinitely many solutions.

Calculation:

Given:  2x – 3y = 0 and 2x + αy = 0

These equations can be written as: A X = B where \(A = \left[ {\begin{array}{*{20}{c}} 2&{ - 3}\\ 2&\alpha \end{array}} \right],\;X = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right]\;and\;B = \left[ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right]\)

As we know that, the given system is a homogenous system of equation. So, in order to say that the system has infinitely many solutions: |A| = 0.

⇒ |A| = 2α + 6 = 0 ⇒ α = -3.

Calculation

\(D=\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 1 & 5 & λ \end{array}\right|=0\)

⇒ λ = 17

\(D_{z}=\left|\begin{array}{lll} 1 & 1 & 6 \\ 1 & 2 & 9 \\ 1 & 5 & μ \end{array}\right| ≠ 0\)

⇒ μ ≠ 18

Hence option 1 is correct

Concept:

Let us consider a system of equations in three variables:

a₁ × x + b₁ × y + c₁ × z = d₁

a₂ × x + b₂ × y + c₂ × z = d₂

a₃ × x + b₃ × y + c₃ × z = d₃

Then, \({\rm{\Delta }} = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right|,\;\;{{\rm{\Delta }}_1} = \left| {\begin{array}{*{20}{c}} {{d_1}}&{{b_1}}&{{c_1}}\\ {{d_2}}&{{b_2}}&{{c_2}}\\ {{d_3}}&{{b_3}}&{{c_3}} \end{array}} \right|,\;{{\rm{\Delta }}_2} = \;\left| {\begin{array}{*{20}{c}} {{a_1}}&{{d_1}}&{{c_1}}\\ {{a_2}}&{{d_2}}&{{c_2}}\\ {{a_3}}&{{d_3}}&{{c_3}} \end{array}} \right|\;and\;{{\rm{\Delta }}_3} = \;\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{d_1}}\\ {{a_2}}&{{b_2}}&{{d_2}}\\ {{a_3}}&{{b_3}}&{{d_3}} \end{array}} \right|\)

By cramer’s rule:

I. If Δ ≠ 0, then the system of equation has unique solution and it is given by: \(x = \frac{{{{\rm{\Delta }}_1}}}{{\rm{\Delta }}},\;y = \frac{{{{\rm{\Delta }}_2}}}{{\rm{\Delta }}}\;and\;z = \frac{{{{\rm{\Delta }}_3}}}{{\rm{\Delta }}}\)

II. If Δ = 0 and atleast one of the determinants Δ, Δ₁, Δ₂ and Δ₃ is non-zero, then the given system is inconsistent.

III. If Δ = 0 and Δ₁ =  Δ₂ =  Δ₃ = 0, then the system is consistent and has infinitely many solutions.

Calculation:

Given: x + y + z = 3, x – y + 2z = 6 and x + y + α z = β.

As we know that,

\({\rm{\Delta }} = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right|,\;{{\rm{\Delta }}_1} = \left| {\begin{array}{*{20}{c}} {{d_1}}&{{b_1}}&{{c_1}}\\ {{d_2}}&{{b_2}}&{{c_2}}\\ {{d_3}}&{{b_3}}&{{c_3}} \end{array}} \right|,\;{{\rm{\Delta }}_2} = \;\left| {\begin{array}{*{20}{c}} {{a_1}}&{{d_1}}&{{c_1}}\\ {{a_2}}&{{d_2}}&{{c_2}}\\ {{a_3}}&{{d_3}}&{{c_3}} \end{array}} \right|\;and\;{{\rm{\Delta }}_3} = \;\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{d_1}}\\ {{a_2}}&{{b_2}}&{{d_2}}\\ {{a_3}}&{{b_3}}&{{d_3}} \end{array}} \right|\)

\( \Rightarrow {\rm{\Delta }} = \left| {\begin{array}{*{20}{c}} 1&1&1\\ 1&{ - 1}&2\\ 1&1&\alpha \end{array}} \right|,\;{{\rm{\Delta }}_1} = \left| {\begin{array}{*{20}{c}} 3&1&1\\ 6&{ - 1}&2\\ \beta &1&\alpha \end{array}} \right|,\;{{\rm{\Delta }}_2} = \;\left| {\begin{array}{*{20}{c}} 1&3&1\\ 1&6&2\\ 1&\beta &\alpha \end{array}} \right|\;and\;{{\rm{\Delta }}_3} = \;\left| {\begin{array}{*{20}{c}} 1&1&3\\ 1&{ - 1}&6\\ 1&1&\beta \end{array}} \right|$\)

⇒ Δ = 2 – 2α, Δ₁ = 3β – 9α, Δ₂ =  3α - β and Δ₃ = 6 – 2β.

As we know that, for the given system of equation to have infinitely many solutions according to cramer’s rule: Δ = 0 and Δ₁ =  Δ₂ =  Δ₃ = 0

⇒ Δ = 2 – 2α = 0 ⇒ α = 1.

For α = 1 , we have: Δ₁ = 3β – 9, Δ₂ =  3 – β and Δ₃ =  6 – 2β

So, in order to have infinitely many solutions, Δ₁ =  Δ₂ =  Δ₃ = 0.

⇒ β = 3.

Concept:

Let us consider a system of equations in two variables:

a₁ × x + b₁ × y = c₁

a₂ × x + b₂ × y = c₂

Then, \({\rm{\Delta }} = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}\\ {{a_2}}&{{b_2}} \end{array}} \right|,\;{{\rm{\Delta }}_1} = \left| {\begin{array}{*{20}{c}} {{c_1}}&{{b_1}}\\ {{c_2}}&{{b_2}} \end{array}} \right|\;and\;{{\rm{\Delta }}_2} = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{c_1}}\\ {{a_2}}&{{c_2}} \end{array}} \right|\)

By Cramer's rule, the solution of a system of the equation has a unique solution if Δ ≠ 0 and the solution is given by: \(x = \frac{{{{\rm{\Delta }}_1}}}{{\rm{\Delta }}}\;and\;y = \frac{{{{\rm{\Delta }}_2}}}{{\rm{\Delta }}}\)

By cramer’s rule:

I. If Δ ≠ 0, then the system of the equation has a unique solution and it is given by: \(x = \frac{{{{\rm{\Delta }}_1}}}{{\rm{\Delta }}},\;y = \frac{{{{\rm{\Delta }}_2}}}{{\rm{\Delta }}}\;and\;z = \frac{{{{\rm{\Delta }}_3}}}{{\rm{\Delta }}}\)

II. If Δ = 0 and at least one of the determinants Δ, Δ1, Δ2, and Δ3 is non-zero, then the given system is inconsistent.

III. If Δ = 0 and Δ1 =  Δ2 =  Δ3 = 0, then the system is consistent and has infinitely many solutions.

Calculation:

Given: 4x + y = α and βx + 2y = 3

As we know that, the given system of equation will be inconsistent if if Δ = 0 and at least one of the determinants Δ₁ and Δ₂ is non-zero.

\(⇒ {\rm{\Delta }} = \left| {\begin{array}{*{20}{c}} 4&1\\ β &2 \end{array}} \right| = 8 - β \)

∵ The given system has no solution

⇒ Δ = 8 - β = 0 ⇒ β = 8.

\( \Rightarrow {{\rm{\Delta }}_1} = \left| {\begin{array}{*{20}{c}} α &1\\ 3&2 \end{array}} \right| = 2α - 3\;and\;{{\rm{\Delta }}_2} = \left| {\begin{array}{*{20}{c}} 4&α \\ {8}&3 \end{array}} \right| = 12 - 8α \)

So, for α = 3 / 2, we can see that both Δ₁ and Δ₂ are zero.

for α = 3 / 2 and β = 8 the given system is consistent and has infinitely many solutions.

Hence, option 4 is correct.

Calculation

Given:

The system of simultaneous linear equations:

\(x - 2y + 3z = 4\)

\(3x + y - 2z = 7\)

\(2x + 3y + z = 6\)

\(D = \begin{vmatrix} 1 & -2 & 3 \\ 3 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}\)

⇒ \(D = 1(1+6) - (-2)(3+4) + 3(9-2)\)

⇒ \(D = 7 + 14 + 21 = 42\)

Since \(D \neq 0\), the system has a unique solution.

\(D_z = \begin{vmatrix} 1 & -2 & 4 \\ 3 & 1 & 7 \\ 2 & 3 & 6 \end{vmatrix}\)

⇒ \(D_z = 1(6-21) - (-2)(18-14) + 4(9-2)\)

⇒ \(D_z = -15 + 8 + 28 = 21\)

\(z = \frac{D_z}{D} = \frac{21}{42} = \frac{1}{2}\)

Thus, we have a unique solution with \(z = \frac{1}{2}\).

Hence option 4 is correct

Calculation

For non-zero solution,

\(\begin{vmatrix}a^3&(a+1)^3 &(a+2)^3 \\a &(a+1)&(a+2)\\1&1&1 \end{vmatrix}=0\)

\(\Rightarrow \begin{vmatrix} 1 & 1 & 1 \\ a & (a+1) & (a+2) \\ a^{ 3 } & (a+1)^{ 3 } & (a+2)^{ 3 } \end{vmatrix}=0\)

Since \(\begin{vmatrix} 1 & 1 & 1 \\ x& y & z \\ x^{ 3 } & y^{ 3 } & z^{ 3 } \end{vmatrix}=\)(x−y)(y−z)(z−x)(x+y+z)

\(\Rightarrow -(a-a-1)(a+1-a-2)(a+2-a)\times (a+a+1+a+2)=0\)

\(\Rightarrow -2(3a+3)=0\)

\(\Rightarrow a=-1\)

Hence option 3 is correct
$\left[\because \begin{vmatrix} 1 & 1 & 1 \ x&y&z \x^3&y^3&z^3 \end{vmatrix}=(x-y)(y-z)(z-x)(x+y+z)\right]$