Mathematics Test - 3

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.

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 unique solution: |A| ≠ 0.

⇒ |A| = 2α + 6 ≠ 0 ⇒ α ≠ -3.

Explanation:

• A matrix system AX = B will always have a unique solution if A is non-singular or |A| ≠ 0. (Option 1 is true)
• A matrix system AX = B will always be inconsistent if |A| = 0 but (adjA)B ≠ 0 and consistent if |A| = 0 and (adjA)B = 0 and in this case there will be infinite number of solution (Option 2 is false and Option3 is true)
• |A| ≠ 0 is not the only condition for consistency, a matrix system AX = B will also be consistent if |A| = 0 and (adjA)B = 0 (Option 4 is false)

So, the correct answer is option 1.

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 = 0, x – y + z = 0 and 2x + 3y + αz = 0

As we know that,

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

\( \Rightarrow {\rm{\Delta }} = \left| {\begin{array}{*{20}{c}} 1&1&1\\ 1&{ - 1}&1\\ 2&3&\alpha \end{array}} \right|,\;{{\rm{\Delta }}_1} = \;\left| {\begin{array}{*{20}{c}} 0&1&1\\ 0&{ - 1}&1\\ 0&3&\alpha \end{array}} \right|,\;{{\rm{\Delta }}_2} = \;\left| {\begin{array}{*{20}{c}} 1&0&1\\ 1&0&1\\ 2&0&\alpha \end{array}} \right|\;and\;{{\rm{\Delta }}_3} = \;\left| {\begin{array}{*{20}{c}} 1&1&0\\ 1&{ - 1}&0\\ 2&3&0 \end{array}} \right|\)

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

We can see that, Δ₁ =  Δ₂ =  Δ₃ = 0 ∵ they contain one column with all entries 0.

So, we need to make Δ = 0.

⇒ Δ = - 2α + 4 = 0 ⇒ α = 2.

Concept:

A matrix system of form AX = B has a unique solution X = A-1B  if |A| ≠ 0

(kA)^-1 = (1/k)A

Calculation:

Given:

M = \(\begin{bmatrix} 3 & 6 & 9\\6 & 9 & 12\\9 & 12 & 18\end{bmatrix}\) and  M-1 = \(\begin{bmatrix} -2/3 & 0 & 1/3\\0 & 1 & -2/3\\ 1/3& -2/3 & 1/3\end{bmatrix}\)

• The system of equations x + 2y + 3z = 5, 2x + 3y + 4z = 7, 3x + 4y + 6z = 11 has a unique solution.
• So, we can write this as AX = B Where,

A = \(\begin{bmatrix} 1 & 2 & 3\\2 & 3 & 4\\3 & 4 & 6\end{bmatrix}\), B = \(\begin{bmatrix} 5 \\ 7 \\ 11 \end{bmatrix} \), X = \(\begin{bmatrix} x \\ y \\ z \end{bmatrix} \)

By the property of the matrices,

⇒ M = 3\(\begin{bmatrix} 1 & 2 & 3\\2 & 3 & 4\\3 & 4 & 6\end{bmatrix}\) 

⇒ M-1 = (1/3)\(\begin{bmatrix} 1 & 2 & 3\\2 & 3 & 4\\3 & 4 & 6\end{bmatrix} ^{-1}\)

⇒ \(\begin{bmatrix} 1 & 2 & 3\\2 & 3 & 4\\3 & 4 & 6\end{bmatrix} ^{-1}\)= 3 × M-1   

= 3\(\begin{bmatrix} -2/3 & 0 & 1/3\\0 & 1 & -2/3\\ 1/3& -2/3 & 1/3\end{bmatrix}\) 

= \(\begin{bmatrix} -2 & 0 & 1\\0 & 3 & -2\\ 1& -2 & 1\end{bmatrix}\)

∴ X = A^-1B

⇒ X = \(\begin{bmatrix} 1 & 2 & 3\\2 & 3 & 4\\3 & 4 & 6\end{bmatrix} ^{-1}\)\(\begin{bmatrix} 5 \\ 7 \\ 11 \end{bmatrix} \)

⇒ X = \(\begin{bmatrix} -2 & 0 & 1\\0 & 3 & -2\\ 1& -2 & 1\end{bmatrix}\)\(\begin{bmatrix} 5 \\ 7 \\ 11 \end{bmatrix} \)

= \(\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \)

∴ x = 1, y = -1, z = 2

∴ xyz = -2

Concept:

• The system of linear equations has a unique solution if D ≠ 0.
• The system of linear equations has infinite solutions if D = 0, D₁ = 0, D₂ = 0, D₃ = 0.
• The system of linear equations has no solution if D = 0 and at least one of D₁, D2, and D3 is not zero.

For a homogeneous system -

• If D ≠ 0, then only trivial solution x = y = z =0.
• If D = 0, then infinite solutions are there.

Calculation:

Given:

x + ωy + ω2z = 0 

ωx + ω2y + z = 0

ω2x + y + ωz = 0

Where ω is a complex cube root of unity

The given system is homogeneous.

D = \(\begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2& 1 \\\omega^2 & 1 &\omega\end{vmatrix} \)

Applying C₁ → C₁ + C₂ + C₃

⇒ D = \(\begin{vmatrix} 1+\omega +\omega^2 & \omega & \omega^2 \\ 1+\omega +\omega^2 & \omega^2& 1 \\1+\omega +\omega^2 & 1 &\omega\end{vmatrix} \)

∵ 1 + ω + ω2 = 0

⇒  D = \((1+\omega+\omega^2)\begin{vmatrix} 1 & \omega & \omega^2 \\ 1 & \omega^2& 1 \\1 & 1 &\omega\end{vmatrix} \)

⇒ D = 0 , Hence infinite solutions.

Concept:

For a homogeneous system -

•  If D ≠ 0, then only trivial solution x = y = z =0.
•  If D = 0, then infinite solutions are there.

Where D is the determinant of matrices formed by the coefficient of x, y, and z out of the given three equations and D1 is calculated by replacing the first column with constant terms of the given equation.

Calculation:

Given:

The system of equations px - 2y - pz = 0, 2x + py -pz = 0, -px + py + 2z = 0 has a non-trivial solution.

If there is a non-trivial solution, ∴ D must be zero.

⇒ D = 0

⇒ \(\begin{vmatrix} p & -2 & -p\\ 2 & p & -p \\ -p & p & 2\end{vmatrix} \) = 0

⇒ p(p × 2 - (-p) × p) - (-2)(2 × 2 - (-p) × (-p)) + (-p)(2 × p - (-p) × p) = 0

⇒ -2p² + 8 = 0

⇒ p² = 4

⇒ p = ± 2 

∴ sum of possible values = 2 + (-2) =0

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,

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

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

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

Given:

kx + 5y + 3z = 1

x + 3y + 2z = 3

x + 2y + z = 3

Concept:

The system of linear equations has a unique solution if D ≠ 0.

The system of linear equations has infinite solutions if D = 0, D1 = 0, D2 = 0, D3 = 0.

The system of linear equations has no solution if D = 0 and at least one of D1, D2 and D3 is not zero.

Where D is the determinant of matrices formed by the coefficient of x, y, and z out of the given three equations and D₁ is calculated by replacing the first column with constant terms of the given equation.

Calculation:

D = \(\begin{vmatrix} k & 5 & 3\\ 1 & 3 & 2 \\ 1 & 2 & 1\end{vmatrix} \)

⇒ D = k(3 × 1 - 2 × 2) - 5(1 × 1 - 1 × 2) + 3(1 × 2 - 1 × 3)

⇒ D = - k +2

Similarly we can check D₁ ≠ 0

D1 = \(\begin{vmatrix} 1 & 5 & 3\\ 3 & 3 & 2 \\ 3 & 2 & 1\end{vmatrix} \)

⇒ D₁ = 1(3 × 1 - 2 × 2) - 5(3 × 1 - 3 × 2) + 3(3 × 2 - 3 × 3)

= 5 ≠ 0

Now for no solution, D = 0 and atleast one of D1, D2 and D3 is not zero. 

⇒ -k + 2 = 0

⇒ k = 2

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.