Engineering Mathematics Test 1

Concept:

If A is any square matrix of order n, we can form the matrix [A – λI], where I is the n^th order unit matrix. The determinant of this matrix equated to zero i.e. |A – λI| = 0 is called the characteristic equation of A.

The roots of the characteristic equation are called Eigen values or latent roots or characteristic roots of matrix A.

Properties of Eigen values:

The sum of Eigen values of a matrix A is equal to the trace of that matrix A

The product of Eigen values of a matrix A is equal to the determinant of that matrix A

Calculation:

Let \(A = \left[ {\begin{array}{*{20}{c}}3&0&0\\0&2&{ - 3}\\0&1&{ - 2}\end{array}} \right]\)

|A – λI| = 0

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

⇒ (3 – λ) [ (2 – λ) (-2 – λ) – (-3)] = 0

⇒ (3 – λ) [-4 – 2 λ + 2 λ + λ² + 3] = 0

⇒ (3 – λ) [λ² – 1] = 0

⇒ λ = 3, 1, –1

Eigen values of given matrix = 3, 1, –1

Alternate Method:

Sum of Eigen values of given matrix = Trace = 3 + 2 - 2 = 3

Explanation:

For unique solution, the coefficient matrix rank should be 3.

\(\left[ {\begin{array}{*{20}{c}}2&3&5\\7&3&{ - 2}\\2&3&\lambda \end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\8\\7\end{array}} \right]\) 

Thus,

\(\left| {\begin{array}{*{20}{c}}2&3&5\\7&3&{ - 2}\\2&3&\lambda \end{array}} \right| \ne 0\) 

2 (3λ + 6) – 3 (7λ + 4) + 5 (21 - 6) ≠ 0

6λ + 12 – 21λ - 12 + 75 ≠ 0

-15λ + 75 ≠ 0

-15 (λ - 5) ≠ 0

Concept:

Methods to check Linearly dependent or Linearly Independent vectors:

Let x₁, x₂, x₃ ….. x_r are the n-vectors.

Consider A = [x₁, x₂, x₃…. x_r]_{n × r}

General Method:

1) of ρ(A) = number of vector (R), then Linearly Independent.

2) If ρ(A) < number of vector (r), then Linearly Dependent

Matrix method:

If A is a square matrix :

1) |A| ≠ 0, vectors are Linearly Independent.

2) If |A| = 0, vectors are Linearly Dependent.

Calculation:

For linearly dependent vectors, |A| = 0

⇒ \(A = \left[ {\begin{array}{*{20}{c}} 1&2&3\\ 4&\lambda &6\\ 3&4&5 \end{array}} \right]\)

⇒ |A| = 1(5λ - 24) – 2(20 - 18) + 3 (16 – 3λ)

⇒ |A| = 5λ – 24 – 4 + 48 – 9λ ⇒ λ = 5

Hence for the vectors to be linearly dependent λ = 5.

Explanation:

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

Matrix A is said to be singular when |A| = 0

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

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

⇒ x³ – x – x + 1 + 1 – x = 0

⇒ x³ – 3x + 2 = 0

⇒ x = 1, 1, -2.

The number of district real values of x = 2.

Characteristic equation: |A – λI| = 0

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

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

⇒ (1 – λ) (-6 + 3λ – 2λ + λ² + 6) – 1(-15 – 5λ + 12) + 3(-5 + 4 – 2λ) = 0

⇒ λ³ = 0

According to Cayley Hamilton theorem every square matrix satisfies its characteristic polynomial.

⇒ A³ = 0

A¹⁴ + 3A – 2I = A¹² A² + 3A – 2I = 3A – 2I

\(= 3\left[ {\begin{array}{*{20}{c}}1&1&3\\5&2&6\\{ - 2}&{ - 1}&{ - 3}\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3&9\\{15}&4&{18}\\{ - 6}&{ - 3}&{ - 11}\end{array}} \right]\)

Characteristic equation is given by

\(\left| {A - \lambda I} \right| = 0\)

\(\therefore \left| {\begin{array}{*{20}{c}} {1 - \lambda }&p\\ q&{2 - \lambda } \end{array}} \right| = 0\)

⇒ (1 - λ) (2 - λ) – pq = 0

λ² – 3λ + 2 – pq = 0

\(\therefore {\rm{\lambda }} = \frac{{3 \pm \sqrt {9 - 4\left( {2 - pq} \right)} }}{2}\)

Now, for eigen values to be real

9 – 4 (2 - pq) ≥ 0

9 – 8 + 4 pq ≥ 0

⇒ 4 pq ≥ - 1

\(pq \ge - \frac{1}{4}\)

Now, for eigen values to be positive,

λ₁ . λ₂ > 0

⇒ 2 – pq > 0

[Product of roots of equation ax² + bx + c = 0 ⇒ c/a ]

∴ pq ≤ 2.

Now, for eigenvalues to be real and positive,

\({u^T}\) = \(\left( {\begin{array}{*{20}{c}} 1\\ 1\\ 2 \end{array}} \right)\) and \({v^T}\) = (1 2 1)

\(M = {u^T}{v^T}\)

\(M = \left( {\begin{array}{*{20}{c}} 1\\ 1\\ 2 \end{array}} \right){\rm{\;}}\left( {1\;2\;1} \right)\)

\(M = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 1&2&1\\ 1&2&1\\ 2&4&2 \end{array}} \right]\)

Characteristic equation is given

\(\left| {\begin{array}{*{20}{c}} {1 - \lambda }&2&1\\ 1&{2 - \lambda }&1\\ 2&4&{2 - \lambda } \end{array}} \right| = 0\)

\(\lambda ^3 - 5\lambda ^2 =0\)

\(\lambda = 0\;or\;\lambda = 5\)

Therefore largest value is 5.

Concept:

Consider the system of m linear equations

a₁₁ x₁ + a₁₂ x₂ + … + a_1n x_n = b₁

a₂₁ x₁ + a₂₂ x₂ + … + a_2n x_n = b₂

…

a_m1 x₁ + a_m2 x₂ + … + a_mn x_n = b_m

The above equations containing the n unknowns x₁, x₂, …, x_n. To determine whether the above system of equations is consistent or not, we need to find the rank of following matrices.

\(A = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}\\{{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}\\ \ldots & \ldots & \ldots & \ldots \\{{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}}\end{array}} \right]\) and \(\left[ {A{\rm{|}}B} \right] = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}&{{b_1}}\\{{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}&{{b_2}}\\ \ldots & \ldots & \ldots & \ldots & \ldots \\{{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}}&{{b_m}}\end{array}} \right]\)

A is the coefficient matrix and [A|B] is called as augmented matrix of the given system of equations.

We can find consistency of the given system of equations as follows:

(i) If the rank of matrix A is equal to rank of augmented matrix and it is equal to number of unknowns, then system is consistent and there is a unique solution.

Rank of A = Rank of augmented matrix = n

(ii) If the rank of matrix A is equal to rank of augmented matrix and it is less than the number of unknowns, then system is consistent and there are infinite number of solutions.

Rank of A = Rank of augmented matrix < n

(iii) If the rank of matrix A is not equal to rank of augmented matrix, then system is inconsistent, and it has no solution.

Rank of A ≠ Rank of augmented matrix

Calculation:

The given system of equations can be represented in a matrix form as shown below.

\(A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&{ - 3}\\2&0&1\\0&{ - 2}&{ - 7}\end{array}} \right],\;X = \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right],\;B = \left[ {\begin{array}{*{20}{c}}3\\0\\\alpha \end{array}} \right]\)

The Augmented matrix can be written by

\(\left[ {A{\rm{|}}B} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&{ - 3}\\2&0&1\\0&{ - 2}&{ - 7}\end{array}{\rm{|}}\begin{array}{*{20}{c}}3\\0\\\alpha \end{array}} \right]\)

R₂ → R₂ – 2R₁

\(= \left[ {\begin{array}{*{20}{c}}1&{ - 1}&{ - 3}\\0&2&7\\0&{ - 2}&{ - 7}\end{array}{\rm{|}}\begin{array}{*{20}{c}}3\\{ - 6}\\\alpha \end{array}} \right]\)

R₃ → R₃ + R₂

\(= \left[ {\begin{array}{*{20}{c}}1&{ - 1}&{ - 3}\\0&2&7\\0&0&0\end{array}{\rm{|}}\begin{array}{*{20}{c}}3\\{ - 6}\\{\alpha - 6}\end{array}} \right]\)

To have a solution, rank of A should be equal to rank of augmented matrix [A|B]

Concept:

∗ Distinct Eigen values gives district Eigen vectors

∗ Eigen values are roots of the Characteristic Equation which is given by

det (A - λI) = 0

Calculation:

det \(\left( A-\lambda I \right)=\det \left( \begin{array}{*{35}{r}} -\lambda & -1 & -1 \\ 1 & 2-\lambda & 1 \\ 1 & 1 & 2-\lambda \\\end{array} \right)\)

\(-\lambda \left[ {{\left( 2-\lambda \right)}^{2}}-1 \right]+1\left[ \left( 2-\lambda \right)-1 \right]-1\left( 1-\left( 2-\lambda \right) \right)\)

⇒ - λ³ + 4 λ²-5λ + 2

Characteristic equation

⇒ - λ³ + 4 λ² - 5λ + 2 = 0

⇒ λ³ - 4 λ² + 5λ - 2 = 0

⇒ (λ - 1)² (λ - 2) = 0

λ = 1, 1, 2