Engineering Mathematics Test 3

Explanation:

The characteristic equation for the given matrix is

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

\({\lambda ^2} - 5\;\lambda = 0\)

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

The eigenvector corresponding to eigenvalue 0

\(\left[ {\begin{array}{*{20}{c}} {1 - 0}&2\\ 2&{4 - 0} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right]\)

R₂ → R₂ – 2R₁

\(\left[ {\begin{array}{*{20}{c}} 1&2\\ 0&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right]\)

x₁ + 2x₂ = 0

∴ if x₁ = -2

then x₂ = 1

The eigenvector corresponding to eigenvalue 0 is \(\left[ {\begin{array}{*{20}{c}} { - 2}\\ 1 \end{array}} \right]\)

The eigenvector corresponding to eigenvalue 5

\(\left[ {\begin{array}{*{20}{c}} {1 - 5}&2\\ 2&{4 - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}} { - 4}&2\\ 2&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right]\)

R₂ → 2R₂ + R₁

\(\left[ {\begin{array}{*{20}{c}} { - 4}&2\\ 0&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right]\)

-4x₁ + 2x₂ = 0

∴ if x₁ = 1

then x₂ = 2

The eigenvector corresponding to eigenvalue 5 is \(\left[ {\begin{array}{*{20}{c}} 1\\ 2 \end{array}} \right]\)

Tips and Tricks:

Since the given matrix is symmetric, so the product of transpose of one eigenvector to the other eigenvector should be zero. This can be verified from the given options

\({\left[ {\begin{array}{*{20}{c}} { - 2}\\ 1 \end{array}} \right]^T} = \;\left[ {\begin{array}{*{20}{c}} { - 2}&1 \end{array}} \right]\)

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

The eigenvectors of a symmetric matrix A corresponding to different eigenvalues are orthogonal to each other.

Concept:

Trace of matrix X = sum of its eigenvalues

Calculation:

Since it is 3 × 3 matrix, so roots will be 3 means 3 eigenvalues. Two eigenvalues are given as 6 and 10 + √-1

As 3 × 3 is a real matrix, in real matrix roots can never be imaginary. We have to take the complex roots in pair. So,

Three eigenvalues are:

6, (10 + √-1) and (10 - √-1), that is, 5, (10 + i) and (10 - i)

Trace of matrix X = 6 + 10 + i + 10 - i = 26

Important point:

√-1 = i → iota (imaginary number)

The given values are given by the solution of the characteristic equation, i.e.

|A - λI| = 0

λ₁, λ₂ ….. are different eigenvalues.

If Eigenvalues are distinct, then corresponding eigenvectors are linearly independent.

Vectors V₁, V₂ …. V_n are linearly independent.

If C₁V₁ + C₂V₂ + C₃V₃ …… + C_nV_n = 0

Implies C₁ = C₂ = ……. C_n = 0

|A| ≠ 0 implies that ‘O’ is not an eigenvalues

Other eigenvalues may be repeated, i.e. 1, 2, 2

The two same eigenvalues give linearly dependent eigenvectors.

Concept:

Determinant of Matrix  \(M = {\left[ {\begin{array}{*{20}{c}} a\\ b\\ :\\ d \end{array}} \right]_{n × 1}}{\left[ {p\;\;\;q\;\;\;..\;\;s} \right]_{1 × n\;}}\)  is 0

product of Eigenvalues =  Determinant of matrix 

sum of Eigenvalues =  trace of matrix

Explanation:

\(M = {\left[ {\begin{array}{*{20}{c}} 12\\ { 9}\\ \end{array}} \right]_{2 × 1}}{\left[ {6\;\;\;19\;\;} \right]_{1 × 2\;}}\;\)

\(M =\left[ \begin{matrix} 72 & 228 \\ 54 & 171 \\ \end{matrix} \right]\)

\(M =\left| \begin{matrix} 72 & 228 \\ 54 & 171 \\ \end{matrix} \right|\)

\(|M| =12 × 9\left| \begin{matrix} 6 & 19 \\ 6 & 19 \\ \end{matrix} \right|\)

|M| = 12 × 9 × 0 = 0

∴ λ1 × λ2 = 0

Either λ1 = 0 or λ1 = 0 

Trace = 6 + 19 = 25

λ₁ + λ2 = 25

Since trace ≠ 0

if λ1 = 0 then λ2 = 25

The largest eigenvalue of M is  25.

Characteristic equation of given matrix

\(\left| {\begin{array}{*{20}{c}} {3 - \lambda }&1&4\\ 0&{2\; - \;\lambda }&6\\ 0&0&{5 - \;\lambda } \end{array}} \right| = 0\)

\(\left( {3\; - \;\lambda } \right) \times \left( {2\; - \lambda } \right) \times \left( {5\; - \;\lambda } \right)\; = 0\)

Eigenvalues: 2, 3 and 5

Eigenvector corresponds to: λ = 2

\(\left[ {\begin{array}{*{20}{c}} 1&1&4\\ 0&0&6\\ 0&0&5 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right]\) = 0

\(x + y + 4z = 0\)

\(6z = 0\)

\(5z = 0\)

\(\therefore z = \;0\)

\(\therefore x = \; - y\)

if y = -2 then x = 2

Eigenvector: (2, -2, 0)

Eigenvector corresponds to: λ = 3

\(\left[ {\begin{array}{*{20}{c}} 0&1&4\\ 0&{ - 1}&6\\ 0&0&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right]\) = 0

\(y + 4z = 0\)

\(- y + \;6z = 0\)

\(2z = 0\)

\(\therefore z = \;0\)

\(\therefore y = \;0\)

x can take any value

Eigenvector: (5, 0, 0)

Eigen vector corresponds to: λ = 5

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

\(0 \times z = 0\)

∴ z can take any value

Let z = 1

\(- 3y + 6z = 0\)

\(\therefore y = \;2\)

\(- 2x + y + 4z = 0\)

\(\therefore x = \;3\)

Eigenvector: (3, 2, 1)

A² – 4A = 5I

A² – 4A - 5I = 0

Every matrix satisfies its own characteristic equation.

λ² – 4λ - 5 = 0

(λ – 5) × (λ + 1) = 0

∴ λ = 5 or λ = -1

where λ is eigen value

Product of eigen value is equal to determinant of matrix

ad – 4 × 2 = 5 × -1

ad – 8 = -5

ad = 3

The sum of the eigen value of a matrix is equal to its trace

a + d = 5 + (-1)

a + d = 4

(a – d)² = (a + d)² - 4ad

= 16 – 12

= 4

∴ a – d = ± 2

|a – d | = 2

Let \(\text{A}=\left[ \begin{matrix} 2 & -4 \\ 4 & -2 \\\end{matrix} \right]\)

Considering characteristic equation |A - λ I| = 0

∴ Characteristics equation is given by

\(\left| \begin{array}{*{35}{r}} 2-\lambda & -4 \\ 4 & -2-\lambda \\\end{array} \right|=0\)

⇒ λ² + 2λ – 2λ – 4 +16 = 0

⇒ λ² + 12 = 0

\(\text{ }\!\!\lambda\!\!\text{ }=+2\sqrt{3}~i\) (Thus complex eigen value)

Now consider (A - λI) X = 0

\(\left[ \begin{array}{*{35}{r}} 2-\text{ }\!\!\lambda\!\!\text{ } & -4 \\ 4 & -2-\text{ }\!\!\lambda\!\!\text{ } \\\end{array} \right]\left[ \begin{matrix} {{\text{x}}_{1}} \\ {{\text{x}}_{2}} \\\end{matrix} \right]=\left[ \begin{matrix} 0 \\ 0 \\\end{matrix} \right]\)

When \(\text{ }\!\!\lambda\!\!\text{ }=\pm 2\sqrt{3}~i\)

Eigen vector is given by:

\(\left[ \begin{array}{*{35}{r}} -2-2\sqrt{3\text{i}} & -4 \\ 4 & -2\sqrt{3\text{i}} \\\end{array} \right]\left[ \begin{matrix} {{\text{x}}_{1}} \\ {{\text{x}}_{2}} \\\end{matrix} \right]=\left[ \begin{matrix} 0 \\ 0 \\\end{matrix} \right]\)

\(2-2\sqrt{3}~i~{{\text{x}}_{1}}+4\text{ }\!\!~\!\!\text{ }{{\text{x}}_{2}}=0\)

\(\Rightarrow \frac{{{\text{x}}_{1}}}{4}=\frac{{{\text{x}}_{2}}}{2-2\sqrt{3}\text{ }\!\!~\!\!\text{ i}}\)

Thus, Eigen vector \({{\text{X}}_{1}}=\left[ \begin{matrix} {{\text{x}}_{1}} \\ {{\text{x}}_{2}} \\\end{matrix} \right]=\left[ \begin{matrix} 4 \\ 2-2\sqrt{3}~i \\\end{matrix} \right]\)

When \(\text{ }\!\!\lambda\!\!\text{ }=-2\sqrt{3}~i\)

Eigen vector is given by:

\(\left[ \begin{array}{*{35}{r}} -2-2\sqrt{3}~i & -4 \\ 4 & -2+2\sqrt{+3}~i \\\end{array} \right]\left[ \begin{matrix} {{\text{x}}_{1}} \\ {{\text{x}}_{2}} \\\end{matrix} \right]=\left[ \begin{matrix} 0 \\ 0 \\\end{matrix} \right]\)

\(2+2\sqrt{3}~i~{{x}_{1}}-4{{x}_{2}}=0\)

\(\Rightarrow \frac{{{x}_{1}}}{4}=\frac{{{x}_{2}}}{2+2\sqrt{3}~i}\)

Thus Eigen vector \({{\text{X}}_{2}}=\left[ \begin{matrix} {{\text{x}}_{1}} \\ {{\text{x}}_{2}} \\\end{matrix} \right]=\left[ \begin{matrix} 4 \\ 2+2\sqrt{3}\text{ }\!\!~\!\!\text{ i} \\\end{matrix} \right]\)

Hence the eigen vectors are \({{\text{x}}_{1}}=\left[ \begin{matrix} 4 \\ 2+2\sqrt{3}\text{ }\!\!~\!\!\text{ i} \\\end{matrix} \right]\And \text{ }\!\!~\!\!\text{ }{{\text{x}}_{2}}=\left[ \begin{matrix} 4 \\ 2+2\sqrt{3}~i \\\end{matrix} \right]\) are complex.

∴ The matrix \(\left( \begin{matrix} 2 & -4 \\ 4 & -2 \\\end{matrix} \right)\) has complex eigen values and eigen vectors.

 \(M =\left[ {\begin{array}{*{20}{c}} 2&0&0&0&2\\ 0&2&2&2&0\\ 0&2&2&2&0\\ 0&2&2&2&0\\ 2&0&0&0&2 \end{array}} \right]\)  

\(\left| {\rm{M}} \right| = \left| {\begin{array}{*{20}{c}} 2&0&0&0&2\\ 0&2&2&2&0\\ 0&2&2&2&0\\ 0&2&2&2&0\\ 2&0&0&0&2 \end{array}} \right|\)

R5 → R5 – R1

R3 → R3 – R2

R4 → R4 – R2

\(\left| {\rm{M}} \right| = {\rm{\;}}\left| {\begin{array}{*{20}{c}} 2&0&0&0&2\\ 0&2&2&2&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0 \end{array}} \right|\)

 Rank of the matrix is 2

number non-zero value eigen value in M ≤ rank(M)

Therefore, there are 2 or less nonzero eigen values

The characteristic equation of A is| M - λ I | = 0

where λ is eigen value and I is identity matrix

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

R1 → R1 + R5

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

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

R2 → R2 + R3 + R4

\(\left( {4 - \lambda } \right)\left| {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&{6 - \lambda }&{6 - \lambda }&{6 - \lambda }&0\\ 0&2&{2 - \lambda }&2&0\\ 0&2&2&{2 - \lambda }&0\\ 2&0&0&0&{2 - \lambda } \end{array}} \right| = 0\)

\(\left( {4 - \lambda } \right)\left( {6 - \lambda } \right)\left| {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&1&1&0\\ 0&1&{2 - \lambda }&1&0\\ 0&1&1&{2 - \lambda }&0\\ 1&0&0&0&{2 - \lambda } \end{array}} \right| = 0\)

∴ λ = 4 or λ = 6

Therefore, the product of eigen values = 4 × 6 = 24

Concept:

Diagonalization of matrix:

If a square matrix Q of order n has n linearly independent Eigen vectors, then matrix P can be found such that \({P^{ - 1}}QP\) is a diagonal matrix.

Let Q be a square matrix of order 3.

Let λ₁, λ₂, and λ₃ be Eigen values of matrix Q and \({X_1} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{y_1}}\\{{z_1}}\end{array}} \right],\;{X_2} = \left[ {\begin{array}{*{20}{c}}{{x_2}}\\{{y_2}}\\{{z_2}}\end{array}} \right],\;{X_3} = \left[ {\begin{array}{*{20}{c}}{{x_3}}\\{{y_3}}\\{{z_3}}\end{array}} \right]\) be the corresponding Eigen vectors.

Let denote the square matrix \(\left[ {\begin{array}{*{20}{c}}{{X_1}}&{{X_2}}&{{X_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{x_1}}&{{x_2}}&{{x_3}}\\{{y_1}}&{{y_2}}&{{y_3}}\\{{z_1}}&{{z_2}}&{{z_3}}\end{array}} \right]\) by P.

Now, the given matrix A can be diagonalized by \(D = {P^{ - 1}}QP\)

Or the matrix A can be represented by \(Q = PD{P^{ - 1}}\)

Where D is the diagonal matrix and it is represented by \(D = \left[ {\begin{array}{*{20}{c}}{{\lambda _1}}&0&0\\0&{{\lambda _2}}&0\\0&0&{{\lambda _3}}\end{array}} \right]\)

Sum of all the elements of matrix \({P^{ - 1}}QP\) = Sum of all the elements of matrix D = sum of Eigen values of Q

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:

|Q – λI| = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}}{3 - \lambda }&2&4\\2&{ - \lambda }&2\\4&2&{3 - \lambda }\end{array}} \right| = 0\)

⇒ (3 – λ) [(-λ) (3 – λ) – 4] – 2 [2(3-λ) – 8] + 4 [4 – (-4λ)] = 0

⇒ (3 – λ) [-3λ + λ² – 4] – 2 [6 – 2λ – 8] + 4 [4 + 4λ] = 0

⇒ -9 λ + 3 λ² + 3 λ² – λ³ – 12 + 4 λ + 4 + 4 λ + 16 + 16 λ = 0

⇒ λ³ – 6 λ² – 15 λ – 8 = 0

⇒ λ = 8, -1, -1

Eigen values of the matrix Q = 8, -1, -1

Diagonal matrix of Q is given by,

\(D = {P^{ - 1}}QP = \left[ {\begin{array}{*{20}{c}}8&0&0\\0&{ - 1}&0\\0&0&{ - 1}\end{array}} \right]\)

Sum of the absolute values of all the elements = 8 + 1 + 1 = 10

Common Mistakes:

If the question is to find the sum of all the elements of matrix \({P^{ - 1}}QP\) = sum of Eigen values of matrix Q = λ₁ + λ₂ + λ₃ = trace of matrix Q = 3 + 0 + 3 = 6 (one of the options given)

But the actual question is to find the absolute sum of all the elements of matrix \({P^{ - 1}}QP\) = absolute sum of all the Eigen values of matrix = |λ₁| + |λ₂| + |λ₃|

So, we can’t find directly the absolute sum of all the elements of Eigen values of matrix. We need to find the Eigen values of the matrix Q as solved above.

\({\lambda ^3} + \;a{\lambda ^2} + 47\lambda \; - \;60 = 0\)

Since one eigenvalue of matrix is 4 it satisfies the characteristic equation

\({4^3} + \;a{4^2} + 47\left( 4 \right)\; - \;60 = 0\)

\(64 + \;16a + 188\; - \;60 = 0\)

\(a = \; - 12\)

\({\lambda ^3} - 12\;{\lambda ^2} + 47\lambda \; - \;60 = 0\)

Solving this will give

\(\left( {\lambda - \;3} \right)\left( {\lambda \; - \;4} \right)\left( {\lambda \; - \;5} \right) = 0\)

\(\lambda \; = \;3\;or\;\lambda \; = \;4\;or\;\lambda \; = \;5\)

smallest eigenvalue is 3