Engineering Mathematics Test 1

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.

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

Explanation:

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

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

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

Calculation:

Given:

Trace (A) = 14, det(A) = 100

Trace of (A) = sum of diagonal elements

\(\left[ {\begin{array}{*{20}{c}}a&0&3&7\\2&5&1&3\\0&0&2&4\\0&0&0&b\end{array}} \right]\) 

14 = a + 5 + 2 + b

a + b = 7         ---(1)

Now,

\(\det \left( A \right) = 5\left| {\begin{array}{*{20}{c}}a&3&7\\0&2&4\\0&0&b\end{array}} \right|\)

\(5 \times 2 \times a \times b = 100\) 

ab = 10          ---(2)

from equation (1) and (2)

Either,  a = 5, b = 2

Or,        a = 2, b = 5

So,

|a - b| = |5 - 2| = |2 - 5| = 3

Explanation:

Characteristic equation is |A - λI| = 0

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

(3 - λ)(2 - λ)(5 - λ) = 0

λ = 2, 3, 5

Note:

The Eigen values of a triangular matrix are the diagonal elements

Now, highest Eigen value = 5

Eigen vector corresponding to Eigen value, λ = 5

⇒ [A – 5I]X = 0

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

⇒ -2x + y + yz = 0     …1)

& -3y + 6z = 0 ⇒ y = 2z

Put in 1)

-2x + 2z + 4z = 0

x = 3z

Thus,

\(\frac{x}{3} = \frac{y}{2} = \frac{z}{1}\)

Concept:

If |A| ≠ 0 ⇒ It has a unique solution.

If |A| =  0 ⇒ It has either no solution or infinite solution.

ρ(A) ≠ ρ(A|B) ⇒ It has either no solution

ρ(A) = ρ(A|B) r < n⇒ It has infinite solution

Calculation:

For given system of equations to have no solution, the determinant of the coefficient matrix should be equal to 0

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

\(1 \times \left( {4x - 24} \right) - 1 \times \left( {2x - 12} \right) + 1\left( {8 - 8} \right) = 0\)

\(2x - 12 = 0\)

\(\therefore x = 6\)

Augmented matrix is given by

\(\left[ {\begin{array}{*{20}{c}} 1&1&1&{18}\\ 2&4&6&{12}\\ 2&4&6&y \end{array}} \right]\)

\(R3 \to R3 - R2\)

\(\left[ {\begin{array}{*{20}{c}} 1&1&1&{18}\\ 2&4&6&{12}\\ 0&0&0&{y - 12} \end{array}} \right]\)

For given system of equations to have no solution, the rank of the augmented matrix should not be equal to coefficient matrix

ρ(A) = 2

\(\therefore y - 12 ≠ 0\)

\(\therefore y ≠ 12\)

Explanation:

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

\(D = \left[ {\begin{array}{*{20}{c}}{ - 2}&4\\3&{ - 1}\end{array}} \right]\) 

Then, BAC = D

AC = B^-1.D

A = B^-1.D.C^-1

Now,

\({B^{ - 1}} = \frac{{adj\;B}}{{\left| B \right|}} = \frac{1}{1}\left[ {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 3}&2\end{array}} \right]\)

\({C^{ - 1}} = \frac{1}{{9 - 10}}\left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 2}\\{ - 5}&{ - 3}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&2\\5&3\end{array}} \right]\) 

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

\(A = \left[ {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 3}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{14}&8\\4&3\end{array}} \right]\) 

\(\therefore A = \left[ {\begin{array}{*{20}{c}}{24}&{13}\\{ - 34}&{ - 18}\end{array}} \right]\)

Explanation:

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

|A| = -2 + 3 = 1

|A¹⁰⁰⁹ – 5 A¹⁰⁰⁸| = |A¹⁰⁰⁸ (A – 5I)|

|A¹⁰⁰⁹ – 5 A¹⁰⁰⁸| = |A¹⁰⁰⁸| |(A – 5I)|

|A¹⁰⁰⁹ – 5 A¹⁰⁰⁸| = |A|¹⁰⁰⁸ |(A – 5I)|

∴ |A¹⁰⁰⁹ – 5 A¹⁰⁰⁸| = |A – 5I|

Now,

\(A - 5I = \left[ {\begin{array}{*{20}{c}}2&{ - 3}\\1&{ - 1}\end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\) 

\(A - 5I = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 3}\\1&{ - 6}\end{array}} \right]\) 

|A – 5I| = 18 + 3 = 21

Concept:

The rank of a matrix is defined as the maximum number of linearly independent column vectors in the matrix or the maximum number of linearly independent row vectors in the matrix.

In terms of determinant, rank of a matrix can be viewed as m where m is the size of the largest non- zero m × m submatrix with non-zero determinant.

Calculation:  

\(P = \left[ {\begin{array}{*{20}{c}}1&1&{ - 1}\\2&{ - 3}&4\\3&{ - 2}&3\end{array}} \right]\;and\;Q = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}&{ - 1}\\6&{12}&6\\5&{10}&5\end{array}} \right]\)

\(Let\;X = P + Q = \;\left[ {\begin{array}{*{20}{c}}0&{ - 1}&{ - 2}\\8&9&{10}\\8&8&8\end{array}} \right]\)

\(\left| X \right| = \left| {\begin{array}{*{20}{c}}0&{ - 1}&{ - 2}\\8&9&{10}\\8&8&8\end{array}} \right|\)

|X| = 0 (72 – 80) – (–1) (64 – 80) – 2(64 – 72) 

∴ |X| = 0 + (–16) + 16 = 0

Rank of 3 × 3 matrix is zero.

∴ Rank(X) ≠ 3

\(\left| {\begin{array}{*{20}{c}}0&{ - 1}\\8&9\end{array}} \right| = 8 \ne 0\)