Engineering Mathematics Test 2

Concept:

It is case of equal matrix

∴ L. H. S = R. H. S

Calculation:

\(\left[ {\begin{array}{*{20}{c}} 2&5\\ { - 4}&3 \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right\} = \left\{ {\begin{array}{*{20}{c}} 2\\ { - 30} \end{array}} \right\}\)

2x + 5y = 2     ----(I)

-4x + 3y = -30   ----(II)

Multiply eq (1) by 4 and (II) by 2

∴8x + 20y = 8       ---(III)

-8x + 6y = -60       ---(IV)

Adding eq (I) and (II)

26y = -52 ⇒ y = -2

Putting the value of y is eq (I)

2x + 5 (-2) = 2 ⇒ x = 6

Given linear equations 

10a + 3b + 2c = 0

6a + 5b + c = 0

4a + pb + 5c = 0

Corresponding matrix

\(M=\left[ {\begin{array}{*{20}{c}} 10&3&2\\ 6&5&1\\ 4&p&5 \end{array}} \right]\)

linear equations have infinitely many solutions

Therefore determinant of the matrix is 0

\(|M|= 0\)

\(\left| {\begin{array}{*{20}{c}} 10&3&2\\ 6&5&1\\ 4&p&5 \end{array}} \right|= 0\)

10(25 - p) -3(30 - 4) + 2(6p - 20) = 0

250 - 10p - 78 + 12p - 40 = 0

2p = -132 

∴ p = -66

An invertible matrix A has an LU decomposition provided that all it's leading submatrices have non-zero determinants.

All the matrices in the options are invertible because 

1.  \(|A| = 2\) 
2.  \(|A| = 1\)
3.  \(|A| = 1\)
4. \(|A| = -6\)

In option 4

\(A = \left[ {\begin{array}{*{20}{c}} 0&2\\ 3&1 \end{array}} \right]\) in this leading sub matrix \(A = [ 0 ]\)

Hence LU decomposition is not possible

(x + 3) p + 12q = 6x

xp + (x - 5)q = 5x + 8

\(A = \left[ {\begin{array}{*{20}{c}} {x + 3}&12\\ x&{x - 5} \end{array}} \right]B = \left[ {\begin{array}{*{20}{c}} {6x}\\ {5x + 8} \end{array}} \right]\)

The given system of linear equations has infinitely many solutions if ρ(A) = ρ(A|B) < n = 2

\(\left[ {A|B} \right] = \left[ {\begin{array}{*{20}{c}} {x + 3}&12&|~~{6x}\\ x&{x - 5}&~~~~|{5x+8 } \end{array}} \right]\)

To get the rank less than 2, one row should be dependent on another.

(x+3)(x - 5) - 12x = 0

x² - 5x + 3x - 15 - 12x = 0

x2 - 14x - 15 = 0

x2 - 15x + x - 15 = 0

(x-15)(x + 1) = 0

x = 15 and x = -1

For x = 15, system of equation is inconsistent.

For x = - 1, it has infinately many solution. 

2x – 4z + 6y  + 2 = 0

∴2x + 6y – 4z = – 2

6z + 10y + 16 = 0

∴10y + 6z = -16

– 10z + 2x – 4y

∴ 2x – 4y – 10z = 14

The coefficient matrix of the given linear equation is

\(\left[ {\begin{array}{*{20}{c}} 2&6&{ - 4}\\ 0&10&6\\ 2&{ - 4}&{ - 10} \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} x\\ y \end{array}}\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} { - 2}\\ { - 16} \end{array}}\\ 14 \end{array}} \right]\;\)

\(let\;M = \left[ {\begin{array}{*{20}{c}} 2&6&{ - 4}\\ 0&10&6\\ 2&{ - 4}&{ - 10} \end{array}} \right]\)

order = 3

Let the augmented matrix be

\(X = \;\left[ {\begin{array}{*{20}{c}} 2&6&{ - 4}\\ 0&10&6\\ 2&{ - 4}&{ - 10} \end{array}\left| {\begin{array}{*{20}{c}} { - 2}\\ { - 16}\\ 14 \end{array}} \right.} \right]\)

R3 → R3 – R1

\(X = \;\left[ {\begin{array}{*{20}{c}} 2&6&{ - 4}\\ 0&10&6\\ 0&{ - 10}&{ - 6} \end{array}\left| {\begin{array}{*{20}{c}} { - 2}\\ { - 16}\\ 16 \end{array}} \right.} \right]\)

R3 → R3 + R2

\(X = \;\left[ {\begin{array}{*{20}{c}} 2&6&{ - 4}\\ 0&10&6\\ 0&{ 0}&{ 0} \end{array}\left| {\begin{array}{*{20}{c}} { - 2}\\ { - 16}\\ 0 \end{array}} \right.} \right]\)

Since Rank of M = Rank of X < 3

Therefore, the system has infinitely many solutions.

\(\)LU = M

\(\left[ {\begin{array}{*{20}{c}} 1&0\\ a&1 \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} x&y\\ 0&z \end{array}} \right]\;= \left[ {\begin{array}{*{20}{c}} 7&11\\ 5&9 \end{array}} \right]\;\)

\(\left[ {\begin{array}{*{20}{c}} x&y\\ ax+0&ay+z \end{array}} \right]\;= \left[ {\begin{array}{*{20}{c}} 7&11\\ 5&9 \end{array}} \right]\;\)

Comparing:

x = 7, y = 11

ax = 5, ay+z = 9

a = 5/7

ay+z = 9

(5/7) × 11 + z = 9 

z = 8/7

Diagonal of x and z

x × z = 7 × 8/7 = 8

The product of the trace of Matrix U is 8

Set S1 contains homogeneous equations, homogeneous equations are always consistent because it always satisfied a trivial solution i.e. \(x = y = z = 0\)

For S2

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

R2 → R2 - 3R1

\(\left[ {\begin{array}{*{20}{c}} 2&6&0&{ - 11}\\ 0&2&{ - 6}&{30}\\ 0&6&{ - 18}&{ - 1} \end{array}} \right]\)

R2 → 3R2

\(\left[ {\begin{array}{*{20}{c}} 2&6&0&{ - 11}\\ 0&6&{ - 18}&{90}\\ 0&6&{ - 18}&{ - 1} \end{array}} \right]\)

R3 → R3 - R2

\(\left[ {\begin{array}{*{20}{c}} 2&6&0&{ - 11}\\ 0&6&{ - 18}&{90}\\ 0&0&0&{ - 91} \end{array}} \right]\)

Rank Augmented matrix is 3 while rank of coefficient matrix has rank is 2

Hence the S2 is inconsistent

∴ only S1 is consistent

\(\left[ {\begin{array}{*{20}{c}} 2&3&3\\ 3&2&1\\ 3&1&7 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&0&0\\ {{a_{12}}}&{{a_{22}}}&0\\ {{a_{13}}}&{{a_{23}}}&{{a_{33}}} \end{array}} \right]{\left[ {\begin{array}{*{20}{c}} {{a_{11}}}&0&0\\ {{a_{12}}}&{{a_{22}}}&0\\ {{a_{13}}}&{{a_{23}}}&{{a_{33}}} \end{array}} \right]^T}\)

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 2&3&3\\ 3&2&1\\ 3&1&7 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&0&0\\ {{a_{12}}}&{{a_{22}}}&0\\ {{a_{13}}}&{{a_{23}}}&{{a_{33}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ 0&{{a_{22}}}&{{a_{23}}}\\ 0&0&{{a_{33}}} \end{array}} \right]\)

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 2&3&3\\ 3&2&1\\ 3&1&7 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{11}}{a_{12}}}&{{a_{11}}{a_{13}}}\\ {{a_{11}}{a_{12}}}&{a_{12}^2 + a_{22}^2}&{{a_{12}}{a_{13}} + {a_{22}}{a_{23}}}\\ {{a_{11}}{a_{13}}}&{{a_{12}}{a_{13}} + {a_{22}}{a_{23}}}&{a_{13}^2 + a_{23}^2 + a_{33}^2} \end{array}} \right]\)

By comparing on both sides,

a₁₁ = 2

a₁₁ a₁₂ = 3 ⇒ a₁₂ = 3/2

a₁₁ a₁₃ = 3 ⇒ a₁₃ = 3/2

\(a_{12}^2 + a_{22}^2 = 2\)

\( \Rightarrow {\left( {\frac{3}{2}} \right)^2} + a_{22}^2 = 2 \Rightarrow {a_{22}} = \frac{j}{2}\)

a₁₂ a₁₃ + a₂₂ a₂₃ = 1

\( \Rightarrow \left( {\frac{3}{2}} \right)\left( {\frac{3}{2}} \right) + \left( {\frac{j}{2}} \right){a_{23}} = 1 \Rightarrow {a_{23}} = j\frac{5}{2}\)

\(a_{13}^2 + a_{23}^2 + a_{33}^2 = 7\)

\( \Rightarrow {\left( {\frac{3}{2}} \right)^2} + {\left( {j\frac{5}{2}} \right)^2} + a_{33}^2 = 7\)

\( \Rightarrow a_{33}^2 = 7 - \frac{9}{4} + \frac{{25}}{4} = 11 \Rightarrow {a_{33}} = \sqrt {11} \)

Now the lower triangular representation of the given 3 × 3 matrices obtained through the given decomposition becomes

\(\left[ {\begin{array}{*{20}{c}} 2&3&3\\ 3&2&1\\ 3&1&7 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2&0&0\\ {\frac{3}{2}}&{j\frac{1}{2}}&0\\ {\frac{3}{2}}&{j\frac{5}{2}}&{\sqrt {11} } \end{array}} \right]{\left[ {\begin{array}{*{20}{c}} 2&0&0\\ {\frac{3}{2}}&{j\frac{1}{2}}&0\\ {\frac{3}{2}}&{j\frac{5}{2}}&{\sqrt {11} } \end{array}} \right]^T}\)

Therefore, the number of purely real elements = 7

• If c ≠ a, the equations are inconsistent i.e.  no solution possible
• If c = a < n, the equations are consistent with infinite number of solutions
• If c = a =n, the equations are consistent and there is a unique solution.

For given system of equations to have unique solution, the determinant of the coefficient matrix should be not equal to 0

\(\left| {\begin{array}{*{20}{c}}2&a&1\\1&3&4\\1&2&3\end{array}} \right| \ne 0\)

\(2\left( {9 - 8} \right)\; - \;a\left( {3 - 4} \right) + \left( {2\; - \;3} \right) \ne 0\)

\(2 + a\; - \;1\; \ne 0\;\)

\(a\; \ne \; - 1\)

Since Δ of coefficient matrix is not equal to 0 then b and c can take any value

Note:

If coefficient matrix’s Δ ≠ 0 then rank of coefficient matrix is always equal rank of augmented matrix so right-hand side of given equation can take any value