Engineering Mathematics Test 1

If M_{a × b} N_{b × c} then (MN)_{a × c} , that is, if order of M is a × b and order of N is b × c then order of MN is a × c.

Order of A^T: 5 × 7

Order of C^T: 5 × 4

Order of (A^T B) = [AT]_{5 × 7}[B]7 ×5 = 5 × 5 = Order of (A^T B)^-1

Order of [C (A^T B)^-1 ] = [C]4 × 5 [(AT B)-1]5 × 5 = 4 × 5

Order of [C (A^T B)^-1CT] = [C(AT B)-1]4 × 5[CT]5 × 4 = 4 × 4

Order of [C (AT B)-1CT]^T = 4 × 4

Maximum square can be form of order 4 × 4

Rank ≤ 4 since all entries are 1, 

∴|any 4 × 4 submatrices | = | any 3 × 3 submatrices | = |any 2 × 2 submatrices | = 0

only 1 × 1 or single entries are ≠ 0 

∴ rank = 1

Explanation:

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

Since all the rows are equal

R₂ → R2 - R₁

R3 → R3 - R1

R4 → R4 - R1

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

Therefore rank = 1

 [M]^T = [M^-1]

Therefore M is the orthogonal matrix

 ∴ [M] [M]^T = I

 \(\left[ {\begin{array}{*{20}{c}} {\frac{3}{5}}&{\frac{4}{5}}\\ x&{\frac{3}{5}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{3}{5}}&x\\ {\frac{4}{5}}&{\frac{3}{5}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {1}&{\frac{3x}{5}} +\frac{12}{25}\\ \frac{3x}{5}+\frac{12}{25}&x^2+{\frac{9}{25}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

so, \(\frac{{12}}{{25}} + \frac{3}{5}x = 0 \Rightarrow x = \frac{{ - 4}}{5}\)

The value of x is \(\frac{{ - 4}}{5}\)

Concept:

Determinant of a singular matrix is 0

Calculation:

|A| = 0

\(\left| {\begin{array}{*{20}{c}} 8&0&0&0\\ 9&{11}&13&{15}\\ 19&0&15&5\\ {11}&0&a&18 \end{array}} \right| = 0\)

\(8 × \;\left| {\begin{array}{*{20}{c}} {11}&13&{15}\\ 0&15&5\\ 0&a&18 \end{array}} \right| = 0\)

\(\therefore 8 × 11 × \;\left| {\begin{array}{*{20}{c}} 15&5\\ a&18 \end{array}} \right| = 0\)

8 × 11 × (15 × 18 - 5a) = 0

(15 × 18 - 5a) = 0

5a = 15 × 18

a =54

\(X = \;\left[ {\begin{array}{*{20}{c}}5&7&2&1\\1&1&{ - 8}&1\\2&3&5&0\\3&4&{ - 3}&1\end{array}} \right]\)

R₁ ↔ R₂

\(X \sim \;\left[ {\begin{array}{*{20}{c}}1&1&{ - 8}&1\\5&7&2&1\\2&3&5&0\\3&4&{ - 3}&1\end{array}} \right]\)

R₂ → R₂ – 5R₁

R₃ → R₃ – 5R₁

R₄ → R₄ – 5R₁

\(X \sim \;\left[ {\begin{array}{*{20}{c}}1&1&{ - 8}&1\\0&2&{42}&{ - 4}\\0&1&{21}&{ - 2}\\0&1&{21}&{ - 2}\end{array}} \right]\)

R₃ → R₃ – (R₂ ÷ 2)

R₄ → R₄ – (R₂ ÷ 2)

\(X \sim \;\left[ {\begin{array}{*{20}{c}}1&1&{ - 8}&1\\0&2&{42}&{ - 4}\\0&0&0&0\\0&0&0&0\end{array}} \right]\)

Rank(X) = 4 – 2 = 2

or

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

∴ rank = 2

The determinant of the given matrix is

\(\left| {\begin{array}{*{20}{c}} 4&16&1&{64}\\ 5&25&{1}&{125}\\ 8&64&{1}&{512}\\ 10&100&{1}&{1000} \end{array}} \right|\)

C₁ ↔C3

Determinant change by -ve

\(-\left| {\begin{array}{*{20}{c}} 1&16&4&{64}\\ 1&25&{5}&{125}\\ 1&64&{8}&{512}\\ 1&100&{10}&{1000} \end{array}} \right|\)

C2 ↔C3

Determinant change by -ve

\(\left| {\begin{array}{*{20}{c}} 1&4&16&{64}\\ 1&5&{25}&{125}\\ 1&8&{64}&{512}\\ 1&10&{100}&{1000} \end{array}} \right|\)

It is of the form

\(\left| {\begin{array}{*{20}{c}} 1&a&{{a^2}}&{{a^3}}\\ 1&b&{{b^2}}&{{b^3}}\\ 1&c&{{c^2}}&{{c^3}}\\ 1&d&{{d^2}}&{{d^3}} \end{array}} \right|\)

\({\rm{Δ }} = \left( {a\; - \;b} \right)\left( {a\; - \;c} \right)\left( {a\; - \;d} \right)\left( {b\; - \;c} \right)\left( {b\; - \;d} \right)\left( {c\; - \;d} \right)\)

\(\)a = 4, b = 5, c = 8 and d = 10

Δ = (4 - 5)(4 - 8)(4-10)(5-8)(5-10)(8 - 10)

Δ = (-1) × (-4) × (-6) × (-3) × (-5) × (-2)

Let the symmetric matrix be

\(X=~\left[ \begin{matrix} b & a \\ a & c \\ \end{matrix} \right]\)

∴ |X| = bc – a²

Since the given matrix is real, to get maximum determinant value of a should be equal to 0.

b + c = 24

c = 24 – b

∴ |X| = b × (24 – b)

|X| = 24 b – b²

Differentiating with respect to b

|X|’ = 24 – 2b = 0

∴ b = 12

|X|’’ = - 2 < 0

At b = 12 it will have maximum value

a = 24 – 12 =12

Maximum value = 12 × 12 – 0 = 144 

\(A = \left[ {\begin{array}{*{20}{c}} 2&5&7\\ \lambda &{10}&{4\lambda - 2}\\ {\lambda + 4}&{4 + {\lambda ^2}}&{2{\lambda ^2} - 4} \end{array}} \right]\)

Rank = Number of independent rows = 1

That implies that two rows are dependent.

R₃ → R₃ - 2R₁

 \(A = \left[ {\begin{array}{*{20}{c}} 2&5&7\\ \lambda &{10}&{4\lambda - 2}\\ \lambda &{{\lambda ^2} - 6}&{2{\lambda ^2} - 18} \end{array}} \right]\)

R₃ → R₃ - R₂

\(A = \left[ {\begin{array}{*{20}{c}} 2&5&7\\ \lambda &{10}&{4\lambda - 2}\\ \lambda &{{\lambda ^2} - 6}&{2{\lambda ^2} - 4\lambda - 18} \end{array}} \right]\)

R₂ → R₂ - 2R₁

\(A = \left[ {\begin{array}{*{20}{c}} 2&5&7\\ {\lambda - 4}&0&{4\lambda - 16}\\ 0&{{\lambda ^2} - 16}&{2{\lambda ^2} - 4\lambda - 16} \end{array}} \right]\)

For rank = 1, two rows should be zero.

⇒ λ = 4

Characteristic equation of given matrix

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

\(\left( { - \;3\; - \;\lambda } \right)\left( {2 - \lambda } \right) + 4 = 0\)

\({\lambda ^2} + \lambda - 2 = 0\)

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

\({\lambda ^3} = 2\lambda \; - \;{\lambda ^2}\)

\({\lambda ^3} = 2\lambda \; - \;\left( {2\; - \lambda } \right)\)

\({\lambda ^3} = 3\lambda - 2\)

Every matrix satisfies its own characteristic equation

\({M^3} = 3M - 2I\)

\(|A| =\left| {\begin{array}{*{20}{c}} 2&5&3\\ 7&8&{ 1}\\ 0&7&4 \end{array}} \right| = 57\)

\(|B| =\left| {\begin{array}{*{20}{c}} 0&7&4\\ 7&8&{ 1}\\ 2&5&3 \end{array}} \right|\)

R₃ ↔ R1

\(-|B| =\left| {\begin{array}{*{20}{c}} 2&5&3\\ 7&8&{ 1}\\ 0&7&4 \end{array}} \right| \)

-|B| = |A|

∴ |B| = -57

\(x\times y = -57 \times 57 = -3249\)