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:

$A=\left[\begin{array}{cccc}x& x& x& x\\ x& x& x& x\\ x& x& x& x\\ x& x& x& x\end{array}\right]$

Since all the rows are equal

R₂ → R2 - R₁

R3 → R3 - R1

R4 → R4 - R1

$A=\left[\begin{array}{cccc}x& x& x& x\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$

Therefore rank = 1

 [M]^T = [M^-1]

Therefore M is the orthogonal matrix

 ∴ [M] [M]^T = I

 $\left[\begin{array}{cc}\frac{3}{5}& \frac{4}{5}\\ x& \frac{3}{5}\end{array}\right]\left[\begin{array}{cc}\frac{3}{5}& x\\ \frac{4}{5}& \frac{3}{5}\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$=\left[\begin{array}{cc}1& \frac{3x}{5}+\frac{12}{25}\\ \frac{3x}{5}+\frac{12}{25}& x^2+\frac{9}{25}\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

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

$\left|\begin{array}{cccc}8& 0& 0& 0\\ 9& 11& 13& 15\\ 19& 0& 15& 5\\ 11& 0& a& 18\end{array}\right|=0$

$8\times \left|\begin{array}{ccc}11& 13& 15\\ 0& 15& 5\\ 0& a& 18\end{array}\right|=0$

$\therefore 8\times 11\times \left|\begin{array}{cc}15& 5\\ a& 18\end{array}\right|=0$

8 × 11 × (15 × 18 - 5a) = 0

(15 × 18 - 5a) = 0

5a = 15 × 18

a =54

$X=\left[\begin{array}{cccc}5& 7& 2& 1\\ 1& 1& -8& 1\\ 2& 3& 5& 0\\ 3& 4& -3& 1\end{array}\right]$

R₁ ↔ R₂

$X\sim \left[\begin{array}{cccc}1& 1& -8& 1\\ 5& 7& 2& 1\\ 2& 3& 5& 0\\ 3& 4& -3& 1\end{array}\right]$

R₂ → R₂ – 5R₁

R₃ → R₃ – 5R₁

R₄ → R₄ – 5R₁

$X\sim \left[\begin{array}{cccc}1& 1& -8& 1\\ 0& 2& 42& -4\\ 0& 1& 21& -2\\ 0& 1& 21& -2\end{array}\right]$

R₃ → R₃ – (R₂ ÷ 2)

R₄ → R₄ – (R₂ ÷ 2)

$X\sim \left[\begin{array}{cccc}1& 1& -8& 1\\ 0& 2& 42& -4\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$

Rank(X) = 4 – 2 = 2

or

$\left|\begin{array}{cc}1& 1\\ 0& 2\end{array}\right|=2\ne 0$

∴ rank = 2

The determinant of the given matrix is

$\left|\begin{array}{cccc}4& 16& 1& 64\\ 5& 25& 1& 125\\ 8& 64& 1& 512\\ 10& 100& 1& 1000\end{array}\right|$

C₁ ↔C3

Determinant change by -ve

$-\left|\begin{array}{cccc}1& 16& 4& 64\\ 1& 25& 5& 125\\ 1& 64& 8& 512\\ 1& 100& 10& 1000\end{array}\right|$

C2 ↔C3

Determinant change by -ve

$\left|\begin{array}{cccc}1& 4& 16& 64\\ 1& 5& 25& 125\\ 1& 8& 64& 512\\ 1& 10& 100& 1000\end{array}\right|$

It is of the form

$\left|\begin{array}{cccc}1& a& a^2& a^3\\ 1& b& b^2& b^3\\ 1& c& c^2& c^3\\ 1& d& d^2& d^3\end{array}\right|$

$\mathrm{\Delta }=\left(a-b\right)\left(a-c\right)\left(a-d\right)\left(b-c\right)\left(b-d\right)\left(c-d\right)$

$$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

$X=\left[\begin{array}{cc}b& a\\ a& c\end{array}\right]$

∴ |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 

$A=\left[\begin{array}{ccc}2& 5& 7\\ \lambda & 10& 4\lambda -2\\ \lambda +4& 4+{\lambda }^2& 2{\lambda }^2-4\end{array}\right]$

Rank = Number of independent rows = 1

That implies that two rows are dependent.

R₃ → R₃ - 2R₁

 $A=\left[\begin{array}{ccc}2& 5& 7\\ \lambda & 10& 4\lambda -2\\ \lambda & {\lambda }^2-6& 2{\lambda }^2-18\end{array}\right]$

R₃ → R₃ - R₂

$A=\left[\begin{array}{ccc}2& 5& 7\\ \lambda & 10& 4\lambda -2\\ \lambda & {\lambda }^2-6& 2{\lambda }^2-4\lambda -18\end{array}\right]$

R₂ → R₂ - 2R₁

$A=\left[\begin{array}{ccc}2& 5& 7\\ \lambda -4& 0& 4\lambda -16\\ 0& {\lambda }^2-16& 2{\lambda }^2-4\lambda -16\end{array}\right]$

For rank = 1, two rows should be zero.

⇒ λ = 4

Characteristic equation of given matrix

$\left|M\right|=\left|\begin{array}{cc}-3-\lambda & 4\\ -1& 2-\lambda \end{array}\right|=0$ 

$\left(-3-\lambda \right)\left(2-\lambda \right)+4=0$

${\lambda }^2+\lambda -2=0$

${\lambda }^2=2-\lambda$

${\lambda }^3=2\lambda -{\lambda }^2$

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

${\lambda }^3=3\lambda -2$

Every matrix satisfies its own characteristic equation

$M^3=3M-2I$

$|A|=\left|\begin{array}{ccc}2& 5& 3\\ 7& 8& 1\\ 0& 7& 4\end{array}\right|=57$

$|B|=\left|\begin{array}{ccc}0& 7& 4\\ 7& 8& 1\\ 2& 5& 3\end{array}\right|$

R₃ ↔ R1

$-|B|=\left|\begin{array}{ccc}2& 5& 3\\ 7& 8& 1\\ 0& 7& 4\end{array}\right|$

-|B| = |A|

∴ |B| = -57

$x\times y=-57\times 57=-3249$