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Cramer's Rule

If a + b = x,...

Question

If a + b = x, a – b = y and x < y, then b can be equal to

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x \neq a, y \neq b, z \neq c

∣ ∣ ∣ ∣ \begin{matrix} a & y & z x & b & z x & y & c \end{matrix} ∣ ∣ ∣ ∣ = 0

(\frac{x + a}{x - a} + \frac{y + b}{y - b} + \frac{z + c}{z - c})

(i) If a(y+z) = b(z+x) = c(x+y) and out of a, b, c no two of them are equal then show that,

\frac{y - z}{a (b - c)} = \frac{z - x}{b (c - a)} = \frac{x - y}{c (a - b)}

\frac{x}{3 x - y - z} = \frac{y}{3 y - z - x} = \frac{z}{3 z - x - y}

x + y + z \neq 0

then show that the value of each ratio is equal to 1.

(iii) If

\frac{a x + b y}{x + y} = \frac{b x + a z}{x + z} = \frac{a y + b z}{y + z}

x + y + z \neq 0

\frac{a + b}{2}

\frac{y + z}{a} = \frac{z + x}{b} = \frac{x + y}{c}

\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c}

\frac{3 x - 5 y}{5 z + 3 y} = \frac{x + 5 z}{y - 5 x} = \frac{y - z}{x - z}

then show that every ratio =

\frac{x}{y}

a s i n^{2} x + b c o s^{2} x = c,

b s i n^{2} y + a c o s^{2} y = d,

a t a n x = b t a n y,

\frac{a^{2}}{b^{2}}

y = a + \frac{b}{x}

a

b

are constants, and if

y = 1

x = - 1

y = 5

x = - 5

a + b

A

B

are symmetric matrices of the same order and

X = A B + B A

Y = A B - B A

(X Y)^{T}

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