Determinants Test - 16

Since, A and B are invertible matrices, So, we can say that

\((AB)^{-1} = B^{-1}A^{-1} \dots(i)\)

Also, \(A^{-1} = \frac{1}{|A|} \)(adj A)

⇒ adj A = |A|.\(A^{-1} \)  ....(ii)

Also, det\((A)^{-1} = [det(A)]^{-1}\)

⇒ det\((A)^{-1}\) = \(\frac{1}{[det(A)]}\)

⇒ det(A). det\((A)^{-1}\) = 1  ....(iii)

Which is true.

Again, \((A + B)^{-1} = \frac{1}{|(A + B)|} adj(A + B)\)

⇒ \((A + B)^{-1} \neq B^{-1} + A^{-1}\)  ....(iv)

We have, \(\begin{vmatrix} 1 + x&1 & 1 \\[0.3em] 1& 1+y &1 \\[0.3em] 1 & 1& 1+z \end{vmatrix}\) = 0

Applying \(C_1 \rightarrow C_1 - C_3\) and \(C_2 \rightarrow C_2 - C_3\),

⇒ \(\begin{vmatrix} x& 0& 1 \\[0.3em] 0& y &1 \\[0.3em] -z & -z &1+z \end{vmatrix} = 0\)

Expanding along \(R_1\)

x[y(1 + z) + z] - 0 + 1(yz) = 0

⇒ x(y + yz + z) + yz = 0

⇒ xy + xyz + xz + yz = 0

⇒ \(\frac{xy}{xyz} + \frac{xyz}{xyz} + \frac{xz}{xyz} + \frac{yz}{xyz}\) = 0 [on dividing (xyz) from both sides]

⇒ \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 = 0\)

⇒ \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = -1\)

\(\therefore x^{-1} + y^{-1} + z^{-1} = -1\)

We have,

\(\begin{vmatrix} x& x+y& x + 2y \\[0.3em] x + 2y& x & x+y \\[0.3em] x+y & x+2y &x \end{vmatrix}\)

= \(\begin{vmatrix} 3(x+y)& x+y& y \\[0.3em] 3(x + y)& x &y \\[0.3em] 3( x+y) & x+2y &-2y \end{vmatrix}\) \([\because C_1\) → \(C_1 + C_2 + C_3\) and \(C_3\) → \(C_3 - C_2\)]

= 3(x + y)\(\begin{vmatrix} 1& (x+y)& y \\[0.3em] 1& x &y \\[0.3em] 1 & (x+2y) &-2y \end{vmatrix}\) [taking 3(x+y) common from first column]

= 3(x + y)\(\begin{vmatrix} 0& y&0 \\[0.3em] 1& x &y \\[0.3em] 1 & (x+2y) &-2y \end{vmatrix}\) \([\because R_1 \rightarrow R_1 - R_2]\)

Expanding along \(R_1\),

= 3(x + y)[-y(-2y) - y]

= \(3y^2\).3(x + y) = \(9y^2\)(x + y)

\(\Delta = \) \(\begin{vmatrix} 1 & -2& 5 \\[0.3em] 2 & a &-1 \\[0.3em] 0 &4&2a \end{vmatrix}\) = 86

⇒ 1(2\(a^2\) + 4) - 2(-4a - 20) + 0 = 86 [expanding along first column]

⇒ 2\(a^2\) + 4 + 8a + 40 = 86

⇒ 2\(a^2\) + 8a + 44 - 86 = 0

⇒ \(a^2\) + 4a - 21 = 0

⇒ \(a^2\) + 7a - 3a - 21 = 0

⇒ (a + 7)(a - 3) = 0

a = -7 and 3

\(\therefore\)Required sum = -7 + 3 = -4

The characteristic equation is 

\(\begin{vmatrix} -1-\lambda& \frac{3}{2} \\[0.3em] \frac{-1}{2} & \frac{1}{2}-\lambda \\[0.3em] \end{vmatrix} = 0\)

⇒ \(\lambda^2 + \frac{\lambda}{2} + \frac{1}{4} = 0\)

⇒ \(A^2 + \frac{A}{2} + \frac{I}{4} = 0\) gives \(A^2 = \frac{-A}{2} - \frac{I}{4}\)

\(\therefore A^3 = \frac{-A^2}{2} - \frac{A}{4} \)

= \(\frac{-1}{2}(\frac{-A}{2} - \frac{I}{4}) - \frac{A}{4}\)

= \(\frac{A}{4} + \frac{I}{8} - \frac{A}{4} = \frac{I}{8}\)

\(A^2\) = \(\begin{bmatrix} 2& 1 \\[0.3em] -4 & -2\\[0.3em] \end{bmatrix}\)\(\begin{bmatrix} 2& 1 \\[0.3em] -4 & -2\\[0.3em] \end{bmatrix}\)

= \(\begin{bmatrix} 0& 0 \\[0.3em] 0 & 0\\[0.3em] \end{bmatrix} = 0\)

I + 2A + 3\(A^2 + 4A^3\) ......\(\infty\)

= I + 2A (\(\therefore A^2 = 0, A^3 = A^4 = .... = 0)\)

= \(\begin{bmatrix} 5& 2 \\[0.3em] -8 & -3\\[0.3em] \end{bmatrix}\)

I + A + \(A^2 + \dots \infty = (I - A)^{-1}\)

= \(\begin{bmatrix} 3& -3 \\[0.3em] 1& 0 \\[0.3em] \end{bmatrix}^{-1}\)

= \(\frac{1}{3}\begin{bmatrix} 0& 3 \\[0.3em] - 1& 3 \\[0.3em] \end{bmatrix}\)

\(A^2\) = \(\begin{bmatrix} 3& 1 \\[0.3em] -9 & -3 \\[0.3em] \end{bmatrix}\)\(\begin{bmatrix} 3& 1 \\[0.3em] -9 & -3 \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 0& 0 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix} = 0\)

\(\therefore I + 2A + 3A^2 + \dots \infty\)

= \(I + 2A + 0+ \dots \infty\)

= \(\begin{bmatrix} 1& 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix} + \begin{bmatrix} 6& 2 \\[0.3em] -18 & -6 \\[0.3em] \end{bmatrix} \)

= \(\begin{bmatrix} 7& 2 \\[0.3em] -18 & -5 \\[0.3em] \end{bmatrix} \)

\(\Delta =\)\(\begin{vmatrix} a &1 & 1 \\[0.3em] 1 & b & 1\\[0.3em] 1 & 1 & c \end{vmatrix}\) = abc -(a + b+ c)+ 2 > 0

⇒ abc + 2 > a + b + c

abc + 2 > 3\((abc)^{\frac{1}{3}}\)

\((\because \frac{a+b+c}{3} > \) \((abc)^{\frac{1}{3}}\))

\(x^3\) + 2 > 3x where x = \((abc)^{\frac{1}{3}}\)

\(x^3\) - 3x + 2 > 0 

⇒ \((x - 1)^2(x + 2) > 0\) 

= x > -2

\((abc)^{\frac{1}{3}}\) > -2 ⇒ abc > - 8

\(\begin{vmatrix} a& b & c \\[0.3em] b& c & a \\[0.3em] c & a& b \end{vmatrix}\) = 3abc - \(a^3 - b^3 - c^3\)

= -(a + b + c)\((a^2 + b^2 + c^2 - ab - bc - ca)\)

= \(-\frac{1}{2}\)(a + b + c)[\((a - b)^2 + (b - c)^2 + (c - a)^2] < 0\)

\(\therefore\) negative