Let x, y and z be the investments at the rate of interest 6%, 7% and 8% per annum respectively.
Then x + y + z = 5000  ...(1)
Also, \(\frac{6x}{100}+\frac{7y}{100}+\frac{8z}{100}=358\)
\(\Rightarrow\)   6x + 7y + 8z = 35800  ...(2)
And \(\frac{6x}{100}+\frac{7y}{100}=70+\frac{8z}{100}\) (Given)
\(\Rightarrow\)   6x + 7y - 8z = 7000  ...(3)
From (1), (2) and (3),
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5000 \\ 35800 \\ 7000 \end{matrix} \right] \)
AX = B where A = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8 \end{matrix} \right] ,\quad X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,\quad Z=\left[ \begin{matrix} 5000 \\ 35800 \\ 7000 \end{matrix} \right] \) 
 \(\therefore |A|\) \(=\begin{bmatrix} 1&1&1\\6&7&8\\6&7&-8 \end{bmatrix}=1(-56-56)-1(-48-48)1(42-42)\)
\(=-16\neq0\Rightarrow{A}^{-1}\) exists.
A₁₁  = -112, A₁₂ = 96, A₁₃= 0
A₂₁ = 15, A₂₂ = -14  A₂₃ = -1
A₃₁ = 1, A₃₂ = -2, A₃₃ = -1
\(\therefore \ adj\ A=\begin{bmatrix} -112&96&0\\15&-14&-1\\1&-2&1 \end{bmatrix}^{T}=\begin{bmatrix} -112&15&1\\96&-14&-2\\0&-1&1\end{bmatrix}\)
\(\therefore \quad A^{ -1 }=\quad \frac { 1 }{ |A| } adjA=-\frac { 1 }{ 16 } \left[ \begin{matrix} -112 & 15 & 1 \\ 96 & -14 & -2 \\ 0 & -1 & 1 \end{matrix} \right] \)
Hence, the solution is given by
\(X={A}^{-1}B=-\frac{1}{16}\begin{bmatrix} -112&15&1 \\96 &-4&-2\\0&-1&1 \end{bmatrix}\begin{bmatrix} 5000\\35800\\7000 \end{bmatrix}=-{{1}\over{16}}\begin{bmatrix} -560000+537000+7000\\480000-501200-14000\\0-35800+7000 \end{bmatrix}\)
\(X=\begin{bmatrix} 1000\\2200\\1800 \end{bmatrix}\)  [ \(\because\) x = 1000, y = 2200, z = 1800]
Hence, the three investment are of Rs. 1000,Rs. 2200 and Rs. 1800.

Let the number of radios of type P be x and the radios of type Q be y.
Given 2x + 3y = 130 and 3x + 4y = 180
\(\Rightarrow \left( \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 130 \\ 180 \end{matrix} \right) \)
AX = B where A = \(\left( \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right) ,X=\left( \begin{matrix} x \\ y \end{matrix} \right) and\quad B=\left( \begin{matrix} 130 \\ 180 \end{matrix} \right) \)
|A|=\(\left| \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right| =8-9=-1\neq 0\Rightarrow { A }^{ -1 }\) existsadj A=\(\left( \begin{matrix} 4 & -3 \\ -3 & 2 \end{matrix} \right) \)    [\(\because\) A₁₁ = 4, A₁₂ = -3 A₂₁ = -3, A₂₂ = 2]
\(\therefore \ { A }^{ -1 }=\frac { 1 }{ |A| } adj\quad A=\frac { 1 }{ -1 } \left( \begin{matrix} 4 & -3 \\ -3 & 2 \end{matrix} \right) =\left( \begin{matrix} -4 & 3 \\ 3 & -2 \end{matrix} \right) \)
\(\therefore \ X={ A }^{ -1 }B=\left( \begin{matrix} -4 & 3 \\ 3 & -2 \end{matrix} \right) \left( \begin{matrix} 130 \\ 180 \end{matrix} \right) =\left( \begin{matrix} -520 & +540 \\ 390 & -360 \end{matrix} \right) =\left( \begin{matrix} 20 \\ 30 \end{matrix} \right) \)
\(\therefore\) Number of radios of Type P is 20.
Number of radios of Type Q is 30.

Let \(A=\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] and\quad B=\left[ \begin{matrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{matrix} \right] \)
Now, AB =\(\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] \left[ \begin{matrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{matrix} \right] =\left[ \begin{matrix} -2-9+12 & 0-2+2 & 1+3-4 \\ 0+18-18 & 0+4-3 & 0-6+6 \\ -6-18+24 & 0-4+4 & 3+6-8 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\therefore\)  AB = I₃\(\Rightarrow\) A^-1 = B
\(\therefore { A }^{ -1 }=\left[ \begin{matrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{matrix} \right] \)
Now \(A=\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] ,x=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] and\quad c=\left[ \begin{matrix} 1 \\ 1 \\ 2 \end{matrix} \right] \)
\(X={ A }^{ -1 }C=\left[ \begin{matrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 1 \\ 2 \end{matrix} \right] =\left[ \begin{matrix} -2 & +0 & +2 \\ 9 & +2 & -6 \\ 6 & +1 & -4 \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 5 \\ 3 \end{matrix} \right] \)
\(\because\) x = 0, y = 5 and z = 3

Given A = \(\begin{bmatrix} 1&2&2\\2&1&2\\2&2&1 \end{bmatrix}\)
\({A}^{2}=A.A=\begin{bmatrix} 1&2&2 \\2&1&2\\2&2&1 \end{bmatrix}\begin{bmatrix} 1 &2&2 \\ 2 &1&2\\2&2&1 \end{bmatrix}=\begin{bmatrix} 9&8&8 \\ 8 &9&8\\8&8&9 \end{bmatrix}\)
\(4A=4\begin{bmatrix} 1&2&2 \\ 2 & 1 & 1\\2 & 2 & 1\end{bmatrix}=\begin{bmatrix} 4 & 8 & 8\\8 & 4 & 8\\ 8 & 8 & 9\end{bmatrix}\)and \({5I}_{3}=\begin{bmatrix} 5& 0& 0\\0 & 5 & 0\\0 & 0 & 5 \end{bmatrix}\)
\(\therefore\ A^2-4A-5I_3=\begin{bmatrix} 9&8&8\\8&9&8\\8&8&9 \end{bmatrix}-\begin{bmatrix} 4&8&8 \\ 8&4&8\\8&8&4 \end{bmatrix}-\begin{bmatrix} 5&0&0\\0&5&0\\0&0&5 \end{bmatrix}\)
\(=\begin{bmatrix} 9-4-5&8-8-0&8-8-0\\8-8-0&9-4-5&8-8-0\\8-8-0&8-8-0&9-4-5 \end{bmatrix}=\begin{bmatrix} 0&0&0 \\ 0&0&0\\0&0&0 \end{bmatrix}\)
\(\therefore\)  A² - 4A - 5I₃ = 0
\(\Rightarrow\)    A²-4A = 5I₃
Premultiplying throughout by A^-1 we get
A^-1 A² - 4A^-1 A = 5A^-1 I₃
\(\Rightarrow\) (A^-1 A)· A - 4 (A^-1 A) = 5 A^-1 I₃
\(\Rightarrow\) 5A^-1 = A - 4I
\(\Rightarrow\)   \({A}^{-1}=\frac{1}{5}[A-4I]\)
\(\Rightarrow\)  \({ A }^{ -1 }=\frac { 1 }{ 5 } \left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{matrix} \right] =\left[ \begin{matrix} \frac { -3 }{ 5 } & \frac { 2 }{ 5 } & \frac { 2 }{ 5 } \\ \frac { 2 }{ 5 } & \frac { -3 }{ 5 } & \frac { 2 }{ 5 } \\ \frac { 2 }{ 5 } & \frac { 2 }{ 5 } & \frac { -3 }{ 5 } \end{matrix} \right]\)

Let A = \(\left| \begin{matrix} 2y+4 & \quad 5y+7 & 8y+a \\ 3y+5 & 6y+8 & 9y+b \\ 4y+6 & 7y+9 & 10y+c \end{matrix} \right| \)
Applying R₂\(\rightarrow\) 2R₂ and dividing by 2 we get,
A = \(\frac { 1 }{ 2 } \left| \begin{matrix} 2y+4 & 5y+7 & 8y+a \\ 6y+10 & 12y+16 & 18y+2b \\ 4y+6 & 7y+9 & 10y+c \end{matrix} \right| \)
Applying R₂\(\rightarrow\) 2R₂-(R₁ + R₃) we get
\(A=\frac { 1 }{ 2 } \left[ \begin{matrix} 2y+4 & \quad 5y+7 & 8y+a \\ 0 & 0 & 2b-(a+c) \\ 4y+6 & 7y+9 & 10y+c \end{matrix} \right] \)
Given a,b,c in A.P \(\Rightarrow\) 2b = a + c \(\Rightarrow\) 2b - (a + c) = 0
\(\therefore A=\frac { 1 }{ 2 } \left| \begin{matrix} 2y+4 & 5y+7 & 8y+a \\ 0 & 0 & 0 \\ 4y+6 & 7y+9 & 10y+c \end{matrix} \right| =0\)
\(\therefore\) |A| = 0