Exercise 1.1 Solutions – Applications of Matrices and Determinants (12th Maths)

IMPORTANT QUESTIONS

  • 2 Marks : 1 , 2 , 9 ,10 || Example : 1.11 , 1.4
  • 3 Marks : 3,14,15
  • 5 Marks : Nill

1 (i) . Find the adjoint of the following

A =\begin{bmatrix}-3 & 4 \\6 & 2\end{bmatrix}

\text{adj}(A) =\begin{bmatrix}2 & -4 \\-6 & -3\end{bmatrix}

 

1 (ii) . Find the adjoint of the following

A =\begin{bmatrix}2 & 3 & 1 \\3 & 4 & 1 \\3 & 7 & 2\end{bmatrix}

\text{adj}(A) =\begin{bmatrix}+\begin{vmatrix}4 & 1 \\ 7 & 2\end{vmatrix} &-\begin{vmatrix}3 & 1 \\ 3 & 2\end{vmatrix} &+\begin{vmatrix}3 & 4 \\ 3 & 7\end{vmatrix} \\-\begin{vmatrix}3 & 1 \\ 7 & 2\end{vmatrix} &+\begin{vmatrix}2 & 1 \\ 3 & 2\end{vmatrix} &-\begin{vmatrix}2 & 3 \\ 3 & 7\end{vmatrix} \\+\begin{vmatrix}3 & 1 \\ 4 & 1\end{vmatrix} &-\begin{vmatrix}2 & 1 \\ 3 & 1\end{vmatrix} &+\begin{vmatrix}2 & 3 \\ 3 & 4\end{vmatrix}\end{bmatrix}^{T}

=\begin{bmatrix}+(8-7) & -(6-3) & +(21-12) \\-(6-7) & +(4-3) & -(14-9) \\+(3-4) & -(2-3) & +(8-9)\end{bmatrix}^{T}

=\begin{bmatrix}1 & -3 & 9 \\1 & 1 & -5 \\-1 & 1 & -1\end{bmatrix}^{T}

=\begin{bmatrix}1 & 1 & -1 \\-3 & 1 & 1 \\9 & -5 & -1\end{bmatrix}

1 (iii). Find the adjoint of the following

 A = \frac{1}{3}\begin{bmatrix}2 & 2 & 1 \\-2 & 1 & 2 \\1 & -2 & 2\end{bmatrix}

\text{adj}(kA) = k^{n-1}\text{adj}(A), \; n=3

= \left(\frac{1}{3}\right)^2 \text{adj}(A)= \frac{1}{9}\text{adj}(A)

=\frac{1}{9}\begin{bmatrix}+(2-(-4)) & -(-4-2) & +(4-1) \\-(4-(-2)) & +(4-1) & -(-4-2) \\+(4-1) & -(4-(-2)) & +(2-(-4))\end{bmatrix}^{T}

=\frac{1}{9}\begin{bmatrix}6 & 6 & 3 \\-6 & 3 & 6 \\3 & -6 & 6\end{bmatrix}^{T}

=\frac{1}{9}\begin{bmatrix}6 & -6 & 3 \\6 & 3 & -6 \\3 & 6 & 6\end{bmatrix}

=\frac{3}{9}\begin{bmatrix}2 & -2 & 1 \\2 & 1 & -2 \\1 & 2 & 2\end{bmatrix}

=\frac{1}{3}\begin{bmatrix}2 & -2 & 1 \\2 & 1 & -2 \\1 & 2 & 2\end{bmatrix}

2(i) . Find the inverse of the following

A =\begin{bmatrix}-2 & 4 \\1 & -3\end{bmatrix}

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

|A| =\begin{vmatrix}-2 & 4 \\1 & -3\end{vmatrix}= 6 - 4 = 2

\text{adj}(A) =\begin{bmatrix}-3 & -4 \\-1 & -2\end{bmatrix}

A^{-1} = \frac{1}{2}\begin{bmatrix}-3 & -4 \\-1 & -2\end{bmatrix}

 

2(ii) . Find the inverse of the following

A =\begin{bmatrix}5 & 1 & 1 \\1 & 5 & 1 \\1 & 1 & 5\end{bmatrix}

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

|A| =5\begin{vmatrix}5 & 1 \\1 & 5\end{vmatrix}-1\begin{vmatrix}1 & 1 \\1 & 5\end{vmatrix}+1\begin{vmatrix}1 & 5 \\1 & 1\end{vmatrix}

= 5(25-1) - (5-1) + (1-5) = 120 - 4 - 4 = 112

\text{adj}(A) =\begin{bmatrix}+\begin{vmatrix}5 & 1 \\ 1 & 5\end{vmatrix} &-\begin{vmatrix}1 & 1 \\ 1 & 5\end{vmatrix} &+\begin{vmatrix}1 & 5 \\ 1 & 1\end{vmatrix} \\-\begin{vmatrix}1 & 1 \\ 1 & 5\end{vmatrix} &+\begin{vmatrix}5 & 1 \\ 1 & 5\end{vmatrix} &-\begin{vmatrix}5 & 1 \\ 1 & 1\end{vmatrix} \\+\begin{vmatrix}1 & 1 \\ 5 & 1\end{vmatrix} &-\begin{vmatrix}5 & 1 \\ 1 & 1\end{vmatrix} &+\begin{vmatrix}5 & 1 \\ 1 & 5\end{vmatrix}\end{bmatrix}^{T}

=\begin{bmatrix}+(25-1) & -(5-1) & +(1-5) \\-(5-1) & +(25-1) & -(5-1) \\+(1-5) & -(5-1) & +(25-1)\end{bmatrix}^{T}

=\begin{bmatrix}24 & -4 & -4 \\-4 & 24 & -4 \\-4 & -4 & 24\end{bmatrix}^{T}

=\begin{bmatrix}24 & -4 & -4 \\-4 & 24 & -4 \\-4 & -4 & 24\end{bmatrix}

=4\begin{bmatrix}6 & -1 & -1 \\-1 & 6 & -1 \\-1 & -1 & 6\end{bmatrix}

A^{-1} = \frac{1}{112} \times 4\begin{bmatrix}6 & -1 & -1 \\-1 & 6 & -1 \\-1 & -1 & 6\end{bmatrix}

= \frac{1}{28}\begin{bmatrix}6 & -1 & -1 \\-1 & 6 & -1 \\-1 & -1 & 6\end{bmatrix}



2(iii). Find the inverse of the following

A =\begin{bmatrix}2 & 3 & 1 \\3 & 4 & 1 \\3 & 7 & 2\end{bmatrix}

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

|A| =2\begin{vmatrix}4 & 1 \\7 & 2\end{vmatrix}-3\begin{vmatrix}3 & 1 \\3 & 2\end{vmatrix}+1\begin{vmatrix}3 & 4 \\3 & 7\end{vmatrix}

= 2(8-7) - 3(6-3) + (21-12) = 2 - 9 + 9 = 2

\text{adj}(A) =\begin{bmatrix}+\begin{vmatrix}4 & 1 \\ 7 & 2\end{vmatrix} &-\begin{vmatrix}3 & 1 \\ 3 & 2\end{vmatrix} &+\begin{vmatrix}3 & 4 \\ 3 & 7\end{vmatrix} \\-\begin{vmatrix}3 & 1 \\ 7 & 2\end{vmatrix} &+\begin{vmatrix}2 & 1 \\ 3 & 2\end{vmatrix} &-\begin{vmatrix}2 & 3 \\ 3 & 7\end{vmatrix} \\+\begin{vmatrix}3 & 1 \\ 4 & 1\end{vmatrix} &-\begin{vmatrix}2 & 1 \\ 3 & 1\end{vmatrix} &+\begin{vmatrix}2 & 3 \\ 3 & 4\end{vmatrix}\end{bmatrix}^{T}

=\begin{bmatrix}+(8-7) & -(6-3) & +(21-12) \\-(6-7) & +(4-3) & -(14-9) \\+(3-4) & -(2-3) & +(8-9)\end{bmatrix}^{T}

=\begin{bmatrix}1 & -3 & 9 \\1 & 1 & -5 \\-1 & 1 & -1\end{bmatrix}^{T}

=\begin{bmatrix}1 & 1 & -1 \\-3 & 1 & 1 \\9 & -5 & -1\end{bmatrix}

A^{-1} = \frac{1}{2}\begin{bmatrix}1 & 1 & -1 \\-3 & 1 & 1 \\9 & -5 & -1\end{bmatrix}

 

Q.3 : Show that [F(α)]−1=F(−α)[F(\alpha)]^{-1} = F(-\alpha)

F(\alpha) =\begin{bmatrix}\cos\alpha & 0 & \sin\alpha \\0 & 1 & 0 \\-\sin\alpha & 0 & \cos\alpha\end{bmatrix}

\text{Let } F(\alpha) = A,\; \text{then } (F(\alpha))^{-1} = A^{-1}

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

|A| =\begin{vmatrix}\cos\alpha & 0 & \sin\alpha \\0 & 1 & 0 \\-\sin\alpha & 0 & \cos\alpha\end{vmatrix}

= \cos\alpha(\cos\alpha - 0) - 0(0 - 0) + \sin\alpha(0 - (-\sin\alpha))

= \cos^{2}\alpha + \sin^{2}\alpha

= 1 \ne 0

A^{-1} = \frac{1}{|A|}(\text{adj}(A))^{T}

= \frac{1}{1}\begin{bmatrix}+\cos\alpha & 0 & +\sin\alpha \\0 & 1 & 0 \\-\sin\alpha & 0 & +\cos\alpha\end{bmatrix}^{T}

=\begin{bmatrix}\cos\alpha & 0 & -\sin\alpha \\0 & 1 & 0 \\\sin\alpha & 0 & \cos\alpha\end{bmatrix}

F(-\alpha) =\begin{bmatrix}\cos(-\alpha) & 0 & \sin(-\alpha) \\0 & 1 & 0 \\-\sin(-\alpha) & 0 & \cos(-\alpha)\end{bmatrix}

\cos(-\alpha)=\cos\alpha,\; \sin(-\alpha)=-\sin\alpha

F(-\alpha) =\begin{bmatrix}\cos\alpha & 0 & -\sin\alpha \\0 & 1 & 0 \\\sin\alpha & 0 & \cos\alpha\end{bmatrix}

\text{LHS} = \text{RHS}

\therefore [F(\alpha)]^{-1} = F(-\alpha)

Q4 :

\textbf{ If } A=\begin{pmatrix}5 & 3\\ -1 & -2\end{pmatrix}, \text{ show that } A^2-3A-7I_2=O_2. \text{ Hence find } A^{-1}.

\textbf{Given: } A=\begin{pmatrix}5 & 3\\ -1 & -2\end{pmatrix}

\textbf{Step 1: Find } A^2

A^2 = \begin{pmatrix}5 & 3\\ -1 & -2\end{pmatrix}\begin{pmatrix}5 & 3\\ -1 & -2\end{pmatrix}= \begin{pmatrix}25-3 & 15-6\\ -5+2 & -3+4\end{pmatrix}= \begin{pmatrix}22 & 9\\ -3 & 1\end{pmatrix}

\textbf{Step 2: Find } A^2 - 3A - 7I

-3A = \begin{pmatrix}-15 & -9\\ 3 & 6\end{pmatrix}, \quad-7I = \begin{pmatrix}-7 & 0\\ 0 & -7\end{pmatrix}

A^2 - 3A - 7I =\begin{pmatrix}22 & 9\\ -3 & 1\end{pmatrix}+\begin{pmatrix}-15 & -9\\ 3 & 6\end{pmatrix}+\begin{pmatrix}-7 & 0\\ 0 & -7\end{pmatrix}= \begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}= O_2

\textbf{Hence proved that } A^2 - 3A - 7I = O_2

\textbf{Step 3: Find } A^{-1}

A^2 - 3A - 7I = 0

\textbf{Multiply by} A^{-1}

\Rightarrow A - 3I - 7A^{-1} = 0\Rightarrow 7A^{-1} = A - 3I

A - 3I =\begin{pmatrix}5 & 3\\ -1 & -2\end{pmatrix}-\begin{pmatrix}3 & 0\\ 0 & 3\end{pmatrix}= \begin{pmatrix}2 & 3\\ -1 & -5\end{pmatrix}

\textbf{Final Answer: }A^{-1} = \frac{1}{7}\begin{pmatrix}2 & 3\\ -1 & -5\end{pmatrix}

Q5

\textbf{5. If } A=\frac{1}{9}\begin{pmatrix}-8 & 1 & 4 \\4 & 4 & 7 \\1 & -8 & 4\end{pmatrix},\text{ prove that } A^{-1}=A^{T}.

|A|=\frac{1}{9^3}\begin{vmatrix}-8 & 1 & 4 \\4 & 4 & 7 \\1 & -8 & 4\end{vmatrix}

=\frac{1}{729}\left[-8(16+56)-1(16-7)+4(-32-4)\right]

=\frac{1}{729}\left[-8(72)-9+4(-36)\right]

=\frac{1}{729}\left[-576-9-144\right]=\frac{-729}{729}=-1\neq0

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

=\frac{1}{-1}\left(\frac{1}{9}\right)^2\left|\begin{matrix}(16+56) & -(16-7) & +(-32-4) \\-(4+32) & +(16-32) & -(64-1) \\+(7-16) & -(56-16) & +(-32-4)\end{matrix}\right|^{T}

=\frac{1}{-1}\left(\frac{1}{9}\right)^2\left|\begin{matrix}72 & -9 & -36 \\-36 & -16 & -63 \\-9 & -40 & -36\end{matrix}\right|^{T}

=-1\times\frac{1}{9^2}\begin{pmatrix}72 & -36 & -9 \\-9 & -16 & -40 \\-36 & -63 & -36\end{pmatrix}

=-\frac{1}{9}\begin{pmatrix}8 & -4 & -1 \\-1 & -4 & 8 \\-4 & -7 & -4\end{pmatrix}

A^{-1}=\frac{1}{9}\begin{pmatrix}-8 & 4 & 1 \\1 & 4 & -8 \\4 & 7 & 4\end{pmatrix}

A^{T}=\frac{1}{9}\begin{pmatrix}-8 & 4 & 1 \\1 & 4 & -8 \\4 & 7 & 4\end{pmatrix}

\therefore\ A^{-1}=A^{T}

 

 

Q6

\textbf{6. If } A=\begin{pmatrix}8 & -4\\ -5 & 3\end{pmatrix}, \text{ verify that } A(\text{adj }A)=(\text{adj }A)A=|A|I_2.

\text{adj }A=\begin{pmatrix}3 & 4\\ 5 & 8\end{pmatrix}

|A|=24-20=4

A(\text{adj }A)=\begin{pmatrix}8 & -4\\ -5 & 3\end{pmatrix}\begin{pmatrix}3 & 4\\ 5 & 8\end{pmatrix}

=\begin{pmatrix}24-20 & 32-32\\-15+15 & -20+24\end{pmatrix}

=\begin{pmatrix}4 & 0\\0 & 4\end{pmatrix}=4\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=|A|I_2

(\text{adj }A)A=\begin{pmatrix}3 & 4\\ 5 & 8\end{pmatrix}\begin{pmatrix}8 & -4\\ -5 & 3\end{pmatrix}

=\begin{pmatrix}24-20 & -12+12\\40-40 & -20+24\end{pmatrix}

=\begin{pmatrix}4 & 0\\0 & 4\end{pmatrix}=4\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=|A|I_2

\therefore\ A(\text{adj }A)=(\text{adj }A)A=|A|I_2

Q7

\textbf{7. If } A=\begin{pmatrix}3 & 2\\ 7 & 5\end{pmatrix} \textbf{ and } B=\begin{pmatrix}-1 & -3\\ 5 & 2\end{pmatrix},

\textbf{verify\ that } (AB)^{-1}=B^{-1}A^{-1}.

\text{LHS: } AB=\begin{pmatrix}3 & 2\\ 7 & 5\end{pmatrix}\begin{pmatrix}-1 & -3\\ 5 & 2\end{pmatrix}

=\begin{pmatrix}-3+10 & -9+4\\-7+25 & -21+10\end{pmatrix}=\begin{pmatrix}7 & -5\\18 & -11\end{pmatrix}

|AB|=-77+90=13

(AB)^{-1}=\frac{1}{|AB|}\text{adj}(AB)

=\frac{1}{13}\begin{pmatrix}-11 & 5\\ -18 & 7\end{pmatrix}

—

\text{RHS: }

A=\begin{pmatrix}3 & 2\\ 7 & 5\end{pmatrix}

|A|=15-14=1

\text{adj }A=\begin{pmatrix}5 & -2\\ -7 & 3\end{pmatrix}

A^{-1}=\frac{1}{|A|}\text{adj }A

=\frac{1}{1}\begin{pmatrix}5 & -2\\ -7 & 3\end{pmatrix}

=\begin{pmatrix}5 & -2\\ -7 & 3\end{pmatrix}

B=\begin{pmatrix}-1 & -3\\ 5 & 2\end{pmatrix}

|B|=-2+15=13

\text{adj }B=\begin{pmatrix}2 & 3\\ -5 & -1\end{pmatrix}

B^{-1}=\frac{1}{|B|}\text{adj }B

=\frac{1}{13}\begin{pmatrix}2 & 3\\ -5 & -1\end{pmatrix}

B^{-1}A^{-1}=\frac{1}{13}\begin{pmatrix}2 & 3\\ -5 & -1\end{pmatrix}\begin{pmatrix}5 & -2\\ -7 & 3\end{pmatrix}

=\frac{1}{13}\begin{pmatrix}10-21 & -4+9\\-25+7 & 10-3\end{pmatrix}=\frac{1}{13}\begin{pmatrix}-11 & 5\\-18 & 7\end{pmatrix}

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

Q8

\textbf{8. If } \text{adj}(A)=\begin{pmatrix}2 & -4 & 2\\-3 & 12 & -7\\-2 & 0 & 2\end{pmatrix},\text{ find } A.

A=\frac{1}{\sqrt{|\text{adj }A|}}\ \text{adj}(\text{adj }A)

|\text{adj }A|=2(24-0)-(-4)(-6-14)+2(0+24)

=2(24)+4(-20)+2(24)

=48-80+48

|\text{adj }A|=16

\sqrt{|\text{adj }A|}=\sqrt{16}=4

\text{adj}(\text{adj }A)=\begin{pmatrix}+(24-0) & -(-6-14) & +(0+24)\\-(-8-0) & +(4+4) & -(0-8)\\+(28-24) & -(-14+6) & +(24-12)\end{pmatrix}^{T}

=\begin{pmatrix}24 & 20 & 24\\8 & 8 & 8\\4 & 8 & 12\end{pmatrix}^{T}

=\begin{pmatrix}24 & 8 & 4\\20 & 8 & 8\\24 & 8 & 12\end{pmatrix}

=4\begin{pmatrix}6 & 2 & 1\\5 & 2 & 2\\6 & 2 & 3\end{pmatrix}

A=\frac{1}{4}\times 4\begin{pmatrix}6 & 2 & 1\\5 & 2 & 2\\6 & 2 & 3\end{pmatrix}

=\begin{pmatrix}6 & 2 & 1\\5 & 2 & 2\\6 & 2 & 3\end{pmatrix}

Q9

\textbf{9. If } \text{adj}(A)=\begin{pmatrix}0 & -2 & 0\\6 & 2 & -6\\-3 & 0 & 6\end{pmatrix},\text{ find } A^{-1}.

A^{-1}=\pm\frac{1}{\sqrt{|\text{adj }A|}}\times \text{adj }A

|\text{adj }A|=-(-2)(36-18)

=2(18)

=36

\sqrt{|\text{adj }A|}=\sqrt{36}=6

\therefore\ A^{-1}=\pm\frac{1}{6}\begin{pmatrix}0 & -2 & 0\\6 & 2 & -6\\-3 & 0 & 6\end{pmatrix}

Q10

\textbf{10. Find } \text{adj}(\text{adj }A) \text{ if } A=\begin{pmatrix}1 & 0 & 1\\0 & 2 & 0\\-1 & 0 & 1\end{pmatrix}

\text{adj}(\text{adj }A)=\begin{pmatrix}+(2-0) & -(0-0) & +(0+2)\\-(0-0) & +(1+1) & -(0+0)\\+(0-2) & -(0-0) & +(2-0)\end{pmatrix}^{T}

=\begin{pmatrix}2 & 0 & 2\\0 & 2 & 0\\-2 & 0 & 2\end{pmatrix}^{T}

=\begin{pmatrix}2 & 0 & -2\\0 & 2 & 0\\2 & 0 & 2\end{pmatrix}

Q11

\textbf{11. } A=\begin{pmatrix}1 & \tan x\\ -\tan x & 1\end{pmatrix}, \text{ show that } A^{T}A^{-1}=\begin{pmatrix}\cos 2x & -\sin 2x\\\sin 2x & \cos 2x\end{pmatrix}

A^{-1}=\frac{1}{|A|}\times \text{adj }A

A^{T}=\begin{pmatrix}1 & -\tan x\\ \tan x & 1\end{pmatrix}

|A|=(1+\tan^2 x)=\sec^2 x\neq0

A^{-1}=\frac{1}{\sec^2 x}\begin{pmatrix}1 & -\tan x\\\tan x & 1\end{pmatrix}

—

A^{T}A^{-1}=\begin{pmatrix}1 & -\tan x\\\tan x & 1\end{pmatrix}\frac{1}{\sec^2 x}\begin{pmatrix}1 & -\tan x\\\tan x & 1\end{pmatrix}

=\frac{1}{\sec^2 x}\begin{pmatrix}1 & -\tan x\\\tan x & 1\end{pmatrix}\begin{pmatrix}1 & -\tan x\\\tan x & 1\end{pmatrix}

=\frac{1}{\sec^2 x}\begin{pmatrix}1-\tan^2 x & -\tan x-\tan x\\\tan x+\tan x & -\tan^2 x+1\end{pmatrix}

=\frac{1}{\sec^2 x}\begin{pmatrix}1-\tan^2 x & -2\tan x\\2\tan x & 1-\tan^2 x\end{pmatrix}

—

=\cos^2 x\begin{pmatrix}1-\frac{\sin^2 x}{\cos^2 x} & -\frac{2\sin x}{\cos x}\\\frac{2\sin x}{\cos x} & 1-\frac{\sin^2 x}{\cos^2 x}\end{pmatrix}

=\cos^2 x\begin{pmatrix}\frac{\cos^2 x-\sin^2 x}{\cos^2 x} & -\frac{2\sin x}{\cos x}\\\frac{2\sin x}{\cos x} & \frac{\cos^2 x-\sin^2 x}{\cos^2 x}\end{pmatrix}

=\begin{pmatrix}\cos^2 x-\sin^2 x & -2\sin x\cos x\\2\sin x\cos x & \cos^2 x-\sin^2 x\end{pmatrix}

=\begin{pmatrix}\cos 2x & -\sin 2x\\\sin 2x & \cos 2x\end{pmatrix}

\therefore\ \text{hence proved}

Q12

\textbf{12. Find the matrix } A \text{ for which }A\begin{pmatrix}5 & 3\\ -1 & -2\end{pmatrix}=\begin{pmatrix}14 & 7\\ 7 & 7\end{pmatrix}.

\text{Let } A=\begin{pmatrix}a & b\\ c & d\end{pmatrix}

\begin{pmatrix}a & b\\ c & d\end{pmatrix}\begin{pmatrix}5 & 3\\ -1 & -2\end{pmatrix}=\begin{pmatrix}14 & 7\\ 7 & 7\end{pmatrix}

\begin{pmatrix}5a-b & 3a-2b\\5c-d & 3c-2d\end{pmatrix}=\begin{pmatrix}14 & 7\\ 7 & 7\end{pmatrix}

5a-b=14 \quad (1)

3a-2b=7 \quad (2)

5c-d=7 \quad (3)

3c-2d=7 \quad (4)

—

(1)\times2:\quad 10a-2b=28

10a-2b=28

3a-2b=7

7a=21

a=\frac{21}{7}=3

—

\text{Sub } a \text{ in } (1)

5a-b=14

5(3)-b=14

15-b=14

-b=-1\Rightarrow b=1

—

5c-d=7 \quad (3)

3c-2d=7 \quad (4)

(3)\times2:\quad 10c-2d=14

10c-2d=14

3c-2d=7

7c=7

c=\frac{7}{7}=1

—

\text{Sub } c \text{ in } (3)

5c-d=7

5(1)-d=7

5-d=7

-d=2\Rightarrow d=-2

—

\therefore\ A=\begin{pmatrix}3 & 1\\ 1 & -2\end{pmatrix}

Q13

\textbf{13. Given } A=\begin{pmatrix}1 & -1\\ 2 & 0\end{pmatrix},\;B=\begin{pmatrix}3 & -2\\ 1 & 1\end{pmatrix}\text{ and } C=\begin{pmatrix}1 & 1\\ 2 & 2\end{pmatrix},\text{ find a matrix } X \text{ such that } AXB=C.

\text{Pre multiply by } A^{-1} \text{ and post multiply by } B^{-1}

A^{-1}AXB B^{-1}=A^{-1}CB^{-1}

X=A^{-1}CB^{-1}

—

|A|=0+2=2

A^{-1}=\frac{1}{|A|}\text{adj }A

=\frac{1}{2}\begin{pmatrix}0 & 1\\ -2 & 1\end{pmatrix}

—

|B|=3+2=5

B^{-1}=\frac{1}{|B|}\text{adj }B

=\frac{1}{5}\begin{pmatrix}1 & 2\\ -1 & 3\end{pmatrix}

—

\therefore\ X=A^{-1}CB^{-1}=\frac{1}{2}\begin{pmatrix}0 & 1\\ -2 & 1\end{pmatrix}\begin{pmatrix}1 & 1\\ 2 & 2\end{pmatrix}\frac{1}{5}\begin{pmatrix}1 & 2\\ -1 & 3\end{pmatrix}

=\frac{1}{10}\begin{pmatrix}0 & 1\\ -2 & 1\end{pmatrix}\begin{pmatrix}1-1 & 2+3\\2-2 & 4+6\end{pmatrix}

=\frac{1}{10}\begin{pmatrix}0 & 1\\ -2 & 1\end{pmatrix}\begin{pmatrix}0 & 5\\ 0 & 10\end{pmatrix}

=\frac{1}{10}\begin{pmatrix}0+0 & 0+10\\0+0 & -10+10\end{pmatrix}

=\frac{1}{10}\begin{pmatrix}0 & 10\\ 0 & 0\end{pmatrix}

X=\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}

Q14

\textbf{14. If } A=\begin{pmatrix}0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0\end{pmatrix}, \text{ show that } A^{-1}=\frac{1}{2}(A^2-3I).

\text{LHS}

|A|=0(0-1)-1(0-1)+1(1-0)

=-1(-1)+1(1)

=1+1=2

A^{-1}=\frac{1}{|A|}\text{adj }A

=\frac{1}{2}\begin{vmatrix}+(0-1) & -(0-1) & +(1-0)\\-(0-1) & +(0-1) & -(0-1)\\+(1-0) & -(0-1) & +(0-1)\end{vmatrix}^{T}

=\frac{1}{2}\begin{pmatrix}-1 & 1 & 1\\1 & -1 & 1\\1 & 1 & -1\end{pmatrix}^{T}

A^{-1}=\frac{1}{2}\begin{pmatrix}-1 & 1 & 1\\1 & -1 & 1\\1 & 1 & -1\end{pmatrix}

—

\text{RHS}

A^2=\begin{pmatrix}0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0\end{pmatrix}\begin{pmatrix}0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0\end{pmatrix}

=\begin{pmatrix}2 & 1 & 1\\1 & 2 & 1\\1 & 1 & 2\end{pmatrix}

3I=3\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}=\begin{pmatrix}3 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 3\end{pmatrix}

A^2-3I=\begin{pmatrix}2 & 1 & 1\\1 & 2 & 1\\1 & 1 & 2\end{pmatrix}-\begin{pmatrix}3 & 0 & 0\\0 & 3 & 0\\0 & 0 & 3\end{pmatrix}

=\begin{pmatrix}-1 & 1 & 1\\1 & -1 & 1\\1 & 1 & -1\end{pmatrix}

\frac{1}{2}(A^2-3I)=\frac{1}{2}\begin{pmatrix}-1 & 1 & 1\\1 & -1 & 1\\1 & 1 & -1\end{pmatrix}

\therefore\ \text{LHS}=\text{RHS}

Q15

\textbf{15. Decrypt the received encoded message } [2\ -3][20\ 4] \text{ with the encryption matrix }\begin{pmatrix}-1 & -1\\ 2 & 1\end{pmatrix}.

A=\begin{pmatrix}-1 & -1\\ 2 & 1\end{pmatrix}

|A|=-1+2=1

\text{adj }A=\begin{pmatrix}1 & 1\\ -2 & -1\end{pmatrix}

A^{-1}=\frac{1}{|A|}\text{adj }A

=\frac{1}{1}\begin{pmatrix}1 & 1\\ -2 & -1\end{pmatrix}

A^{-1}=\begin{pmatrix}1 & 1\\ -2 & -1\end{pmatrix}

—

\text{Coded row matrix}\quad \text{Decoding matrix } A^{-1}\quad \text{Decoded row}

[2\ -3]\begin{pmatrix}1 & 1\\ -2 & -1\end{pmatrix}=[8\ 5]

[20\ 4]\begin{pmatrix}1 & 1\\ -2 & -1\end{pmatrix}=[12\ 16]

—

\text{The sequence of decoded matrix is } [8\ 5][12\ 16]

\text{The receiver reads this message as HELP}

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