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2.   Draw a rough sketch of a quadrilateral KLMN. State,

             (a) two pairs of opposite sides,

            (b) two pairs of opposite angles,

            (c) two pairs of adjacent sides,

            (d) two pairs of adjacent angles.

The following is the sketch of a quadrilateral KLMN

(a) Two pairs of opposite sides are

    (i) KL and MN.

    (ii) LM and NK 

(b) Two pairs of opposite angles are

    (i) \angle KLM and \angle MNK

    (ii) \angle LMN and \angle NKL

(c) Two pairs of adjacent sides are

    (i) KL and LM

    (ii) LM and MN

(d) Two pairs of adjacent angles are

    (i) \angle KLM and \angle LMN

    (ii) \angle NKL and \angle MNK

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Sayak

1.   Draw a rough sketch of a quadrilateral PQRS. Draw its diagonals. Name them. Is the meeting point of the diagonals in the interior or exterior of the quadrilateral?

Diagonal PR and diagonal  SQ meet  at O which is inside the quadrilaterl

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Rk. Sravani

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Q : 18       If A is an invertible matrix of order 2, then det  \small (A^-^1) is equal to

                (A) \small det(A)       (B)  \small \frac{1}{det (A)}       (C)  \small 1       (D) \small 0

Given that the matrix is invertible hence A^{-1} exists and A^{-1} = \frac{1}{|A|}adjA

Let us assume a matrix of the order of 2;

A = \begin{bmatrix} a &b \\ c &d \end{bmatrix}.

Then |A| = ad-bc.

adjA = \begin{bmatrix} d &-b \\ -c & a \end{bmatrix}    and  |adjA| = ad-bc

Now,

A^{-1} = \frac{1}{|A|}adjA   

Taking determinant both sides;

|A^{-1}| = |\frac{1}{|A|}adjA| = \begin{bmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{bmatrix}

\therefore|A^{-1}| = \begin{vmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{vmatrix} = \frac{1}{|A|^2}\begin{vmatrix} d &-b \\ -c& a \end{vmatrix} = \frac{1}{|A|^2}(ad-bc) =\frac{1}{|A|^2}.|A| = \frac{1}{|A|}

Therefore we get;

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

Hence the correct answer is B.

 

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Divya Prakash Singh

Q : 17        Let A be a nonsingular square matrix of order \small 3\times 3. Then \small |adjA| is equal to

                 (A) \small |A|      (B) \small |A|^2      (C) \small |A|^3      (D) \small 3|A|

We know the identity (adjA)A = |A| I

Hence we can determine the value of |(adjA)|.

Taking both sides determinant value we get,

|(adjA)A| = ||A| I|     or   |(adjA)||A| = ||A||| I|

or taking R.H.S.,

||A||| I| = \begin{vmatrix} |A| & 0&0 \\ 0&|A| &0 \\ 0&0 &|A| \end{vmatrix}

= |A| (|A|^2) = |A|^3

or, we have then |(adjA)||A| = |A|^3    

Therefore |(adjA)| = |A|^2

Hence the correct answer is B.

 

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Divya Prakash Singh

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Q : 16      If   \small A=\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix} , verify that \small A^3-6A^2+9A-4I=O. Hence find \small A^-^1.

Given matrix: \small A=\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix};

To show: \small A^3-6A^2+9A-4I

Finding each term:

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

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

= \begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}

A^{3} =\begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}

= \begin{bmatrix} 12+5+5 &-6-10-5 &6+5+10 \\ -10-6-5 &5+12+5 & -5-6-10 \\ 10+5+6&-5-10-6 &5+5+12 \end{bmatrix}

= \begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}

So now we have, \small A^3-6A^2+9A-4I

=\begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}-6 \begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}+9\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}-4\begin{bmatrix} 1 &0 &0 \\ 0 &1 & 0\\ 0& 0& 1 \end{bmatrix}

=\begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}- \begin{bmatrix} 36 &&-30 &&30 \\ -30 &&36 &&-30 \\30 &&-30 && 36 \end{bmatrix}+\begin{bmatrix} 18 &-9 &9 \\ -9 &18 &-9 \\ 9 &-9 &18 \end{bmatrix}-\begin{bmatrix} 4 &0 &0 \\ 0 &4 & 0\\ 0& 0& 4 \end{bmatrix}

= \begin{bmatrix} 22-36+18-4 &-21+30-9 &21-30+9 \\ -21+30-9&22-36+18-4 &-21+30-9 \\ 21-30+9&-21+30-9 & 22-36+18-4 \end{bmatrix}

= \begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0&0 & 0 \end{bmatrix} = O

Now finding the inverse of A;

Post-multiplying by A^{-1} as, |A| \neq 0

\Rightarrow (AAA)A^{-1}-6(AA)A^{-1} +9AA^{-1}-4IA^{-1} = 0

\Rightarrow AA(AA^{-1})-6A(AA^{-1}) +9(AA^{-1})=4IA^{-1}

\Rightarrow A^{2}-6A +9I=4A^{-1}

A^{-1} = \frac{1}{4}(A^{2}-6A+9I)                                        ...................(1)

Now,

From equation (1) we get;

A^{-1} = \frac{1}{4}(\begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}-6\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}+9\begin{bmatrix} 1 & 0& 0\\ 0&1 &0 \\ 0& 0&1 \end{bmatrix})

A^{-1} = \frac{1}{4} \begin{bmatrix} 6-12+9 &&-5+6 &&5-6 \\ -5+6 &&6-12+9 &&-5+6 \\ 5-6 &&-5+6 && 6-12+9 \end{bmatrix}

Hence inverse of A is :

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

 

 

 

 

 

 

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Divya Prakash Singh

Q : 15     For the matrix  \small A=\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}   Show that \small A^3-6A^2+5A+11I=O     Hence, find \small A^-^1.
.

Given matrix: \small A=\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix};

To show: \small A^3-6A^2+5A+11I=O

Finding each term:

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

= \begin{bmatrix} 1+1+2 &&1+2-1 &&1-3+3 \\ 1+2-6 &&1+4+3 &&1-6-9 \\ 2-1+6 &&2-2-3 && 2+3+9 \end{bmatrix}

= \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}

A^{3} = \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}

= \begin{bmatrix} 4+2+2 &4+4-1 &4-6+3 \\ -3+8-28 &-3+16+14 & -3-24-42 \\ 7-3+28&7-6-14 &7+9+42 \end{bmatrix}

= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}

So now we have, \small A^3-6A^2+5A+11I

= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}-6\begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}+5\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}+11\begin{bmatrix} 1 &0 &0 \\ 0 &1 & 0\\ 0& 0& 1 \end{bmatrix}

= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}-\begin{bmatrix} 24 &&12 &&6 \\ -18 &&48 &&-84 \\ 42 &&-18 && 84 \end{bmatrix}+\begin{bmatrix} 5 &5 &5 \\ 5 &10 &-15 \\ 10 &-5 &15 \end{bmatrix}+\begin{bmatrix} 11 &0 &0 \\ 0 &11 & 0\\ 0& 0& 11 \end{bmatrix}

= \begin{bmatrix} 8-24+5+11 &7-12+5 &1-6+5 \\ -23+18+5&27-48+10+11 &-69+84-15 \\ 32-42+10&-13+18-5 & 58-84+15+11 \end{bmatrix}

= \begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0&0 & 0 \end{bmatrix} = 0

Now finding the inverse of A;

Post-multiplying by A^{-1} as, |A| \neq 0

\Rightarrow (AAA)A^{-1}-6(AA)A^{-1} +5AA^{-1}+11IA^{-1} = 0

\Rightarrow AA(AA^{-1})-6A(AA^{-1}) +5(AA^{-1})=- 11IA^{-1}

\Rightarrow A^{2}-6A +5I=- 11A^{-1}

A^{-1} = \frac{-1}{11}(A^{2}-6A+5I)                                        ...................(1)

Now,

From equation (1) we get;

A^{-1} = \frac{-1}{11}( \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}-6\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}+5\begin{bmatrix} 1 & 0& 0\\ 0&1 &0 \\ 0& 0&1 \end{bmatrix})

 

A^{-1} = \frac{-1}{11}( \begin{bmatrix} 4-6+5 &&2-6 &&1-6 \\ -3-6 &&8-12+5 &&-14+18 \\ 7-12 &&-3+6 && 14-18+5 \end{bmatrix}

 

A^{-1} = \frac{-1}{11}( \begin{bmatrix} 3 &&-4 &&-5 \\ -9 &&1 &&4 \\ -5 &&3 && 1 \end{bmatrix}

 

 

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Divya Prakash Singh

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Q : 14      For the matrix \small A=\begin{bmatrix} 3 &2 \\ 1 & 1 \end{bmatrix} , find the numbers \small a and \small b such that A^2+aA+bI=0.

Given \small A=\begin{bmatrix} 3 &2 \\ 1 & 1 \end{bmatrix} then we have the relation A^2+aA+bI=O 

So, calculating each term;

A^2 = \begin{bmatrix} 3& 2\\ 1& 1 \end{bmatrix}\begin{bmatrix} 3&2 \\ 1& 1 \end{bmatrix} = \begin{bmatrix} 9+2 &6+2 \\ 3+1&2+1 \end{bmatrix} = \begin{bmatrix} 11 &8 \\ 4& 3 \end{bmatrix}

therefore  A^2+aA+bI=O;

=\begin{bmatrix}11 &8 \\ 4& 3 \end{bmatrix} + a\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix} + b \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}

\begin{bmatrix} 11+3a+b & 8+2a \\ 4+a & 3+a+b \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}

So, we have equations; 

11+3a+b = 0,\ 8+2a = 0   and 4+a = 0,and\ \ 3+a+b = 0

We get a = -4\ and\ b= 1.

 

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Divya Prakash Singh

Q : 13            If   \small A=\begin{bmatrix} 3 &1 \\ -1 &2 \end{bmatrix}​ , show that  A^2-5A+7I=O. Hence find \small A^-^1

Given \small A=\begin{bmatrix} 3 &1 \\ -1 &2 \end{bmatrix} then we have to show the relation A^2-5A+7I=0

So, calculating each term;

A^2 = \begin{bmatrix} 3& 1\\ -1& 2 \end{bmatrix}\begin{bmatrix} 3&1 \\ -1& 2 \end{bmatrix} = \begin{bmatrix} 9-1 &3+2 \\ -3-2&-1+4 \end{bmatrix} = \begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix}

therefore  A^2-5A+7I;

=\begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix} - 5\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix} + 7 \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}

=\begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix} - \begin{bmatrix} 15 &5 \\ -5& 10 \end{bmatrix} + \begin{bmatrix} 7 &0 \\ 0 & 7 \end{bmatrix}

\begin{bmatrix} 8-15+7 &&5-5+0 \\ -5+5+0 && 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 &&0 \\ 0 && 0 \end{bmatrix}

Hence A^2-5A+7I = 0.

\therefore A.A -5A = -7I

\Rightarrow A.A(A^{-1}) - 5AA^{-1} = -7IA^{-1} 

[Post multiplying by A^{-1}, also |A| \neq 0]

\Rightarrow A(AA^{-1}) - 5I = -7A^{-1}

\Rightarrow AI - 5I = -7A^{-1}

\Rightarrow -\frac{1}{7}(AI - 5I)= \frac{1}{7}(5I-A)

\therefore A^{-1} = \frac{1}{7}(5\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}-\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}) = \frac{1}{7}\begin{bmatrix} 2 &-1 \\ 1& 3 \end{bmatrix}

 

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Divya Prakash Singh

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Q : 12        Let   \small A=\begin{bmatrix} 3 &7 \\ 2 & 5 \end{bmatrix}   and \small B=\begin{bmatrix} 6 &8 \\ 7 & 9 \end{bmatrix} .  Verify that  \small (AB)^-^1=B^{-1}A^{-1}.

We have \small A=\begin{bmatrix} 3 &7 \\ 2 & 5 \end{bmatrix} and \small B=\begin{bmatrix} 6 &8 \\ 7 & 9 \end{bmatrix}.

then calculating;

AB = \begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix}\begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}

=\begin{bmatrix} 18+49 &24+63 \\ 12+35 & 16+45 \end{bmatrix} = \begin{bmatrix} 67 &87 \\ 47& 61 \end{bmatrix}

Finding the inverse of AB.

Calculating the cofactors fo AB:

AB_{11}=(-1)^{1+1}(61) = 61        AB_{12}=(-1)^{1+2}(47) = -47

AB_{21}=(-1)^{2+1}(87) = -87     AB_{22}=(-1)^{2+2}(67) = 67

Then we have adj(AB):

adj(AB) = \begin{bmatrix} 61 &-87 \\ -47& 67 \end{bmatrix}

and |AB| = 61(67) - (-87)(-47) = 4087-4089 = -2

Therefore we have inverse: 

(AB)^{-1}=\frac{1}{|AB|}adj(AB) = -\frac{1}{2} \begin{bmatrix} 61 &-87 \\ -47 & 67 \end{bmatrix}

= \begin{bmatrix} \frac{-61}{2} &\frac{87}{2} \\ \\ \frac{47}{2} & \frac{-67}{2} \end{bmatrix}                                                      .....................................(1)

Now, calculating inverses of A and B.

|A| = 15-14 = 1   and |B| = 54- 56 = -2

adjA = \begin{bmatrix} 5 &-7 \\ -2 & 3 \end{bmatrix}     and    adjB = \begin{bmatrix} 9 &-8 \\ -7 & 6 \end{bmatrix}

therefore we have 

A^{-1} = \frac{1}{|A|}adjA= \frac{1}{1} \begin{bmatrix} 5&-7 \\ -2& 3 \end{bmatrix}    and  B^{-1} = \frac{1}{|B|}adjB= \frac{1}{-2} \begin{bmatrix} 9&-8 \\ -7& 6 \end{bmatrix}= \begin{bmatrix} \frac{-9}{2} & 4 \\ \\ \frac{7}{2} & -3 \end{bmatrix}

Now calculating B^{-1}A^{-1}.

B^{-1}A^{-1} =\begin{bmatrix} \frac{-9}{2} & 4 \\ \\ \frac{7}{2} & -3 \end{bmatrix}\begin{bmatrix} 5&-7 \\ -2& 3 \end{bmatrix}

=\begin{bmatrix} \frac{-45}{2}-8 && \frac{63}{2}+12 \\ \\ \frac{35}{2}+6 && \frac{-49}{2}-9 \end{bmatrix} = \begin{bmatrix} \frac{-61}{2} && \frac{87}{2} \\ \\ \frac{47}{2} && \frac{-67}{2} \end{bmatrix}                ........................(2)

From (1) and (2) we get 

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

Hence proved.

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Divya Prakash Singh

Q : 11       Find the inverse of each of the matrices (if it exists).

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

Given the matrix :  \small \begin{bmatrix} 1 & 0&0 \\ 0 &\cos \alpha &\sin \alpha \\ 0 &\sin \alpha &-\cos \alpha \end{bmatrix} =A

To find the inverse we have to first find adjA then as we know the relation:

A^{-1} = \frac{1}{|A|}adjA

So, calculating |A| :

|A| = 1(-\cos^2 \alpha-\sin^2 \alpha)+0(0-0)+0(0-0)

=-(\cos^2 \alpha + \sin^2 \alpha) = -1

Now, calculating the cofactors terms and then adjA.

A_{11} = (-1)^{1+1} (-\cos^2 \alpha - \sin^2 \alpha) = -1                  A_{12} = (-1)^{1+2} (0-0) = 0

A_{13} = (-1)^{1+3} (0-0) =0                      A_{21} = (-1)^{2+1} (0-0) = 0

A_{22} = (-1)^{2+2} (-\cos \alpha-0) = -\cos \alpha             A_{23} = (-1)^{2+1} (\sin \alpha-0) = -\sin \alpha

A_{31} = (-1)^{3+1} (0-0) = 0             A_{32} = (-1)^{3+2} (\sin \alpha-0) =-\sin \alpha

A_{33} = (-1)^{3+3} (\cos \alpha - 0) = \cos \alpha               

So, we have adjA = \begin{bmatrix} -1 &0 &0 \\ 0& -\cos \alpha &-\sin \alpha \\ 0& -\sin \alpha &\cos \alpha \end{bmatrix}

Therefore inverse of A will be:

A^{-1} = \frac{1}{|A|}adjA

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

 

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Divya Prakash Singh

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