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# NCERT solutions for class 12 maths exercise 3.2 of chapter 3-Matrices

## NCERT solutions for class 12 maths exercise 3.2 of chapter 3-Matrices

NCERT solutions for class 12 maths exercise 3.2 of chapter 3-Matrices are very important not only to the class 12 CBSE students but also for the students who are going to appear in the entrance exam of engineering, NDA, and CDS. NCERT solutions for class 12 maths exercise 3.2 of chapter 3-Matrices are created here for clearing the doubts of students in chapter 3-Matrices for a better understanding of the concept which is required to solve the questions in the exam.NCERT solutions for class 12 maths exercise 3.2 of chapter 3-Matrices are solved by an expert of the subject.

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lutions for class 12 maths exercise 3.2 of chapter 3-Matrices are the solutions of all unsolved questions of exercise 3.2 of chapter 3-Matrices of the NCERT maths textbook. Matrix has huge applications in the field of business, economics, architect, geology, engineering and science, and technology. Here in this exercise, you will learn multiplication, division, addition, and subtraction of Matrices.

NCERT solutions for class 12 maths exercise 3.1 of chapter 3-Matrices

### NCERT solutions of class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

Study notes of Maths and Science NCERT and CBSE from class 9 to 12

### NCERT solutions for class 12 maths exercise 3.2 of chapter 3-Matrices

Q1.

$\fn_cm Let\: A= \begin{bmatrix}2 &4 \\ 3& 2 \end{bmatrix},B=\begin{bmatrix}1 & 3\\ -2&5 \end{bmatrix},C=\begin{bmatrix}-2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following.

(i) A + B      (ii)  A – B     (iii)  3A – C    (iv) AB   (v) BA

Ans. (i) A + B

$\fn_cm =\begin{bmatrix}2 &4 \\ 3& 2 \end{bmatrix}+\begin{bmatrix}1 & 3\\ -2&5 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}2+1 & 4+3\\ 3-2&2+5 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}3 & 7\\ 1&7 \end{bmatrix}$

(ii)  A – B

$\fn_cm =\begin{bmatrix}2 & 4\\ 3&2 \end{bmatrix}-\begin{bmatrix}1 & 3\\ -2&5 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}2-1 & 4-3\\ 3+2&2-5 \end{bmatrix}$

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

(iii)  3A – C

$\fn_cm =\begin{bmatrix}2 &4 \\ 3 & 2 \end{bmatrix}\begin{bmatrix}1 & 3\\ -2 & 5 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}6 & 12\\ 9& 6 \end{bmatrix}-\begin{bmatrix}-2 &5 \\ 3& 4 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}6+2 & 12-5\\ 9-3& 6-4 \end{bmatrix}$

(iv) AB

$\fn_cm =\begin{bmatrix}2 &4 \\ 3 & 2 \end{bmatrix}\begin{bmatrix}1 & 3\\ -2 & 5 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}2\times 1+4\times -2 & 2\times 3+4\times 5\\ 3\times 1+2\times -2 & 3\times 3+2\times 5 \end{bmatrix}$

$\fn_cm =\begin{bmatrix} -6 & 26\\ -1 & 19 \end{bmatrix}$

(v) BA

$\fn_cm =\begin{bmatrix} 1 & 3\\ -2 & 5 \end{bmatrix}\begin{bmatrix}2 &4 \\ 3 & 2 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}1\times 2+3\times 3 &1\times 4+3\times 2 \\ -2\times 2+5\times 3& -2\times 4+5\times 2 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}11 &10 \\ 11& 2 \end{bmatrix}$

Q2. Compute the following

$\fn_cm \left ( i \right )\begin{bmatrix}a &b \\ -b& a \end{bmatrix}+\begin{bmatrix}a & b\\ b& a \end{bmatrix}$            $\fn_cm \left ( ii \right )\begin{bmatrix}a^{2}+b^{2} &b^{2}+c^{2} \\ a^{2}+c^{2}& a^{2}+b^{2} \end{bmatrix}+\begin{bmatrix}2ab & 2bc\\ -2ac& -2ab \end{bmatrix}$

$\fn_cm \left ( iii \right )\begin{bmatrix}-1 &4 & -6\\ 8 & 5 & 16\\ 2 & 8 & 5 \end{bmatrix}+\begin{bmatrix}12 &7 &6 \\ 8&0 & 5\\ 3 & 2 & 4 \end{bmatrix}$          $\fn_cm \left ( iv \right )\begin{bmatrix}cos^{2}x &sin^{2}x \\ sin^{2}x& cos^{2}x \end{bmatrix}+\begin{bmatrix}sin^{2}x &cos^{2}x \\ cos^{2}x& sin^{2}x \end{bmatrix}$

Ans.

$\fn_cm \left ( i \right )\begin{bmatrix}a &b \\ -b& a \end{bmatrix}+\begin{bmatrix}a & b\\ b& a \end{bmatrix}$

$\fn_cm =\begin{bmatrix}a+a &b+b \\ -b+b& a+a \end{bmatrix}=\begin{bmatrix}2a & 2b\\ 0 & 2a \end{bmatrix}$

$\fn_cm \left ( ii \right )\begin{bmatrix}a^{2}+b^{2} &b^{2}+c^{2} \\ a^{2}+c^{2}& a^{2}+b^{2} \end{bmatrix}+\begin{bmatrix}2ab & 2bc\\ -2ac& -2ab \end{bmatrix}$

$\fn_cm =\begin{bmatrix}a^{2}+b^{2}+2ab & b^{2}+c^{2}+2bc\\ a^{2}+c^{2}-2ac& a^{2}+b^{2}-2ab \end{bmatrix}=\begin{bmatrix}\left ( a+b \right )^{2} &\left ( b+c \right )^{2} \\ \left ( a-c \right )^{2} & \left ( a-b \right )^{2} \end{bmatrix}$

$\fn_cm \left ( iii \right )\begin{bmatrix}-1 &4 & -6\\ 8 & 5 & 16\\ 2 & 8 & 5 \end{bmatrix}+\begin{bmatrix}12 &7 &6 \\ 8&0 & 5\\ 3 & 2 & 4 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}-1+12 &4+7 & -6+6\\ 8+8 & 5+0 & 16+5\\ 2+3 &8+2 & 5+4 \end{bmatrix}=\begin{bmatrix}11 & 11 & 0\\ 16 & 5 & 21\\ 5 & 10 & 9 \end{bmatrix}$

$\fn_cm \left ( iv \right )\begin{bmatrix}cos^{2}x &sin^{2}x \\ sin^{2}x& cos^{2}x \end{bmatrix}+\begin{bmatrix}sin^{2}x &cos^{2}x \\ cos^{2}x& sin^{2}x \end{bmatrix}$

$\fn_cm =\begin{bmatrix}cos^{2}x+sin^{2}x &sin^{2}x+cos^{2}x \\ sin^{2}x+cos^{2}x& cos^{2}x+sin^{2}x \end{bmatrix}=\begin{bmatrix}1 &1 \\ 1& 1 \end{bmatrix}$

Q3. Compute the indicated products

$\fn_cm \left ( i\right )\begin{bmatrix}a &b \\ -b & a \end{bmatrix}\begin{bmatrix}a & -b\\ b& a \end{bmatrix}$     $\fn_cm \left ( ii \right )\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}\begin{bmatrix}2 & 3 & 4 \end{bmatrix}$

$\fn_cm \left ( iii \right )\begin{bmatrix}1 & -2\\ 2& 3 \end{bmatrix}\begin{bmatrix}1 & 2 & 3\\ 2& 3& 1 \end{bmatrix}$    $\fn_cm \left ( iv \right )\begin{bmatrix}2 & 3 & 4\\ 3&4 &5 \\ 4& 5& 6 \end{bmatrix}\begin{bmatrix}1 & -3 & 5\\ 0& 2 & 4\\ 3& 0 & 5 \end{bmatrix}$

$\fn_cm \left ( v \right )\begin{bmatrix}2 &1 \\ 3&2 \\ -1& 1 \end{bmatrix}\begin{bmatrix}1 & 0 & 1\\ -1& 2 & 1 \end{bmatrix}$        $\fn_cm \left ( vi \right )\begin{bmatrix}3 & -1 &3 \\ -1& 0& 2 \end{bmatrix}\begin{bmatrix}2 &-3 \\ 1 & 0\\ 3& 1 \end{bmatrix}$

Ans.

$\fn_cm \left ( i\right )\begin{bmatrix}a &b \\ -b & a \end{bmatrix}\begin{bmatrix}a & -b\\ b& a \end{bmatrix}$

$\fn_cm =\begin{bmatrix}a^{2}+b^{2} & -ab+ba\\ -ab+ab& b^{2}+a^{2} \end{bmatrix}=\begin{bmatrix}a^{2}+b^{2} &0 \\ 0& b^{2}+a^{2} \end{bmatrix}$

Ans.

$\fn_cm \left ( i\right )\begin{bmatrix}a &b \\ -b & a \end{bmatrix}\begin{bmatrix}a & -b\\ b& a \end{bmatrix}$

$\fn_cm =\begin{bmatrix}a^{2}+b^{2} & -ab+ba\\ -ab+ab& b^{2}+a^{2} \end{bmatrix}=\begin{bmatrix}a^{2}+b^{2} &0 \\ 0& b^{2}+a^{2} \end{bmatrix}$

$\fn_cm \left ( ii \right )\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}\begin{bmatrix}2 & 3 & 4 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}1\times 2 & 1\times 3 & 1\times 4\\ 2\times 2& 2\times 3 &2\times 4 \\ 3\times 2 &3\times 3 & 3\times 4 \end{bmatrix}=\begin{bmatrix}2 &3 & 4\\ 4& 6 & 8\\ 6& 9 & 12 \end{bmatrix}$

$\fn_cm \left ( iii \right )\begin{bmatrix}1 & -2\\ 2& 3 \end{bmatrix}\begin{bmatrix}1 & 2 & 3\\ 2& 3& 1 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}1\times 1-2\times 2 & 1\times 2 -2\times 3&1\times 3-2\times 1 \\ 2\times 1+3\times 2&2\times 2+3\times 3 &2\times 3+3\times 1 \end{bmatrix}=\begin{bmatrix}-3 &-4 & 1\\ 8 &13 & 9 \end{bmatrix}$

$\fn_cm \left ( iv \right )\begin{bmatrix}2 & 3 & 4\\ 3&4 &5 \\ 4& 5& 6 \end{bmatrix}\begin{bmatrix}1 & -3 & 5\\ 0& 2 & 4\\ 3& 0 & 5 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}2\times 1+3\times 0+4\times 3 & 2\times -3 +3\times 2+4\times 0&2\times 5+3\times 4+4\times 5 \\ 3\times 1+4\times0 +5\times 3 & 3\times -3+4\times 2+5\times 0 &3\times 5+4\times 4+5\times 5 \\ 4\times 1+5\times 0+6\times 3 &4\times -3+5\times 2+6\times 0 &4\times 5+5\times 4+6\times 5 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}14 &0 &42 \\ 18& -1 &56 \\ 22 & -2 & 70 \end{bmatrix}$

$\fn_cm \left ( v \right )\begin{bmatrix}2 &1 \\ 3&2 \\ -1& 1 \end{bmatrix}\begin{bmatrix}1 & 0 & 1\\ -1& 2 & 1 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}2\times 1+1\times -1 & 2\times 0+1\times 2 & 2\times 1+1\times 1\\ 3\times 1+2\times -1&3\times 0+2\times 2 &3\times 1+2\times 1 \\ -1\times 1+1\times -1&-1\times 0+1\times 2 & -1\times 1+1\times 1 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}1 & 2 &3 \\ 1& 4 & 5\\ -2& 2 & 0 \end{bmatrix}$

$\fn_cm \left ( vi \right )\begin{bmatrix}3 & -1 &3 \\ -1& 0& 2 \end{bmatrix}\begin{bmatrix}2 &-3 \\ 1 & 0\\ 3& 1 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}3\times 2-1\times 1+3\times 3 &3\times -3 -1\times 0+3\times 1\\ -1\times 2+0\times 1+2\times 3& -1\times -3+0\times 0+2\times 1 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}14 &-6 \\ 4& 5 \end{bmatrix}$

Q4. If

$\fn_cm A=\begin{bmatrix}1 &2 &-3 \\ 5&0 &2 \\ 1 & -1 & 1 \end{bmatrix},B=\begin{bmatrix}3 & -1 & 2\\ 4& 2&5 \\ 2& 0 & 3 \end{bmatrix}\: and\:C= \begin{bmatrix}4 & 1& 2\\ 0& 3 & 2\\ 1 & -2 & 3 \end{bmatrix}$

Then compute ,(A +B) and (B – C). Also verify that A + (B – C) = (A + B) – C

Ans.

A + B

$\fn_cm =\begin{bmatrix}1 & 2 &-3 \\ 5& 0 & 2\\ 1& -1 & 1 =\end{bmatrix}+\begin{bmatrix}3 &-1 &2 \\ 4&2 &5 \\ 2&0 & 3 \end{bmatrix}=\begin{bmatrix}1+3 &2-1 &-3+2 \\ 5+4& 0+2 &2+5 \\ 1+2 &-1+0 &1+ 3 \end{bmatrix}=\begin{bmatrix}4 & 1 &-1 \\ 9& 2 & 7\\ 3& -1 & 4 \end{bmatrix}$

(B – C)

$\fn_cm =\begin{bmatrix}3 & -1 &2 \\ 4& 2 &5\\ 2& 0 & 3 \end{bmatrix}-\begin{bmatrix}4 & 1& 2\\ 0& 3& 2\\ 1& -2& 3 \end{bmatrix}=\begin{bmatrix}3-4 &-1-1 &2-2 \\ 4-0& 2-3 &5-2 \\ 2-1& 0+2& 3-3 \end{bmatrix}=\begin{bmatrix}-1 & -2 & 0\\ 4& -1 & 3\\ 1& 2&0 \end{bmatrix}$

Now,we have  to verify that A + (B – C) = (A + B) – C

LHS , A + (B – C) , Putting the value of A and (B – C)

$\fn_cm =\begin{bmatrix}1 & 2 &-3 \\ 5& 0 &2\\ 1& -1 & 1 \end{bmatrix}+\begin{bmatrix}-1 & -2 & 0\\ 4& -1 & 3\\ 1& 2&0 \end{bmatrix}=\begin{bmatrix}1-1 & 2-2 &-3+0 \\ 5+4& 0-1 & 2+3\\ 1+1&-1+2 & 1+0 \end{bmatrix}=\begin{bmatrix}0 & 0 & -3\\ 9& -1 &5 \\ 2& 1 & 1 \end{bmatrix}$

RHS , (A + B) – C , Putting the value of C and (A+B )

$\fn_cm \begin{bmatrix}4 & 1 &-1 \\ 9 & 2 &7 \\ 3 & -1& 4 \end{bmatrix}-\begin{bmatrix}4 & 1 &2 \\ 0& 3 & 2\\ 1& -2 & 3 \end{bmatrix}=\begin{bmatrix}4-4 & 1-1 & -1-2\\ 9-0& 2-3 &7-2 \\ 3-1& -1+2 & 4-3 \end{bmatrix}=\begin{bmatrix}0 & 0 &-3 \\ 9 & -1 & 5\\ 2&1 & 1 \end{bmatrix}$

Therefore , LHS = RHS, Verified

$\fn_cm Q5.If\: A=\begin{bmatrix}2/3 & 1 &5/3 \\ 1/3 &2/3 & 4/3\\ 7/3& 2& 2/3 \end{bmatrix}\: and\: B=\begin{bmatrix}2/5 & 3/5 &1 \\ 1/5& 2/5 & 4/5\\ 7/5& 6/5 & 2/5 \end{bmatrix}$

Then compute 3A – 5B

Ans.  We have to get the value of 3A – 5B

$\fn_cm \: 3A-5B=3\begin{bmatrix}2/3 & 1 &5/3 \\ 1/3 &2/3 & 4/3\\ 7/3& 2& 2/3 \end{bmatrix}\: -\: 5\begin{bmatrix}2/5 & 3/5 &1 \\ 1/5& 2/5 & 4/5\\ 7/5& 6/5 & 2/5 \end{bmatrix}$

$\fn_cm =\begin{bmatrix}2 & 3 &5 \\ 1 &2 & 4\\ 7& 6& 2 \end{bmatrix}\: -\: \begin{bmatrix}2 & 3 &5 \\ 1& 2& 4\\ 7& 6 & 2 \end{bmatrix}=\begin{bmatrix}2-2 &3-3 &5-5 \\ 1-1&2-2 &4-4 \\ 7-7 & 6-6 & 2-2 \end{bmatrix}=\begin{bmatrix}0 &0 &0 \\ 0 &0 & 0\\ 0 &0 & 0 \end{bmatrix}=0$

Q6. Simplify

$\fn_cm cos\Theta \begin{bmatrix}cos\Theta &sin\Theta \\ -sin\Theta & cos\Theta \end{bmatrix}+sin\Theta \begin{bmatrix}sin\Theta & -cos\Theta \\ cos\Theta & sin\Theta \end{bmatrix}$

Ans.  We are given the expression

$\fn_cm cos\Theta \begin{bmatrix}cos\Theta &sin\Theta \\ -sin\Theta & cos\Theta \end{bmatrix}+sin\Theta \begin{bmatrix}sin\Theta & -cos\Theta \\ cos\Theta & sin\Theta \end{bmatrix}$

Simplifying it

$\fn_cm =\begin{bmatrix}cos^{2}\Theta &sin\Theta.cos\Theta \\ -sin\Theta.cos\Theta & cos^{2}\Theta \end{bmatrix}+ \begin{bmatrix}sin^{2}\Theta & -cos\Theta.sin\Theta \\ cos\Theta .sin\Theta & sin^{2}\Theta \end{bmatrix}$

$\fn_cm =\begin{bmatrix}cos^{2}\Theta &sin\Theta.cos\Theta \\ -sin\Theta.cos\Theta & cos^{2}\Theta \end{bmatrix}+ \begin{bmatrix}sin^{2}\Theta & -cos\Theta.sin\Theta \\ cos\Theta .sin\Theta & sin^{2}\Theta \end{bmatrix}$

$\fn_cm \begin{bmatrix}cos^{2}\Theta +sin^{2}\Theta & sin\Theta .cos\Theta -cos\Theta .sin\Theta \\ -sin\Theta .cos\Theta+cos\Theta .sin\Theta & cos^{2}\Theta +sin^{2}\Theta \end{bmatrix}$

$\fn_cm =\begin{bmatrix}1 &0 \\ 0& 1 \end{bmatrix}=I$

Where I mean unit matrix

Q7. Find X and Y if

$\fn_cm \left ( i \right )X +Y=\begin{bmatrix}7 &0 \\ 2& 5 \end{bmatrix}\: and\: X-Y=\begin{bmatrix}3 & 0\\ 0& 3 \end{bmatrix}$

$\fn_cm \left ( ii \right )2X +3Y=\begin{bmatrix}2 &3 \\ 4& 0 \end{bmatrix}\: and\: 3X+2Y=\begin{bmatrix}2 & -2\\ -1& 5 \end{bmatrix}$

Ans.

(i) We are given

$\fn_cm X +Y=\begin{bmatrix}7 &0 \\ 2& 5 \end{bmatrix}...\left ( 1 \right ) , X-Y=\begin{bmatrix}3 & 0\\ 0& 3 \end{bmatrix}....\left ( 2 \right )$

Adding both equations (i) and (ii),we have

$\fn_cm 2X =\begin{bmatrix}7 &0 \\ 2& 5 \end{bmatrix}+ \begin{bmatrix}3 & 0\\ 0& 3 \end{bmatrix}=\begin{bmatrix}10 & 0\\ 2 &8 \end{bmatrix}$

$\fn_cm X= \frac{1}{2}\begin{bmatrix}10 & 0\\ 2& 8 \end{bmatrix}$

$\fn_cm X= \begin{bmatrix}5 & 0\\ 1& 4 \end{bmatrix}$

Putting the value of X in equation (i)

$\fn_cm \begin{bmatrix}5 &0 \\ 1& 4 \end{bmatrix}+Y=\begin{bmatrix}7 & 0\\ 2& 5 \end{bmatrix}$

$\fn_cm Y=\begin{bmatrix}7 & 0\\ 2& 5 \end{bmatrix}-\begin{bmatrix}5 &0 \\ 1& 4 \end{bmatrix}=\begin{bmatrix}2 & 0\\ 1&1 \end{bmatrix}$

(ii) We are given

$\fn_cm 2X+3Y=\begin{bmatrix}2 & 3\\ 4& 0 \end{bmatrix}...\left ( 1 \right ),3X+2Y=\begin{bmatrix}2 &-2 \\ -1& 5 \end{bmatrix}....\left ( 2 \right )$

Multiplying equation (1) by 3 and equation (2) by 2, we get equation (3) and equation (4)

(i) We are given

$\fn_cm 6X+9Y=\begin{bmatrix}6 & 9\\ 12& 0 \end{bmatrix}...\left ( 3 \right ),6X+4Y=\begin{bmatrix}4 &-4 \\ -2& 10 \end{bmatrix}....\left ( 4 \right )$

Subtracting equation   (4) from equation (3)

$\fn_cm 5y=\begin{bmatrix}2 &13 \\ 14& -10 \end{bmatrix}$

$\fn_cm y=\begin{bmatrix}2/5 &13/5 \\ 14/5& -2 \end{bmatrix}$

Putting the value of Y in equation (i)

$\fn_cm 2X +3\begin{bmatrix}2/5 &13/5 \\ 14/5 & -2 \end{bmatrix}=\begin{bmatrix}2 & 3\\ 4& 0 \end{bmatrix}$

$\fn_cm 2X +\begin{bmatrix}6/5 &39/5 \\ 42/5 & -6 \end{bmatrix}=\begin{bmatrix}2 & 3\\ 4& 0 \end{bmatrix}$

$\fn_cm 2X =\begin{bmatrix}2 & 3\\ 4& 0 \end{bmatrix}-\begin{bmatrix}6/5 &39/5 \\ 42/5 & -6 \end{bmatrix}$

$\fn_cm 2X =\begin{bmatrix}4/5 &-24/5 \\ -22/5 & 6 \end{bmatrix}$

$\fn_cm X =\begin{bmatrix}2/5 &-12/5 \\ -11/5 & 3 \end{bmatrix}$

Q8. Find X, if

$\fn_cm Y=\begin{bmatrix}3 &2 \\ 1 & 4 \end{bmatrix}\: and\: 2X +Y=\begin{bmatrix}1 &0 \\ -3& 2 \end{bmatrix}$

Ans. We are given

$\fn_cm 2X +Y=\begin{bmatrix}1 &0 \\ -3& 2 \end{bmatrix}....\left ( 1 \right )$

$\fn_cm Y=\begin{bmatrix}3 &2 \\ 1 & 4 \end{bmatrix}\:$

Putting the value of Y  in equation  (i)

$\fn_cm 2X+ \begin{bmatrix}3 & 2\\ 1& 4 \end{bmatrix}=\begin{bmatrix}1 &0 \\ -3& 2 \end{bmatrix}$

$\fn_cm 2X =\begin{bmatrix}1 &0 \\ -3& 2 \end{bmatrix}-\begin{bmatrix}3 & 2\\ 1& 4 \end{bmatrix}=\begin{bmatrix}-2 & -2\\ -4 & -2 \end{bmatrix}$

$\fn_cm X=\begin{bmatrix}-1 & -1\\ -2 & -1 \end{bmatrix}$

Q9. Find X and Y, if

$\fn_cm 2\begin{bmatrix}1 & 3\\ 0 & x \end{bmatrix}+\begin{bmatrix}y &0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix}5 &6 \\ 1& 8 \end{bmatrix}$

$\fn_cm \begin{bmatrix}2 & 6\\ 0 & 2x \end{bmatrix}+\begin{bmatrix}y &0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix}5 &6 \\ 1& 8 \end{bmatrix}$

$\fn_cm \begin{bmatrix}2+y & 6\\ 1 & 2x+2 \end{bmatrix}=\begin{bmatrix}5 &6 \\ 1& 8 \end{bmatrix}$

Two matrices are equal indicates that their corresponding elements would be equal.

2+ y = 5⇒ y = 5 – 2 = 3

2x + 2 = 8

2x = 8 – 2 = 6

x = 3

Hence x = 3 and y = 3

Q10. Solve the equation for x,y,z and t,if

$\fn_cm 2\begin{bmatrix}x & z\\ y & t \end{bmatrix}+3\begin{bmatrix}1 & -1\\ 0& 2 \end{bmatrix}=3\begin{bmatrix}3 & 5\\ 4 & 6 \end{bmatrix}$

Ans. We are given that

$\fn_cm 2\begin{bmatrix}x & z\\ y & t \end{bmatrix}+3\begin{bmatrix}1 & -1\\ 0& 2 \end{bmatrix}=3\begin{bmatrix}3 & 5\\ 4 & 6 \end{bmatrix}$

Simplifying LHS and RHS

$\fn_cm \begin{bmatrix}2x & 2z\\ 2y & 2t \end{bmatrix}+\begin{bmatrix}3 & -3\\ 0& 6 \end{bmatrix}=\begin{bmatrix}9 & 15\\ 12 & 18 \end{bmatrix}$

$\fn_cm \begin{bmatrix}2x+3 & 2z-3\\ 2y & 2t+6 \end{bmatrix}=\begin{bmatrix}9 & 15\\ 12 & 18 \end{bmatrix}$

Two matrices are equal indicates that their corresponding elements would be equal.

2x + 3 = 9 ⇒2 x = 9- 3 = 6 ⇒x = 3

2z – 3 = 15⇒ 2z = 15+ 3= 18  ⇒ z = 9

2y = 12 ⇒ y = 6

2t + 6 = 18 ⇒ 2t = 18 – 6 = 12⇒ t = 6

Hence value of x= 3, y= 6,z= 9 and t = 6

## NCERT Solutions of Science and Maths for Class 9,10,11 and 12

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### NCERT Solutions for class 11 maths

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CBSE Class 11-Question paper of maths 2015

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### NCERT solutions for class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

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