Samacheer Kalvi 10th Maths Exercise 3.18 Solutions

10th Maths Exercise 3.18 solutions (Chapter 3 Algebra) with step-by-step answers for all 11 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance8 answers · full steps below

Q3 $X=\begin{pmatrix}5&0\\\frac{3}{2}&\frac{9}{2}\end{pmatrix},\quad Y=\begin{pmatrix}2&0\\\frac{3}{2}&\frac{1}{2}\end{pmatrix}$
Q4 (i) $B-5A=\begin{pmatrix}7&-17&-37\\-39&-11&-26\end{pmatrix}$
Q4 (ii) $3A-9B=\begin{pmatrix}-63&-15&-45\\15&-27&-60\end{pmatrix}$
Q5 (i) $x=4,\ y=-10,\ z=12$
Q5 (ii) $x=-10,\ y=14,\ z=10$
Q6 $x=4,\ y=6$
Q7 $x=4$
Q8 $x=5$ or $x=-1$ ; $y=4$ or $y=-2$

Q1. If $A=\begin{pmatrix}1&9\\3&4\\8&-3\end{pmatrix}$ , $B=\begin{pmatrix}5&7\\3&3\\1&0\end{pmatrix}$ then verify that

If $A=\begin{pmatrix}1&9\\3&4\\8&-3\end{pmatrix}$ , $B=\begin{pmatrix}5&7\\3&3\\1&0\end{pmatrix}$ then verify that

(i)

$A+B=B+A$

$A+B=\begin{pmatrix}1&9\\3&4\\8&-3\end{pmatrix}+\begin{pmatrix}5&7\\3&3\\1&0\end{pmatrix}$
Add the corresponding entries.
$A+B=\begin{pmatrix}1+5&9+7\\3+3&4+3\\8+1&-3+0\end{pmatrix}=\begin{pmatrix}6&16\\6&7\\9&-3\end{pmatrix}$
$B+A=\begin{pmatrix}5+1&7+9\\3+3&3+4\\1+8&0-3\end{pmatrix}=\begin{pmatrix}6&16\\6&7\\9&-3\end{pmatrix}$
Both matrices are equal.

Answer

$A+B=B+A=\begin{pmatrix}6&16\\6&7\\9&-3\end{pmatrix}$ , hence matrix addition is commutative.

(ii)

$A+(-A)=(-A)+A=O$ .

$-A=\begin{pmatrix}-1&-9\\-3&-4\\-8&3\end{pmatrix}$
$A+(-A)=\begin{pmatrix}1-1&9-9\\3-3&4-4\\8-8&-3+3\end{pmatrix}=\begin{pmatrix}0&0\\0&0\\0&0\end{pmatrix}=O$
$(-A)+A=\begin{pmatrix}-1+1&-9+9\\-3+3&-4+4\\-8+8&3-3\end{pmatrix}=\begin{pmatrix}0&0\\0&0\\0&0\end{pmatrix}=O$

Answer

$A+(-A)=(-A)+A=O$ , so $-A$ is the additive inverse of $A$ .

Q2. If $A=\begin{pmatrix}4&3&1\\2&3&-8\\1&0&-4\end{pmatrix}$ , $B=\begin{pmatrix}2&3&4\\1&9&2\\-7&1&-1\end{pmatrix}$ and $C=\begin{pmatrix}8&3&4\\1&-2&3\\2&4&-1\end{pmatrix}$ then verify that $A+(B+C)=(A+B)+C$ .

$B+C=\begin{pmatrix}2+8&3+3&4+4\\1+1&9-2&2+3\\-7+2&1+4&-1-1\end{pmatrix}=\begin{pmatrix}10&6&8\\2&7&5\\-5&5&-2\end{pmatrix}$
$A+(B+C)=\begin{pmatrix}4+10&3+6&1+8\\2+2&3+7&-8+5\\1-5&0+5&-4-2\end{pmatrix}=\begin{pmatrix}14&9&9\\4&10&-3\\-4&5&-6\end{pmatrix}$
$A+B=\begin{pmatrix}6&6&5\\3&12&-6\\-6&1&-5\end{pmatrix}$
$(A+B)+C=\begin{pmatrix}6+8&6+3&5+4\\3+1&12-2&-6+3\\-6+2&1+4&-5-1\end{pmatrix}=\begin{pmatrix}14&9&9\\4&10&-3\\-4&5&-6\end{pmatrix}$
Both results are the same.

Answer

$A+(B+C)=(A+B)+C=\begin{pmatrix}14&9&9\\4&10&-3\\-4&5&-6\end{pmatrix}$ , hence matrix addition is associative.

Q3. Find $X$ and $Y$ if $X+Y=\begin{pmatrix}7&0\\3&5\end{pmatrix}$ and $X-Y=\begin{pmatrix}3&0\\0&4\end{pmatrix}$

Adding the two equations: $2X=\begin{pmatrix}7&0\\3&5\end{pmatrix}+\begin{pmatrix}3&0\\0&4\end{pmatrix}=\begin{pmatrix}10&0\\3&9\end{pmatrix}$
$X=\dfrac{1}{2}\begin{pmatrix}10&0\\3&9\end{pmatrix}=\begin{pmatrix}5&0\\\frac{3}{2}&\frac{9}{2}\end{pmatrix}$
Subtracting: $2Y=\begin{pmatrix}7&0\\3&5\end{pmatrix}-\begin{pmatrix}3&0\\0&4\end{pmatrix}=\begin{pmatrix}4&0\\3&1\end{pmatrix}$
$Y=\dfrac{1}{2}\begin{pmatrix}4&0\\3&1\end{pmatrix}=\begin{pmatrix}2&0\\\frac{3}{2}&\frac{1}{2}\end{pmatrix}$

Answer

$X=\begin{pmatrix}5&0\\\frac{3}{2}&\frac{9}{2}\end{pmatrix},\quad Y=\begin{pmatrix}2&0\\\frac{3}{2}&\frac{1}{2}\end{pmatrix}$

Q4. If $A=\begin{pmatrix}0&4&9\\8&3&7\end{pmatrix}$ , $B=\begin{pmatrix}7&3&8\\1&4&9\end{pmatrix}$ find the value of

If $A=\begin{pmatrix}0&4&9\\8&3&7\end{pmatrix}$ , $B=\begin{pmatrix}7&3&8\\1&4&9\end{pmatrix}$ find the value of

(i)

$B-5A$

$5A=5\begin{pmatrix}0&4&9\\8&3&7\end{pmatrix}=\begin{pmatrix}0&20&45\\40&15&35\end{pmatrix}$
$B-5A=\begin{pmatrix}7&3&8\\1&4&9\end{pmatrix}-\begin{pmatrix}0&20&45\\40&15&35\end{pmatrix}$
$=\begin{pmatrix}7-0&3-20&8-45\\1-40&4-15&9-35\end{pmatrix}$

Answer

$B-5A=\begin{pmatrix}7&-17&-37\\-39&-11&-26\end{pmatrix}$

(ii)

$3A-9B$

$3A=\begin{pmatrix}0&12&27\\24&9&21\end{pmatrix}$
$9B=\begin{pmatrix}63&27&72\\9&36&81\end{pmatrix}$
$3A-9B=\begin{pmatrix}0-63&12-27&27-72\\24-9&9-36&21-81\end{pmatrix}$

Answer

$3A-9B=\begin{pmatrix}-63&-15&-45\\15&-27&-60\end{pmatrix}$

Q5. Find the values of $x,\ y,\ z$ if

(i)

$\begin{pmatrix}x-3&3x-z\\x+y+7&x+y+z\end{pmatrix}=\begin{pmatrix}1&0\\1&6\end{pmatrix}$

Equal matrices have equal corresponding entries.
$x-3=1\ \Rightarrow\ x=4$
$3x-z=0\ \Rightarrow\ z=3(4)=12$
$x+y+7=1\ \Rightarrow\ 4+y+7=1\ \Rightarrow\ y=-10$
Check: $x+y+z=4-10+12=6$ ✓

Answer

$x=4,\ y=-10,\ z=12$

(ii)

$(x\quad y-z\quad z+3)+(y\quad 4\quad 3)=(4\quad 8\quad 16)$

Adding the matrices: $(x+y\quad y-z+4\quad z+6)=(4\quad 8\quad 16)$
$z+6=16\ \Rightarrow\ z=10$
$y-z+4=8\ \Rightarrow\ y-10+4=8\ \Rightarrow\ y=14$
$x+y=4\ \Rightarrow\ x+14=4\ \Rightarrow\ x=-10$

Answer

$x=-10,\ y=14,\ z=10$

Q6. Find $x$ and $y$ if $x\begin{pmatrix}4\\-3\end{pmatrix}+y\begin{pmatrix}-2\\3\end{pmatrix}=\begin{pmatrix}4\\6\end{pmatrix}$

Show steps (6)

$\begin{pmatrix}4x\\-3x\end{pmatrix}+\begin{pmatrix}-2y\\3y\end{pmatrix}=\begin{pmatrix}4\\6\end{pmatrix}$
$\begin{pmatrix}4x-2y\\-3x+3y\end{pmatrix}=\begin{pmatrix}4\\6\end{pmatrix}$
$4x-2y=4\ \Rightarrow\ 2x-y=2$ ...(1)
$-3x+3y=6\ \Rightarrow\ -x+y=2$ ...(2)
Adding (1) and (2): $x=4$
From (2): $y=2+x=6$

Answer

$x=4,\ y=6$

Q7. Find the non-zero values of $x$ satisfying the matrix equation $x\begin{pmatrix}2x&2\\3&x\end{pmatrix}+2\begin{pmatrix}8&5x\\4&4x\end{pmatrix}=2\begin{pmatrix}x^2+8&24\\10&6x\end{pmatrix}$

Show steps (6)

$\begin{pmatrix}2x^2&2x\\3x&x^2\end{pmatrix}+\begin{pmatrix}16&10x\\8&8x\end{pmatrix}=\begin{pmatrix}2x^2+16&48\\20&12x\end{pmatrix}$
$\begin{pmatrix}2x^2+16&12x\\3x+8&x^2+8x\end{pmatrix}=\begin{pmatrix}2x^2+16&48\\20&12x\end{pmatrix}$
Equating corresponding entries: $12x=48\ \Rightarrow\ x=4$
$3x+8=20\ \Rightarrow\ x=4$
$x^2+8x=12x\ \Rightarrow\ x^2-4x=0\ \Rightarrow\ x(x-4)=0$
Since $x\neq 0$ , we get $x=4$ .

Answer

$x=4$

Q8. Solve for $x,\ y$ : $\begin{pmatrix}x^2\\y^2\end{pmatrix}+2\begin{pmatrix}-2x\\-y\end{pmatrix}=\begin{pmatrix}5\\8\end{pmatrix}$

$\begin{pmatrix}x^2\\y^2\end{pmatrix}+\begin{pmatrix}-4x\\-2y\end{pmatrix}=\begin{pmatrix}5\\8\end{pmatrix}$
$\begin{pmatrix}x^2-4x\\y^2-2y\end{pmatrix}=\begin{pmatrix}5\\8\end{pmatrix}$
$x^2-4x-5=0\ \Rightarrow\ (x-5)(x+1)=0\ \Rightarrow\ x=5,\ -1$
$y^2-2y-8=0\ \Rightarrow\ (y-4)(y+2)=0\ \Rightarrow\ y=4,\ -2$

Answer

$x=5$ or $x=-1$ ; $y=4$ or $y=-2$

Quick check

Find the matrix $X$ if $2X + \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix}$
1. (1) $\begin{pmatrix} -2 & -2 \\ 2 & -1 \end{pmatrix}$
2. (2) $\begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}$
3. (3) $\begin{pmatrix} 1 & 2 \\ 2 & 2 \end{pmatrix}$
4. (4) $\begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix}$
Show answer
Answer: (2) $\begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}$
1. $2X = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix} - \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix}$
2. $X = \dfrac{1}{2}\begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}$
Question 1 of 5
The values of $a$ and $b$ if $4x^4 - 24x^3 + 76x^2 + ax + b$ is a perfect square are
1. (1) $100,120$
2. (2) $10,12$
3. (3) $-120,100$
4. (4) $12,10$
Show answer
Answer: (3) $-120,100$
1. Let $4x^4 - 24x^3 + 76x^2 + ax + b = (2x^2 - 6x + c)^2$ .
2. Expanding: $4x^4 - 24x^3 + (36 + 4c)x^2 - 12cx + c^2$ .
3. Compare $x^2$ terms: $36 + 4c = 76 \Rightarrow c = 10$ .
4. So $a = -12(10) = -120$ and $b = 10^2 = 100$ .
Question 2 of 5
The solution of the system $x + y - 3z = -6$ , $-7y + 7z = 7$ , $3z = 9$ is
1. (1) $x = 1,\ y = 2,\ z = 3$
2. (2) $x = -1,\ y = 2,\ z = 3$
3. (3) $x = -1,\ y = -2,\ z = 3$
4. (4) $x = 1,\ y = -2,\ z = 3$
Show answer
Answer: (1) $x = 1,\ y = 2,\ z = 3$
1. From $3z = 9$ , $z = 3$ .
2. Put $z=3$ in $-7y + 7z = 7$ : $-7y + 21 = 7 \Rightarrow y = 2$ .
3. Put $y=2,\ z=3$ in $x + y - 3z = -6$ : $x + 2 - 9 = -6 \Rightarrow x = 1$ .
Question 3 of 5
If $A$ is a $2 \times 3$ matrix and $B$ is a $3 \times 4$ matrix, how many columns does $AB$ have
1. (1) $3$
2. (2) $4$
3. (3) $2$
4. (4) $5$
Show answer
Answer: (2) $4$
1. $A_{2\times 3} B_{3\times 4}$ gives order $2 \times 4$ .
2. So $AB$ has $4$ columns.
Question 4 of 5
The square root of $\dfrac{256 x^8 y^4 z^{10}}{25 x^6 y^6 z^6}$ is equal to
1. (1) $\dfrac{16}{5}\left|\dfrac{x^2 z^4}{y^2}\right|$
2. (2) $16\left|\dfrac{y^2}{x^2 z^4}\right|$
3. (3) $\dfrac{16}{5}\left|\dfrac{y}{xz^2}\right|$
4. (4) $\dfrac{16}{5}\left|\dfrac{xz^2}{y}\right|$
Show answer
Answer: (4) $\dfrac{16}{5}\left|\dfrac{xz^2}{y}\right|$
1. $\dfrac{256x^8y^4z^{10}}{25x^6y^6z^6} = \dfrac{256\,x^2 z^4}{25\,y^2}$
2. Taking square root: $\dfrac{16}{5}\left|\dfrac{x z^2}{y}\right|$
Question 5 of 5

Chapter 3 formulas

Roots of a quadratic equation
$ax^2 + bx + c = 0,\; (a \neq 0) \;\Rightarrow\; x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Sum of the roots
$\alpha + \beta = \dfrac{-b}{a} = \dfrac{-\text{Co-efficient of } x}{\text{Co-efficient of } x^2}$
Product of the roots
$\alpha\beta = \dfrac{c}{a} = \dfrac{\text{Constant term}}{\text{Co-efficient of } x^2}$
Quadratic equation formed from its roots
$x^2 - (\alpha + \beta)x + \alpha\beta = 0$
Discriminant
$\Delta = b^2 - 4ac$
Nature of roots: real and unequal
$\Delta > 0 \;\Rightarrow\; \text{the roots are real and unequal}$
Nature of roots: real and equal
$\Delta = 0 \;\Rightarrow\; \text{the roots are real and equal}$
Nature of roots: no real roots
$\Delta < 0 \;\Rightarrow\; \text{there are no real roots}$
Cases for a system of linear equations in three variables
$\text{(i) Unique solution}\quad \text{(ii) Infinitely many solutions}\quad \text{(iii) No solution}$
LCM of algebraic expressions
$\text{LCM} = \text{expression of lowest degree that each given expression exactly divides}$
Quadratic polynomial and its zeroes
$p(x) = ax^2 + bx + c \;(a \neq 0) \;\Rightarrow\; p(x) \text{ has atmost two zeroes, which lie on the } x\text{-axis}$
Order of a matrix
$\text{Order of } A = (\text{Number of rows}) \times (\text{Number of columns}) = m \times n$
Row matrix (row vector)
$A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \end{pmatrix}_{1 \times n}$
Column matrix (column vector)
$A = \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{pmatrix}_{m \times 1}$
Square matrix
$\text{Number of rows} = \text{Number of columns},\; \text{order } n \times n$
Zero matrix (null matrix)
$a_{ij} = 0 \text{ for all } i,\, j$
Transpose of a matrix
$\left(A^{T}\right)_{ij} = a_{ji} \quad (\text{rows and columns of } A \text{ interchanged})$
Diagonal matrix
$a_{ij} = 0 \text{ for } i \neq j \quad (\text{square matrix})$
Scalar matrix
$a_{ij} = \begin{cases} k, & i = j \\ 0, & i \neq j \end{cases}$
Identity matrix (unit matrix)
$a_{ij} = \begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases}$
Lower triangular matrix
$a_{ij} = 0 \text{ for } i < j \quad (\text{all entries above the leading diagonal are zero})$
Upper triangular matrix
$a_{ij} = 0 \text{ for } i > j \quad (\text{all entries below the leading diagonal are zero})$
Equality of matrices
$A = B \iff \text{same order and } a_{ij} = b_{ij} \text{ for all } i,\, j$

Answers checked against the Samacheer Kalvi textbook · Updated 28 September 2026

Spotted a mistake? Email us at [email protected] and we will fix it.

Q1. If A=(19348−3)A=\begin{pmatrix}1&9\\3&4\\8&-3\end{pmatrix}A=​138​94−3​​ , B=(573310)B=\begin{pmatrix}5&7\\3&3\\1&0\end{pmatrix}B=​531​730​​ then verify that

(i)

(ii)

Q3. Find XXX and YYY if X+Y=(7035)X+Y=\begin{pmatrix}7&0\\3&5\end{pmatrix}X+Y=(73​05​) and X−Y=(3004)X-Y=\begin{pmatrix}3&0\\0&4\end{pmatrix}X−Y=(30​04​)

Q4. If A=(049837)A=\begin{pmatrix}0&4&9\\8&3&7\end{pmatrix}A=(08​43​97​) , B=(738149)B=\begin{pmatrix}7&3&8\\1&4&9\end{pmatrix}B=(71​34​89​) find the value of

(i)

(ii)

Q5. Find the values of x, y, zx,\ y,\ zx, y, z if

(i)

(ii)

Q6. Find xxx and yyy if x(4−3)+y(−23)=(46)x\begin{pmatrix}4\\-3\end{pmatrix}+y\begin{pmatrix}-2\\3\end{pmatrix}=\begin{pmatrix}4\\6\end{pmatrix}x(4−3​)+y(−23​)=(46​)

Q7. Find the non-zero values of xxx satisfying the matrix equation x(2x23x)+2(85x44x)=2(x2+824106x)x\begin{pmatrix}2x&2\\3&x\end{pmatrix}+2\begin{pmatrix}8&5x\\4&4x\end{pmatrix}=2\begin{pmatrix}x^2+8&24\\10&6x\end{pmatrix}x(2x3​2x​)+2(84​5x4x​)=2(x2+810​246x​)

Q8. Solve for x, yx,\ yx, y : (x2y2)+2(−2x−y)=(58)\begin{pmatrix}x^2\\y^2\end{pmatrix}+2\begin{pmatrix}-2x\\-y\end{pmatrix}=\begin{pmatrix}5\\8\end{pmatrix}(x2y2​)+2(−2x−y​)=(58​)

Quick check

Q1. If $A=\begin{pmatrix}1&9\\3&4\\8&-3\end{pmatrix}$ , $B=\begin{pmatrix}5&7\\3&3\\1&0\end{pmatrix}$ then verify that

Q3. Find $X$ and $Y$ if $X+Y=\begin{pmatrix}7&0\\3&5\end{pmatrix}$ and $X-Y=\begin{pmatrix}3&0\\0&4\end{pmatrix}$

Q4. If $A=\begin{pmatrix}0&4&9\\8&3&7\end{pmatrix}$ , $B=\begin{pmatrix}7&3&8\\1&4&9\end{pmatrix}$ find the value of

Q5. Find the values of $x,\ y,\ z$ if

Q6. Find $x$ and $y$ if $x\begin{pmatrix}4\\-3\end{pmatrix}+y\begin{pmatrix}-2\\3\end{pmatrix}=\begin{pmatrix}4\\6\end{pmatrix}$

Q7. Find the non-zero values of $x$ satisfying the matrix equation $x\begin{pmatrix}2x&2\\3&x\end{pmatrix}+2\begin{pmatrix}8&5x\\4&4x\end{pmatrix}=2\begin{pmatrix}x^2+8&24\\10&6x\end{pmatrix}$

Q8. Solve for $x,\ y$ : $\begin{pmatrix}x^2\\y^2\end{pmatrix}+2\begin{pmatrix}-2x\\-y\end{pmatrix}=\begin{pmatrix}5\\8\end{pmatrix}$