Samacheer Kalvi 10th Maths Exercise 3.19 Solutions

10th Maths Exercise 3.19 solutions (Chapter 3 Algebra) with step-by-step answers for all 19 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance8 answers · full steps below

Q1 (i) $3 \times 3$
Q1 (ii) $4 \times 2$
Q1 (iii) $4 \times 2$
Q1 (iv) $4 \times 1$
Q1 (v) $1 \times 3$
Q2 $AB$ is of order $p\times r$ ; $BA$ is not defined (it would exist only in the special case $r=p$ , when it is of order $q\times q$ )
Q3 $a = 7,\ b = 10$
Q4 $AB = \begin{pmatrix} 12 & 19 \\ 10 & 3 \end{pmatrix}$ , $BA = \begin{pmatrix} -10 & -4 \\ 24 & 25 \end{pmatrix}$ ; hence $AB \neq BA$

Q1. Find the order of the product matrix $AB$ if

(i)

Order of $A$ is $3\times3$ and order of $B$ is $3\times3$ .

	(i)	(ii)	(iii)	(iv)	(v)
Orders of A	3×3	4×3	4×2	4×5	1×1
Orders of B	3×3	3×2	2×2	5×1	1×3

$A$ is $3\times 3$ , $B$ is $3\times 3$ .
Number of columns of $A$ = number of rows of $B$ = $3$ , so $AB$ exists.
Order of $AB$ = (rows of $A$ ) × (columns of $B$ ).

Answer

$3 \times 3$

(ii)

Order of $A$ is $4\times3$ and order of $B$ is $3\times2$ .

	(i)	(ii)	(iii)	(iv)	(v)
Orders of A	3×3	4×3	4×2	4×5	1×1
Orders of B	3×3	3×2	2×2	5×1	1×3

$A$ is $4\times 3$ , $B$ is $3\times 2$ .
Columns of $A$ = rows of $B$ = $3$ , so $AB$ exists.
Order of $AB$ = (rows of $A$ ) × (columns of $B$ ).

Answer

$4 \times 2$

(iii)

Order of $A$ is $4\times2$ and order of $B$ is $2\times2$ .

	(i)	(ii)	(iii)	(iv)	(v)
Orders of A	3×3	4×3	4×2	4×5	1×1
Orders of B	3×3	3×2	2×2	5×1	1×3

$A$ is $4\times 2$ , $B$ is $2\times 2$ .
Columns of $A$ = rows of $B$ = $2$ , so $AB$ exists.
Order of $AB$ = (rows of $A$ ) × (columns of $B$ ).

Answer

$4 \times 2$

(iv)

Order of $A$ is $4\times5$ and order of $B$ is $5\times1$ .

	(i)	(ii)	(iii)	(iv)	(v)
Orders of A	3×3	4×3	4×2	4×5	1×1
Orders of B	3×3	3×2	2×2	5×1	1×3

$A$ is $4\times 5$ , $B$ is $5\times 1$ .
Columns of $A$ = rows of $B$ = $5$ , so $AB$ exists.
Order of $AB$ = (rows of $A$ ) × (columns of $B$ ).

Answer

$4 \times 1$

(v)

Order of $A$ is $1\times1$ and order of $B$ is $1\times3$ .

	(i)	(ii)	(iii)	(iv)	(v)
Orders of A	3×3	4×3	4×2	4×5	1×1
Orders of B	3×3	3×2	2×2	5×1	1×3

$A$ is $1\times 1$ , $B$ is $1\times 3$ .
Columns of $A$ = rows of $B$ = $1$ , so $AB$ exists.
Order of $AB$ = (rows of $A$ ) × (columns of $B$ ).

Answer

$1 \times 3$

Q2. If $A$ is of order $p \times q$ and $B$ is of order $q \times r$ …

If $A$ is of order $p \times q$ and $B$ is of order $q \times r$ what is the order of $AB$ and $BA$ ?

$A$ is of order $p\times q$ and $B$ is of order $q\times r$ .
For $AB$ : number of columns of $A$ $=q=$ number of rows of $B$ , so $AB$ is defined.
Order of $AB$ = (rows of $A$ ) $\times$ (columns of $B$ ) $= p\times r$ .
For $BA$ : number of columns of $B$ $=r$ , number of rows of $A$ $=p$ , and in general $r\neq p$ .
Hence $BA$ is not defined. (Only in the special case $r=p$ would $BA$ exist, and then it would be of order $q\times q$ .)

Answer

$AB$ is of order $p\times r$ ; $BA$ is not defined (it would exist only in the special case $r=p$ , when it is of order $q\times q$ )

Q3. $A$ has ' $a$ ' rows and ' $a + 3$ ' columns. $B$ has ' $b$ ' …

$A$ has ' $a$ ' rows and ' $a + 3$ ' columns. $B$ has ' $b$ ' rows and ' $17-b$ ' columns, and if both products $AB$ and $BA$ exist, find $a,\ b$ ?

Show steps (6)

$A$ is of order $a\times(a+3)$ and $B$ is of order $b\times(17-b)$ .
$AB$ exists $\Rightarrow$ columns of $A$ = rows of $B$ , so $a+3 = b$ … (1)
$BA$ exists $\Rightarrow$ columns of $B$ = rows of $A$ , so $17-b = a$ … (2)
Substituting (1) in (2): $17-(a+3) = a$
$14 - a = a \Rightarrow 2a = 14 \Rightarrow a = 7$
From (1), $b = 7+3 = 10$ .

Answer

$a = 7,\ b = 10$

Q4. If $A = \begin{pmatrix} 2 & 5 \\ 4 & 3 \end{pmatrix}$ , $B = \begin{pmatrix} 1 & -3 \\ 2 & 5 \end{pmatrix}$ find $AB,\ BA$ and verify $AB = BA$ ?

$AB = \begin{pmatrix} 2 & 5 \\ 4 & 3 \end{pmatrix}\begin{pmatrix} 1 & -3 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix} 2+10 & -6+25 \\ 4+6 & -12+15 \end{pmatrix}$
$AB = \begin{pmatrix} 12 & 19 \\ 10 & 3 \end{pmatrix}$
$BA = \begin{pmatrix} 1 & -3 \\ 2 & 5 \end{pmatrix}\begin{pmatrix} 2 & 5 \\ 4 & 3 \end{pmatrix} = \begin{pmatrix} 2-12 & 5-9 \\ 4+20 & 10+15 \end{pmatrix}$
$BA = \begin{pmatrix} -10 & -4 \\ 24 & 25 \end{pmatrix}$
The corresponding entries are not equal.

Answer

$AB = \begin{pmatrix} 12 & 19 \\ 10 & 3 \end{pmatrix}$ , $BA = \begin{pmatrix} -10 & -4 \\ 24 & 25 \end{pmatrix}$ ; hence $AB \neq BA$

Q5. Given that $A = \begin{pmatrix} 1 & 3 \\ 5 & -1 \end{pmatrix}$ , $B = \begin{pmatrix} 1 & -1 & 2 \\ 3 & 5 & 2 \end{pmatrix}$ , $C = \begin{pmatrix} 1 & 3 & 2 \\ -4 & 1 & 3 \end{pmatrix}$ verify that $A(B + C) = AB + AC$ .

Show steps (9)

LHS $= A(B+C)$
$B+C = \begin{pmatrix} 1+1 & -1+3 & 2+2 \\ 3-4 & 5+1 & 2+3 \end{pmatrix} = \begin{pmatrix} 2 & 2 & 4 \\ -1 & 6 & 5 \end{pmatrix}$
$A(B+C) = \begin{pmatrix} 1 & 3 \\ 5 & -1 \end{pmatrix}\begin{pmatrix} 2 & 2 & 4 \\ -1 & 6 & 5 \end{pmatrix} = \begin{pmatrix} 2-3 & 2+18 & 4+15 \\ 10+1 & 10-6 & 20-5 \end{pmatrix}$
$A(B+C) = \begin{pmatrix} -1 & 20 & 19 \\ 11 & 4 & 15 \end{pmatrix}$ … (1)
RHS $= AB + AC$
$AB = \begin{pmatrix} 1+9 & -1+15 & 2+6 \\ 5-3 & -5-5 & 10-2 \end{pmatrix} = \begin{pmatrix} 10 & 14 & 8 \\ 2 & -10 & 8 \end{pmatrix}$
$AC = \begin{pmatrix} 1-12 & 3+3 & 2+9 \\ 5+4 & 15-1 & 10-3 \end{pmatrix} = \begin{pmatrix} -11 & 6 & 11 \\ 9 & 14 & 7 \end{pmatrix}$
$AB+AC = \begin{pmatrix} -1 & 20 & 19 \\ 11 & 4 & 15 \end{pmatrix}$ … (2)
From (1) and (2), LHS = RHS.

Answer

$A(B+C) = AB + AC = \begin{pmatrix} -1 & 20 & 19 \\ 11 & 4 & 15 \end{pmatrix}$ , verified

Q6. Show that the matrices $A = \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}$ , $B = \begin{pmatrix} 1 & -2 \\ -3 & 1 \end{pmatrix}$ satisfy commutative property $AB = BA$

$AB = \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 1 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 1-6 & -2+2 \\ 3-3 & -6+1 \end{pmatrix} = \begin{pmatrix} -5 & 0 \\ 0 & -5 \end{pmatrix}$ … (1)
$BA = \begin{pmatrix} 1 & -2 \\ -3 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} 1-6 & 2-2 \\ -3+3 & -6+1 \end{pmatrix} = \begin{pmatrix} -5 & 0 \\ 0 & -5 \end{pmatrix}$ … (2)
From (1) and (2), $AB = BA$ .

Answer

$AB = BA = \begin{pmatrix} -5 & 0 \\ 0 & -5 \end{pmatrix}$ , so the commutative property holds

Q7. Let $A = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}$ , $B = \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix}$ , $C = \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix}$ Show that

Let $A = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}$ , $B = \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix}$ , $C = \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix}$ Show that

(i)

$A(BC) = (AB)C$

Show steps (7)

LHS $= A(BC)$
$BC = \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 8+0 & 0+0 \\ 2+5 & 0+10 \end{pmatrix} = \begin{pmatrix} 8 & 0 \\ 7 & 10 \end{pmatrix}$
$A(BC) = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} 8 & 0 \\ 7 & 10 \end{pmatrix} = \begin{pmatrix} 8+14 & 0+20 \\ 8+21 & 0+30 \end{pmatrix} = \begin{pmatrix} 22 & 20 \\ 29 & 30 \end{pmatrix}$ … (1)
RHS $= (AB)C$
$AB = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix} = \begin{pmatrix} 4+2 & 0+10 \\ 4+3 & 0+15 \end{pmatrix} = \begin{pmatrix} 6 & 10 \\ 7 & 15 \end{pmatrix}$
$(AB)C = \begin{pmatrix} 6 & 10 \\ 7 & 15 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 12+10 & 0+20 \\ 14+15 & 0+30 \end{pmatrix} = \begin{pmatrix} 22 & 20 \\ 29 & 30 \end{pmatrix}$ … (2)
From (1) and (2), $A(BC) = (AB)C$ .

Answer

$A(BC) = (AB)C = \begin{pmatrix} 22 & 20 \\ 29 & 30 \end{pmatrix}$ , hence shown

(ii)

$(A - B)C = AC - BC$

Show steps (6)

$A - B = \begin{pmatrix} 1-4 & 2-0 \\ 1-1 & 3-5 \end{pmatrix} = \begin{pmatrix} -3 & 2 \\ 0 & -2 \end{pmatrix}$
LHS $= (A-B)C = \begin{pmatrix} -3 & 2 \\ 0 & -2 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} -6+2 & 0+4 \\ 0-2 & 0-4 \end{pmatrix} = \begin{pmatrix} -4 & 4 \\ -2 & -4 \end{pmatrix}$ … (1)
$AC = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 2+2 & 0+4 \\ 2+3 & 0+6 \end{pmatrix} = \begin{pmatrix} 4 & 4 \\ 5 & 6 \end{pmatrix}$
$BC = \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 8 & 0 \\ 7 & 10 \end{pmatrix}$
RHS $= AC - BC = \begin{pmatrix} 4-8 & 4-0 \\ 5-7 & 6-10 \end{pmatrix} = \begin{pmatrix} -4 & 4 \\ -2 & -4 \end{pmatrix}$ … (2)
From (1) and (2), $(A-B)C = AC - BC$ .

Answer

$(A-B)C = AC - BC = \begin{pmatrix} -4 & 4 \\ -2 & -4 \end{pmatrix}$ , hence shown

(iii)

$(A - B)^T = A^T - B^T$

$A - B = \begin{pmatrix} -3 & 2 \\ 0 & -2 \end{pmatrix}$
LHS $= (A-B)^T = \begin{pmatrix} -3 & 0 \\ 2 & -2 \end{pmatrix}$ … (1)
$A^T = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}$ , $B^T = \begin{pmatrix} 4 & 1 \\ 0 & 5 \end{pmatrix}$
RHS $= A^T - B^T = \begin{pmatrix} 1-4 & 1-1 \\ 2-0 & 3-5 \end{pmatrix} = \begin{pmatrix} -3 & 0 \\ 2 & -2 \end{pmatrix}$ … (2)
From (1) and (2), $(A-B)^T = A^T - B^T$ .

Answer

$(A-B)^T = A^T - B^T = \begin{pmatrix} -3 & 0 \\ 2 & -2 \end{pmatrix}$ , hence shown

Q8. If $A = \begin{pmatrix} \cos\theta & 0 \\ 0 & \cos\theta \end{pmatrix}$ , $B = \begin{pmatrix} \sin\theta & 0 \\ 0 & \sin\theta \end{pmatrix}$ then show that $A^2 + B^2 = I$ .

$A^2 = \begin{pmatrix} \cos\theta & 0 \\ 0 & \cos\theta \end{pmatrix}\begin{pmatrix} \cos\theta & 0 \\ 0 & \cos\theta \end{pmatrix} = \begin{pmatrix} \cos^2\theta & 0 \\ 0 & \cos^2\theta \end{pmatrix}$
$B^2 = \begin{pmatrix} \sin\theta & 0 \\ 0 & \sin\theta \end{pmatrix}\begin{pmatrix} \sin\theta & 0 \\ 0 & \sin\theta \end{pmatrix} = \begin{pmatrix} \sin^2\theta & 0 \\ 0 & \sin^2\theta \end{pmatrix}$
$A^2 + B^2 = \begin{pmatrix} \cos^2\theta+\sin^2\theta & 0 \\ 0 & \cos^2\theta+\sin^2\theta \end{pmatrix}$
Since $\cos^2\theta + \sin^2\theta = 1$ ,
$A^2 + B^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$ .

Answer

$A^2 + B^2 = I$ , hence shown

Q9. If $A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$ prove that $AA^T = I$ .

$A^T = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$
$AA^T = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$
$= \begin{pmatrix} \cos^2\theta+\sin^2\theta & -\cos\theta\sin\theta+\sin\theta\cos\theta \\ -\sin\theta\cos\theta+\cos\theta\sin\theta & \sin^2\theta+\cos^2\theta \end{pmatrix}$
$= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$

Answer

$AA^T = I$ , hence proved

Q10. Verify that $A^2 = I$ when $A = \begin{pmatrix} 5 & -4 \\ 6 & -5 \end{pmatrix}$

$A^2 = \begin{pmatrix} 5 & -4 \\ 6 & -5 \end{pmatrix}\begin{pmatrix} 5 & -4 \\ 6 & -5 \end{pmatrix}$
$= \begin{pmatrix} 25-24 & -20+20 \\ 30-30 & -24+25 \end{pmatrix}$
$= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$

Answer

$A^2 = I$ , verified

Q11. If $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ show that $A^2 - (a + d)A = (bc - ad)I_2$

$A^2 = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a^2+bc & ab+bd \\ ac+cd & bc+d^2 \end{pmatrix}$
$(a+d)A = \begin{pmatrix} a^2+ad & ab+bd \\ ac+cd & ad+d^2 \end{pmatrix}$
$A^2 - (a+d)A = \begin{pmatrix} a^2+bc-a^2-ad & 0 \\ 0 & bc+d^2-ad-d^2 \end{pmatrix}$
$= \begin{pmatrix} bc-ad & 0 \\ 0 & bc-ad \end{pmatrix}$
$= (bc-ad)\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = (bc-ad)I_2$

Answer

$A^2 - (a+d)A = (bc-ad)I_2$ , hence shown

Q12. If $A = \begin{pmatrix} 5 & 2 & 9 \\ 1 & 2 & 8 \end{pmatrix}$ , $B = \begin{pmatrix} 1 & 7 \\ 1 & 2 \\ 5 & -1 \end{pmatrix}$ verify that $(AB)^T = B^T A^T$

Show steps (6)

$AB = \begin{pmatrix} 5 & 2 & 9 \\ 1 & 2 & 8 \end{pmatrix}\begin{pmatrix} 1 & 7 \\ 1 & 2 \\ 5 & -1 \end{pmatrix} = \begin{pmatrix} 5+2+45 & 35+4-9 \\ 1+2+40 & 7+4-8 \end{pmatrix}$
$AB = \begin{pmatrix} 52 & 30 \\ 43 & 3 \end{pmatrix}$
LHS $= (AB)^T = \begin{pmatrix} 52 & 43 \\ 30 & 3 \end{pmatrix}$ … (1)
$B^T = \begin{pmatrix} 1 & 1 & 5 \\ 7 & 2 & -1 \end{pmatrix}$ , $A^T = \begin{pmatrix} 5 & 1 \\ 2 & 2 \\ 9 & 8 \end{pmatrix}$
RHS $= B^TA^T = \begin{pmatrix} 5+2+45 & 1+2+40 \\ 35+4-9 & 7+4-8 \end{pmatrix} = \begin{pmatrix} 52 & 43 \\ 30 & 3 \end{pmatrix}$ … (2)
From (1) and (2), $(AB)^T = B^TA^T$ .

Answer

$(AB)^T = B^TA^T = \begin{pmatrix} 52 & 43 \\ 30 & 3 \end{pmatrix}$ , verified

Q13. If $A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}$ show that $A^2 - 5A + 7I_2 = 0$

$A^2 = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{pmatrix} = \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix}$
$5A = \begin{pmatrix} 15 & 5 \\ -5 & 10 \end{pmatrix}$ , $7I_2 = \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix}$
$A^2 - 5A + 7I_2 = \begin{pmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{pmatrix}$
$= \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = 0$

Answer

$A^2 - 5A + 7I_2 = 0$ , hence shown

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

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Q1. Find the order of the product matrix ABABAB if

(i)

(ii)

(iii)

(iv)

(v)

Q2. If AAA is of order p×qp \times qp×q and BBB is of order q×rq \times rq×r …

Q3. AAA has 'aaa' rows and 'a+3a + 3a+3' columns. BBB has 'bbb' …

Q4. If A=(2543)A = \begin{pmatrix} 2 & 5 \\ 4 & 3 \end{pmatrix}A=(24​53​) , B=(1−325)B = \begin{pmatrix} 1 & -3 \\ 2 & 5 \end{pmatrix}B=(12​−35​) find AB, BAAB,\ BAAB, BA and verify AB=BAAB = BAAB=BA ?

Q6. Show that the matrices A=(1231)A = \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}A=(13​21​) , B=(1−2−31)B = \begin{pmatrix} 1 & -2 \\ -3 & 1 \end{pmatrix}B=(1−3​−21​) satisfy commutative property AB=BAAB = BAAB=BA

Q7. Let A=(1213)A = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}A=(11​23​) , B=(4015)B = \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix}B=(41​05​) , C=(2012)C = \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix}C=(21​02​) Show that

(i)

(ii)

(iii)

Q8. If A=(cos⁡θ00cos⁡θ)A = \begin{pmatrix} \cos\theta & 0 \\ 0 & \cos\theta \end{pmatrix}A=(cosθ0​0cosθ​) , B=(sin⁡θ00sin⁡θ)B = \begin{pmatrix} \sin\theta & 0 \\ 0 & \sin\theta \end{pmatrix}B=(sinθ0​0sinθ​) then show that A2+B2=IA^2 + B^2 = IA2+B2=I .

Q9. If A=(cos⁡θsin⁡θ−sin⁡θcos⁡θ)A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}A=(cosθ−sinθ​sinθcosθ​) prove that AAT=IAA^T = IAAT=I .

Q10. Verify that A2=IA^2 = IA2=I when A=(5−46−5)A = \begin{pmatrix} 5 & -4 \\ 6 & -5 \end{pmatrix}A=(56​−4−5​)

Q11. If A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}A=(ac​bd​) and I=(1001)I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}I=(10​01​) show that A2−(a+d)A=(bc−ad)I2A^2 - (a + d)A = (bc - ad)I_2A2−(a+d)A=(bc−ad)I2​

Q12. If A=(529128)A = \begin{pmatrix} 5 & 2 & 9 \\ 1 & 2 & 8 \end{pmatrix}A=(51​22​98​) , B=(17125−1)B = \begin{pmatrix} 1 & 7 \\ 1 & 2 \\ 5 & -1 \end{pmatrix}B=​115​72−1​​ verify that (AB)T=BTAT(AB)^T = B^T A^T(AB)T=BTAT

Q13. If A=(31−12)A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}A=(3−1​12​) show that A2−5A+7I2=0A^2 - 5A + 7I_2 = 0A2−5A+7I2​=0

Quick check

Q1. Find the order of the product matrix $AB$ if

Q2. If $A$ is of order $p \times q$ and $B$ is of order $q \times r$ …

Q3. $A$ has ' $a$ ' rows and ' $a + 3$ ' columns. $B$ has ' $b$ ' …

Q4. If $A = \begin{pmatrix} 2 & 5 \\ 4 & 3 \end{pmatrix}$ , $B = \begin{pmatrix} 1 & -3 \\ 2 & 5 \end{pmatrix}$ find $AB,\ BA$ and verify $AB = BA$ ?

Q6. Show that the matrices $A = \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}$ , $B = \begin{pmatrix} 1 & -2 \\ -3 & 1 \end{pmatrix}$ satisfy commutative property $AB = BA$

Q7. Let $A = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}$ , $B = \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix}$ , $C = \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix}$ Show that

Q8. If $A = \begin{pmatrix} \cos\theta & 0 \\ 0 & \cos\theta \end{pmatrix}$ , $B = \begin{pmatrix} \sin\theta & 0 \\ 0 & \sin\theta \end{pmatrix}$ then show that $A^2 + B^2 = I$ .

Q9. If $A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$ prove that $AA^T = I$ .

Q10. Verify that $A^2 = I$ when $A = \begin{pmatrix} 5 & -4 \\ 6 & -5 \end{pmatrix}$

Q11. If $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ show that $A^2 - (a + d)A = (bc - ad)I_2$

Q12. If $A = \begin{pmatrix} 5 & 2 & 9 \\ 1 & 2 & 8 \end{pmatrix}$ , $B = \begin{pmatrix} 1 & 7 \\ 1 & 2 \\ 5 & -1 \end{pmatrix}$ verify that $(AB)^T = B^T A^T$

Q13. If $A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}$ show that $A^2 - 5A + 7I_2 = 0$