Search

Samacheer Kalvi 10th Maths Exercise 3.17 Solutions

10th Maths Exercise 3.17 solutions (Chapter 3 Algebra) with step-by-step answers for all 12 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance11 answers · full steps below
  1. Q1 (i)1616 elements
  2. Q1 (ii)Order of AA is 4×44\times 4
  3. Q1 (iii)a22=7,  a23=32,  a24=5,  a34=0,  a43=−11,  a44=1a_{22}=\sqrt{7},\; a_{23}=\dfrac{\sqrt{3}}{2},\; a_{24}=5,\; a_{34}=0,\; a_{43}=-11,\; a_{44}=1
  4. Q218 elements: 1×18, 18×1, 2×9, 9×2, 3×6, 6×31\times18,\,18\times1,\,2\times9,\,9\times2,\,3\times6,\,6\times3; 6 elements: 1×6, 6×1, 2×3, 3×21\times6,\,6\times1,\,2\times3,\,3\times2
  5. Q3 (i)A=(135024113)A=\begin{pmatrix}1&3&5\\0&2&4\\1&1&3\end{pmatrix}
  6. Q3 (ii)A=(83964396431253643125372)A=\begin{pmatrix}\dfrac{8}{3}&9&\dfrac{64}{3}\\[4pt]9&\dfrac{64}{3}&\dfrac{125}{3}\\[4pt]\dfrac{64}{3}&\dfrac{125}{3}&72\end{pmatrix}
  7. Q4AT=(5134−78392)A^{T}=\begin{pmatrix}5&1&3\\4&-7&8\\3&9&2\end{pmatrix}
  8. Q5(−A)T=(−75−33−25)(-A)^{T}=\begin{pmatrix}-\sqrt{7}&\sqrt{5}&-\sqrt{3}\\ 3&-2&5\end{pmatrix}
  9. Q7 (i)x=3,  y=12,  z=3x=3,\; y=12,\; z=3
  10. Q7 (ii)z=0z=0 and x=2, y=4x=2,\,y=4 (or) x=4, y=2x=4,\,y=2
  11. Q7 (iii)x=2,  y=4,  z=3x=2,\; y=4,\; z=3

Q1. In the matrix A=(8943−17325143068−111)A = \begin{pmatrix} 8 & 9 & 4 & 3 \\ -1 & \sqrt{7} & \dfrac{\sqrt{3}}{2} & 5 \\ 1 & 4 & 3 & 0 \\ 6 & 8 & -11 & 1 \end{pmatrix} , write

In the matrix A=(8943−17325143068−111)A = \begin{pmatrix} 8 & 9 & 4 & 3 \\ -1 & \sqrt{7} & \dfrac{\sqrt{3}}{2} & 5 \\ 1 & 4 & 3 & 0 \\ 6 & 8 & -11 & 1 \end{pmatrix}, write

(i)

The number of elements

  1. The matrix has 44 rows and 44 columns.

  2. Number of elements =4×4=16=4\times 4=16.

Answer

1616 elements

(ii)

The order of the matrix

  1. Order == (number of rows) ×\times (number of columns).

  2. There are 44 rows and 44 columns.

Answer

Order of AA is 4×44\times 4

(iii)

Write the elements a22,a23,a24,a34,a43,a44a_{22}, a_{23}, a_{24}, a_{34}, a_{43}, a_{44}.

Show steps (7)

  1. aija_{ij} is the element in the ithi^{th} row and jthj^{th} column.

  2. a22a_{22} = 2nd row, 2nd column =7=\sqrt{7}

  3. a23a_{23} = 2nd row, 3rd column =32=\dfrac{\sqrt{3}}{2}

  4. a24a_{24} = 2nd row, 4th column =5=5

  5. a34a_{34} = 3rd row, 4th column =0=0

  6. a43a_{43} = 4th row, 3rd column =−11=-11

  7. a44a_{44} = 4th row, 4th column =1=1

Answer

a22=7,  a23=32,  a24=5,  a34=0,  a43=−11,  a44=1a_{22}=\sqrt{7},\; a_{23}=\dfrac{\sqrt{3}}{2},\; a_{24}=5,\; a_{34}=0,\; a_{43}=-11,\; a_{44}=1

Q2. If a matrix has 1818 elements, what are the possible orders it …

If a matrix has 1818 elements, what are the possible orders it can have? What if it has 66 elements?

Show steps (6)

  1. A matrix of order m×nm\times n has mnmn elements.

  2. For 1818 elements, find ordered pairs with product 1818:

  3. (1,18),(18,1),(2,9),(9,2),(3,6),(6,3)(1,18),(18,1),(2,9),(9,2),(3,6),(6,3)

  4. So the orders are 1×18,  18×1,  2×9,  9×2,  3×6,  6×31\times18,\;18\times1,\;2\times9,\;9\times2,\;3\times6,\;6\times3 (6 orders).

  5. For 66 elements, ordered pairs with product 66:

  6. (1,6),(6,1),(2,3),(3,2)(1,6),(6,1),(2,3),(3,2)

Answer

18 elements: 1×18, 18×1, 2×9, 9×2, 3×6, 6×31\times18,\,18\times1,\,2\times9,\,9\times2,\,3\times6,\,6\times3; 6 elements: 1×6, 6×1, 2×3, 3×21\times6,\,6\times1,\,2\times3,\,3\times2

Q3. Construct a 3×33 \times 3 matrix whose elements are given by

(i)

aij=∣i−2j∣a_{ij} = |i - 2j|

  1. aij=∣i−2j∣a_{ij}=|i-2j|, with i,j=1,2,3i,j=1,2,3.

  2. a11=∣1−2∣=1,a12=∣1−4∣=3,a13=∣1−6∣=5a_{11}=|1-2|=1,\quad a_{12}=|1-4|=3,\quad a_{13}=|1-6|=5

  3. a21=∣2−2∣=0,a22=∣2−4∣=2,a23=∣2−6∣=4a_{21}=|2-2|=0,\quad a_{22}=|2-4|=2,\quad a_{23}=|2-6|=4

  4. a31=∣3−2∣=1,a32=∣3−4∣=1,a33=∣3−6∣=3a_{31}=|3-2|=1,\quad a_{32}=|3-4|=1,\quad a_{33}=|3-6|=3

Answer

A=(135024113)A=\begin{pmatrix}1&3&5\\0&2&4\\1&1&3\end{pmatrix}

(ii)

aij=(i+j)33a_{ij} = \dfrac{(i+j)^3}{3}

  1. aij=(i+j)33a_{ij}=\dfrac{(i+j)^3}{3}, with i,j=1,2,3i,j=1,2,3.

  2. a11=233=83,a12=333=9,a13=433=643a_{11}=\dfrac{2^3}{3}=\dfrac{8}{3},\quad a_{12}=\dfrac{3^3}{3}=9,\quad a_{13}=\dfrac{4^3}{3}=\dfrac{64}{3}

  3. a21=333=9,a22=433=643,a23=533=1253a_{21}=\dfrac{3^3}{3}=9,\quad a_{22}=\dfrac{4^3}{3}=\dfrac{64}{3},\quad a_{23}=\dfrac{5^3}{3}=\dfrac{125}{3}

  4. a31=433=643,a32=533=1253,a33=633=72a_{31}=\dfrac{4^3}{3}=\dfrac{64}{3},\quad a_{32}=\dfrac{5^3}{3}=\dfrac{125}{3},\quad a_{33}=\dfrac{6^3}{3}=72

Answer

A=(83964396431253643125372)A=\begin{pmatrix}\dfrac{8}{3}&9&\dfrac{64}{3}\\[4pt]9&\dfrac{64}{3}&\dfrac{125}{3}\\[4pt]\dfrac{64}{3}&\dfrac{125}{3}&72\end{pmatrix}

Q4. If A=(5431−79382)A = \begin{pmatrix} 5 & 4 & 3 \\ 1 & -7 & 9 \\ 3 & 8 & 2 \end{pmatrix} then find the transpose of AA .

  1. Transpose is got by interchanging rows and columns.

  2. Row 1 (5,4,3)(5,4,3) becomes column 1.

  3. Row 2 (1,−7,9)(1,-7,9) becomes column 2.

  4. Row 3 (3,8,2)(3,8,2) becomes column 3.

Answer

AT=(5134−78392)A^{T}=\begin{pmatrix}5&1&3\\4&-7&8\\3&9&2\end{pmatrix}

Q5. If A=(7−3−523−5)A = \begin{pmatrix} \sqrt{7} & -3 \\ -\sqrt{5} & 2 \\ \sqrt{3} & -5 \end{pmatrix} then find the transpose of −A-A .

  1. First write −A-A by changing the sign of each element.

  2. −A=(−735−2−35)-A=\begin{pmatrix}-\sqrt{7}&3\\ \sqrt{5}&-2\\ -\sqrt{3}&5\end{pmatrix}

  3. Now interchange rows and columns of −A-A.

Answer

(−A)T=(−75−33−25)(-A)^{T}=\begin{pmatrix}-\sqrt{7}&\sqrt{5}&-\sqrt{3}\\ 3&-2&5\end{pmatrix}

Q6. If A=(522−170.752831)A = \begin{pmatrix} 5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \dfrac{5}{2} \\ 8 & 3 & 1 \end{pmatrix} then verify (AT)T=A(A^{T})^{T} = A

  1. Interchange rows and columns of AA:

  2. AT=(5−17820.732521)A^{T}=\begin{pmatrix}5&-\sqrt{17}&8\\ 2&0.7&3\\ 2&\dfrac{5}{2}&1\end{pmatrix}

  3. Now transpose ATA^{T} again (interchange its rows and columns):

  4. (AT)T=(522−170.752831)=A(A^{T})^{T}=\begin{pmatrix}5&2&2\\ -\sqrt{17}&0.7&\dfrac{5}{2}\\ 8&3&1\end{pmatrix}=A

Answer

(AT)T=A(A^{T})^{T}=A is verified

Q7. Find the values of xx , yy and zz from the following …

Find the values of xx, yy and zz from the following equations

(i)

(123x53)=(yz35)\begin{pmatrix} 12 & 3 \\ x & \dfrac{5}{3} \end{pmatrix} = \begin{pmatrix} y & z \\ 3 & 5 \end{pmatrix}

  1. Equal matrices have equal corresponding elements.

  2. y=12y=12

  3. z=3z=3

  4. x=3x=3

Answer

x=3,  y=12,  z=3x=3,\; y=12,\; z=3

(ii)

(x+y25+zxy)=(6258)\begin{pmatrix} x+y & 2 \\ 5+z & xy \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 5 & 8 \end{pmatrix}

Show steps (7)

  1. Equate corresponding elements:

  2. x+y=6x+y=6 … (1)

  3. 5+z=5⇒z=05+z=5\Rightarrow z=0

  4. xy=8xy=8 … (2)

  5. From (1), y=6−xy=6-x. Substitute in (2): x(6−x)=8x(6-x)=8

  6. x2−6x+8=0⇒(x−2)(x−4)=0x^{2}-6x+8=0\Rightarrow (x-2)(x-4)=0

  7. x=2x=2 or x=4x=4; correspondingly y=4y=4 or y=2y=2.

Answer

z=0z=0 and x=2, y=4x=2,\,y=4 (or) x=4, y=2x=4,\,y=2

(iii)

(x+y+zx+zy+z)=(957)\begin{pmatrix} x+y+z \\ x+z \\ y+z \end{pmatrix} = \begin{pmatrix} 9 \\ 5 \\ 7 \end{pmatrix}

Show steps (7)

  1. Equate corresponding elements:

  2. x+y+z=9x+y+z=9 … (1)

  3. x+z=5x+z=5 … (2)

  4. y+z=7y+z=7 … (3)

  5. (1) −- (2): y=9−5=4y=9-5=4

  6. From (3): 4+z=7⇒z=34+z=7\Rightarrow z=3

  7. From (2): x+3=5⇒x=2x+3=5\Rightarrow x=2

Answer

x=2,  y=4,  z=3x=2,\; y=4,\; z=3

Quick check

  1. Find the matrix XX if 2X+(1357)=(5795)2X + \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix}

    1. (1) (−2−22−1)\begin{pmatrix} -2 & -2 \\ 2 & -1 \end{pmatrix}
    2. (2) (222−1)\begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}
    3. (3) (1222)\begin{pmatrix} 1 & 2 \\ 2 & 2 \end{pmatrix}
    4. (4) (2122)\begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix}
    Show answer

    Answer: (2) (222−1)\begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}

    1. 2X=(5795)−(1357)=(444−2)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=12(444−2)=(222−1)X = \dfrac{1}{2}\begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}

    Question 1 of 5

  2. The values of aa and bb if 4x4−24x3+76x2+ax+b4x^4 - 24x^3 + 76x^2 + ax + b is a perfect square are

    1. (1) 100,120100,120
    2. (2) 10,1210,12
    3. (3) −120,100-120,100
    4. (4) 12,1012,10
    Show answer

    Answer: (3) −120,100-120,100

    1. Let 4x4−24x3+76x2+ax+b=(2x2−6x+c)24x^4 - 24x^3 + 76x^2 + ax + b = (2x^2 - 6x + c)^2.

    2. Expanding: 4x4−24x3+(36+4c)x2−12cx+c24x^4 - 24x^3 + (36 + 4c)x^2 - 12cx + c^2.

    3. Compare x2x^2 terms: 36+4c=76⇒c=1036 + 4c = 76 \Rightarrow c = 10.

    4. So a=−12(10)=−120a = -12(10) = -120 and b=102=100b = 10^2 = 100.

    Question 2 of 5

  3. The solution of the system x+y−3z=−6x + y - 3z = -6, −7y+7z=7-7y + 7z = 7, 3z=93z = 9 is

    1. (1) x=1, y=2, z=3x = 1,\ y = 2,\ z = 3
    2. (2) x=−1, y=2, z=3x = -1,\ y = 2,\ z = 3
    3. (3) x=−1, y=−2, z=3x = -1,\ y = -2,\ z = 3
    4. (4) x=1, y=−2, z=3x = 1,\ y = -2,\ z = 3
    Show answer

    Answer: (1) x=1, y=2, z=3x = 1,\ y = 2,\ z = 3

    1. From 3z=93z = 9, z=3z = 3.

    2. Put z=3z=3 in −7y+7z=7-7y + 7z = 7: −7y+21=7⇒y=2-7y + 21 = 7 \Rightarrow y = 2.

    3. Put y=2, z=3y=2,\ z=3 in x+y−3z=−6x + y - 3z = -6: x+2−9=−6⇒x=1x + 2 - 9 = -6 \Rightarrow x = 1.

    Question 3 of 5

  4. If AA is a 2×32 \times 3 matrix and BB is a 3×43 \times 4 matrix, how many columns does ABAB have

    1. (1) 33
    2. (2) 44
    3. (3) 22
    4. (4) 55
    Show answer

    Answer: (2) 44

    1. A2×3B3×4A_{2\times 3} B_{3\times 4} gives order 2×42 \times 4.

    2. So ABAB has 44 columns.

    Question 4 of 5

  5. The square root of 256x8y4z1025x6y6z6\dfrac{256 x^8 y^4 z^{10}}{25 x^6 y^6 z^6} is equal to

    1. (1) 165∣x2z4y2∣\dfrac{16}{5}\left|\dfrac{x^2 z^4}{y^2}\right|
    2. (2) 16∣y2x2z4∣16\left|\dfrac{y^2}{x^2 z^4}\right|
    3. (3) 165∣yxz2∣\dfrac{16}{5}\left|\dfrac{y}{xz^2}\right|
    4. (4) 165∣xz2y∣\dfrac{16}{5}\left|\dfrac{xz^2}{y}\right|
    Show answer

    Answer: (4) 165∣xz2y∣\dfrac{16}{5}\left|\dfrac{xz^2}{y}\right|

    1. 256x8y4z1025x6y6z6=256 x2z425 y2\dfrac{256x^8y^4z^{10}}{25x^6y^6z^6} = \dfrac{256\,x^2 z^4}{25\,y^2}

    2. Taking square root: 165∣xz2y∣\dfrac{16}{5}\left|\dfrac{x z^2}{y}\right|

    Question 5 of 5
Chapter 3 formulas
  • Roots of a quadratic equation

    ax2+bx+c=0,  (a≠0)  ⇒  x=−b±b2−4ac2aax^2 + bx + c = 0,\; (a \neq 0) \;\Rightarrow\; x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

  • Sum of the roots

    α+β=−ba=−Co-efficient of xCo-efficient of x2\alpha + \beta = \dfrac{-b}{a} = \dfrac{-\text{Co-efficient of } x}{\text{Co-efficient of } x^2}

  • Product of the roots

    αβ=ca=Constant termCo-efficient of x2\alpha\beta = \dfrac{c}{a} = \dfrac{\text{Constant term}}{\text{Co-efficient of } x^2}

  • Quadratic equation formed from its roots

    x2−(α+β)x+αβ=0x^2 - (\alpha + \beta)x + \alpha\beta = 0

  • Discriminant

    Δ=b2−4ac\Delta = b^2 - 4ac

  • Nature of roots: real and unequal

    Δ>0  ⇒  the roots are real and unequal\Delta > 0 \;\Rightarrow\; \text{the roots are real and unequal}

  • Nature of roots: real and equal

    Δ=0  ⇒  the roots are real and equal\Delta = 0 \;\Rightarrow\; \text{the roots are real and equal}

  • Nature of roots: no real roots

    Δ<0  ⇒  there are no real roots\Delta < 0 \;\Rightarrow\; \text{there are no real roots}

  • Cases for a system of linear equations in three variables

    (i) Unique solution(ii) Infinitely many solutions(iii) No solution\text{(i) Unique solution}\quad \text{(ii) Infinitely many solutions}\quad \text{(iii) No solution}

  • LCM of algebraic expressions

    LCM=expression of lowest degree that each given expression exactly divides\text{LCM} = \text{expression of lowest degree that each given expression exactly divides}

  • Quadratic polynomial and its zeroes

    p(x)=ax2+bx+c  (a≠0)  ⇒  p(x) has atmost two zeroes, which lie on the x-axisp(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

    Order of A=(Number of rows)×(Number of columns)=m×n\text{Order of } A = (\text{Number of rows}) \times (\text{Number of columns}) = m \times n

  • Row matrix (row vector)

    A=(a11a12⋯a1n)1×nA = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \end{pmatrix}_{1 \times n}

  • Column matrix (column vector)

    A=(a11a21⋮am1)m×1A = \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{pmatrix}_{m \times 1}

  • Square matrix

    Number of rows=Number of columns,  order n×n\text{Number of rows} = \text{Number of columns},\; \text{order } n \times n

  • Zero matrix (null matrix)

    aij=0 for all i, ja_{ij} = 0 \text{ for all } i,\, j

  • Transpose of a matrix

    (AT)ij=aji(rows and columns of A interchanged)\left(A^{T}\right)_{ij} = a_{ji} \quad (\text{rows and columns of } A \text{ interchanged})

  • Diagonal matrix

    aij=0 for i≠j(square matrix)a_{ij} = 0 \text{ for } i \neq j \quad (\text{square matrix})

  • Scalar matrix

    aij={k,i=j0,i≠ja_{ij} = \begin{cases} k, & i = j \\ 0, & i \neq j \end{cases}

  • Identity matrix (unit matrix)

    aij={1,i=j0,i≠ja_{ij} = \begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases}

  • Lower triangular matrix

    aij=0 for i<j(all entries above the leading diagonal are zero)a_{ij} = 0 \text{ for } i < j \quad (\text{all entries above the leading diagonal are zero})

  • Upper triangular matrix

    aij=0 for i>j(all entries below the leading diagonal are zero)a_{ij} = 0 \text{ for } i > j \quad (\text{all entries below the leading diagonal are zero})

  • Equality of matrices

    A=B  ⟺  same order and aij=bij for all i, jA = B \iff \text{same order and } a_{ij} = b_{ij} \text{ for all } i,\, j

Answers checked against the Samacheer Kalvi textbook · Updated

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