12th NCERT/CBSE Matrices Exercise 3.1 Questions 10

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Question (1)

In the matrix $A = \left[ {\begin{array}{*{20}{c}}2&5&{19}&{ - 7}\\{35}&{ - 2}&{\frac{5}{2}}&{12}\\{\sqrt 3 }&1&{ - 5}&{17}\end{array}} \right]$, write
(i) The order of the matrix
(ii) The number of elements
(iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃,

Solution

(i) Order of matrix is 3×4
(ii) The number of elements = 12
(iii) a₁₃ = 19
a₂₁ = +35
, a₃₃ = -5
a₂₄ = 12
, a₂₃= ${\frac{5}{2}}$

Question (2)

If a matrix has 24 elements, what are the possible order it can have? What, if it has 13 elemts?

Solution

The possible order are 24×1, 12×2, 3×8, 4×6, 6×4, 8×3, 12×2, 24×1
If there are 13 elements orders are 1×13, 13×1

Question (3)

If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

Solution

Matrix has 18 elemnts, possible orders are 1×18, 2×9, 3×6, 6×3, 9×2, 18×1

Question (4)

Construct a 2×2 matrix, A = [a_ij], whose elements are given by:
(i) ${a_{ij}} = \frac{{{{\left( {i + j} \right)}^2}}}{2}$
(ii) ${a_{ij}} = \frac{i}{j}$
(iii) ${a_{ij}} = \frac{{{{\left( {i + 2j} \right)}^2}}}{2}$

Solution

construct 2×2 , A = [a_ij]
(i) $[{a_{ij}} = \frac{{{{\left( {i + j} \right)}^2}}}{2}$
${a_{11}} = \frac{{{{\left( {1 + 1} \right)}^2}}}{2} = \frac{4}{2} = 2$
${a_{21}} = \frac{{{{\left( {2 + 1} \right)}^2}}}{2} = \frac{9}{2}$
${a_{12}} = \frac{{{{\left( {1 + 2} \right)}^2}}}{2} = \frac{9}{2}$
${a_{22}} = \frac{{{{\left( {2 + 2} \right)}^2}}}{2} = 8$ \[A = \left[ {\begin{array}{*{20}{c}}2&{\frac{9}{2}}\\{\frac{9}{2}}&8\end{array}} \right]\]
(ii) ${a_{ij}} = \frac{i}{j}$
${a_{11}} = \frac{1}{1} = 1$,    ${a_{12}} = \frac{1}{2}$     ${a_{21}} = \frac{2}{1} = 2$    ${a_{22}} = \frac{2}{2} = 1$
$A = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\2&1\end{array}} \right]$
(iii) ${a_{ij}} = \frac{{{{\left( {i + 2j} \right)}^2}}}{2}$
${a_{11}} = \frac{{{{\left( {1 + 2} \right)}^2}}}{2} = \frac{9}{2}$
${a_{12}} = \frac{{{{\left( {1 + 4} \right)}^2}}}{2} = \frac{{25}}{2}$
${a_{21}} = \frac{{{{\left( {2 + 2} \right)}^2}}}{2} = 8$
${a_{22}} = \frac{{{{\left( {2 + 4} \right)}^2}}}{2} = 18$
$A = \left[ {\begin{array}{*{20}{c}}{\frac{9}{2}}&{\frac{{25}}{2}}\\8&{18}\end{array}} \right]$

Question (5)

Construct a 3 × 4 matrix, whose elements are given by:
(i) ${a_{ij}} = \frac{1}{2}\left| { - 3i + j} \right|$
(ii) ${a_{ij}} = 2i - j$

Solution

(i) ${a_{ij}} = \frac{1}{2}\left| { - 3i + j} \right|$ ${a_{11}} = \frac{1}{2}\left| { - 3 + 1} \right| = 1$; ${a_{12}} = \frac{1}{2}\left| { - 3 + 2} \right| = \frac{1}{2}$; ${a_{13}} = \frac{1}{2}\left| { - 3 + 3} \right| = 0$; ${a_{14}} = \frac{1}{2}\left| { - 3 + 4} \right| = \frac{1}{2}$
${a_{21}} = \frac{1}{2}\left| { - 6 + 1} \right| = \frac{5}{2}$; ${a_{22}} = \frac{1}{2}\left| { - 6 + 2} \right| = 2$; ${a_{23}} = \frac{1}{2}\left| { - 6 + 3} \right| = \frac{3}{2}$; ${a_{24}} = \frac{1}{2}\left| { - 6 + 4} \right| = 1$
${a_{31}} = \frac{1}{2}\left| { - 9 + 1} \right| = 4$; ${a_{32}} = \frac{1}{2}\left| { - 9 + 2} \right| = \frac{7}{2}$; ${a_{33}} = \frac{1}{2}\left| { - 9 + 3} \right| = 3$; ${a_{34}} = \frac{1}{2}\left| { - 9 + 4} \right| = \frac{5}{2}$
$A = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{2}}&0&{\frac{1}{2}}\\{\frac{5}{2}}&2&{\frac{3}{2}}&1\\4&{\frac{7}{2}}&3&{\frac{5}{2}}\end{array}} \right]$
(ii) ${a_{ij}} = 2i - j$
a₁₂=1, a₁₂=2-2=0, a₁₃= 2-3 = -1, a₁₄=2-4 =-2
a₂₁=4-1=3,  a₂₂=4-2=2,  a₂₃= 4-3=1,  a₂₄= 4-4 = 0
a₃₁= 6-1 =5,  a₃₂=6-2=4, a₃₃= 6-3=3,  a₃₄=2
$A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&{ - 2}\\3&2&1&0\\5&4&3&2\end{array}} \right]$

Question (6)

Find the values of x, y and z from the following equations
(i) $\left[ {\begin{array}{*{20}{c}}4&3\\x&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}y&z\\1&5\end{array}} \right]$
(ii) $\left[ {\begin{array}{*{20}{c}}{x + y}&2\\{5 + z}&{xy}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}6&2\\5&8\end{array}} \right]$
(iii) $\left[ {\begin{array}{*{20}{c}}{x + y + z}\\{x + z}\\{y + z}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\5\\7\end{array}} \right]$

Solution

(i) $\left[ {\begin{array}{*{20}{c}}4&3\\x&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}y&z\\1&5\end{array}} \right]$
⇒ y = 4, z =3, x = 1

(ii) $\left[ {\begin{array}{*{20}{c}}{x + y}&2\\{5 + z}&{xy}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}6&2\\5&8\end{array}} \right]$
⇒ x + y = 6 ∴ y = 6 - x
5 +z = 5 ∴ z = 0
xy = 8
Substituting value of y
x(6-x) = 8
6x - x² = 8
x² -6x + 8 = 0
(x-4)(x-2) = 0
x = 4 or x = 2
If x = 4, y = 2
If x = 2 , y = 4

(iii) $\left[ {\begin{array}{*{20}{c}}{x + y + z}\\{x + z}\\{y + z}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\5\\7\end{array}} \right]$
⇒ x + y + z = 9 ---(1)
x+z = 5 ---(2)
y+z = 7 ---(3)
Replace value of y+z in (1)
x+7=9 ⇒ x = 2
Replace x = 2 in (2)
2 + z = 5
z = 3
Replace z = 3 in (3)
y + 3 = 7
y = 4
∴ x = 2, y = 4, z = 3

Question (7)

Find the value of a, b, c and d from the equation:
$\left[ {\begin{array}{*{20}{c}}{a - b}&{2a + c}\\{2a - b}&{3c + d}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}&5\\0&{13}\end{array}} \right]$

Solution

$\left[ {\begin{array}{*{20}{c}}{a - b}&{2a + c}\\{2a - b}&{3c + d}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}&5\\0&{13}\end{array}} \right]$
⇒ a - b = -1 ---(1)
2a + c = 5 --- (2)
2a-b=0 ---(3) 3c+d = 13 ---(4)
From (1) -(3) a-b-2a+b = -1-0
∴ a = 1
Replace a = 1 in (1)
1-b= -1 ⇒ b = 2
Replace a in (2)
2 +c = 5 ⇒ c = 3
Replace c = 3 in (4)
9 +d = 13
d = 4
∴ a =1, b = 2, c =3 and d=4

Question (8)

A = [a_ij]_m×n is square natrix if
(A) m < n    (B) m > n    (C) m = n    (D) None of these

Solution

A = [a_ij]_m×n is square natrix if m = n
So correct answr is option (C)

Question (9)

Which of the given values of x and y makes the following pair of matries equal
\[\left[ {\begin{array}{*{20}{c}}{3x + 7}&5\\{y + 1}&{2 - 3x}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}0&{y - 2}\\8&4\end{array}} \right]\] (A) $x = \frac{{ - 1}}{3}$     (B) Not possible to find
(C) y=7, $x = \frac{{ - 2}}{3}$     (d) $x = \frac{{ - 1}}{3}$, $y = \frac{{ - 2}}{3}$

Solution

$\left[ {\begin{array}{*{20}{c}}{3x + 7}&5\\{y + 1}&{2 - 3x}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}0&{y - 2}\\8&4\end{array}} \right]$
⇒ 3x+7= 0 ∴ $x = \frac{{ - 7}}{3}$
5 = y - 2 ∴ y = 7
y+1=8 ∴ y = 7
2 - 3x = 4 ∴ $x = \frac{{ - 2}}{3}$
We get two different values of x and y =7
So not possible value , Option "B" is correct answer

Question (10)

The number of possible matrices of prder 3 × 3 with each entry 0 or 1 is
(A) 27    (B) 18    (C) 81    (D) 512

Solution

Number of elements = 3×3= 9
Possible matrix = 2⁹ = 512
∴ Option D is correct answer