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NCERT Solutions · Class 12 Mathematics Matrices

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EXERCISE 3.1 1–10 (part 1 of 6)

  1. Exercise 1

    In the matrix \(\displaystyle \mathrm{A}=\left[\begin{array}{cccc}2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17\end{array}\right]\), write:
    (i)
    The order of the matrix,
    (ii)
    The number of elements,
    (iii)
    Write the elements \(\displaystyle a_{13}, a_{21}, a_{33}, a_{24}, a_{23}\).

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    NCERT’s answer
    (i)
    $\displaystyle 3$ × $\displaystyle 4$ (ii) $\displaystyle 12$ (iii) \(\displaystyle 19,35,-5,12, \frac{5}{2}\)
    Use the definition of the order of a matrix: a matrix with \(\displaystyle m\) horizontal rows and \(\displaystyle n\) vertical columns is said to be of order \(\displaystyle m \times n\), written rows first, columns second.
    (i)
    Count the rows and the columns of \(\displaystyle \mathrm{A}\). There are \(\displaystyle 3\) rows, and each row lists \(\displaystyle 4\) entries, so there are \(\displaystyle 4\) columns. Hence
    \[\text{order of }\mathrm{A}=3\times 4.\]
    (ii)
    Every position \(\displaystyle (i,j)\) with \(\displaystyle 1\le i\le 3\), \(\displaystyle 1\le j\le 4\) carries exactly one element, so the number of elements is the product of the two dimensions:
    \[3\times 4=12.\]
    (iii)
    By definition \(\displaystyle a_{ij}\) is the entry standing in the \(\displaystyle i\)-th ROW and the \(\displaystyle j\)-th COLUMN — the first subscript is the row. Reading \(\displaystyle \mathrm{A}\) with that convention:
    \(\displaystyle a_{13}\) — row \(\displaystyle 1\), column \(\displaystyle 3\): \(\displaystyle a_{13}=19\).
    \(\displaystyle a_{21}\) — row \(\displaystyle 2\), column \(\displaystyle 1\): \(\displaystyle a_{21}=35\).
    \(\displaystyle a_{33}\) — row \(\displaystyle 3\), column \(\displaystyle 3\): \(\displaystyle a_{33}=-5\).
    \(\displaystyle a_{24}\) — row \(\displaystyle 2\), column \(\displaystyle 4\): \(\displaystyle a_{24}=12\).
    \(\displaystyle a_{23}\) — row \(\displaystyle 2\), column \(\displaystyle 3\): \(\displaystyle a_{23}=\dfrac{5}{2}\).
    Final answer: order \(\displaystyle 3\times 4\); \(\displaystyle 12\) elements; \(\displaystyle a_{13}=19,\ a_{21}=35,\ a_{33}=-5,\ a_{24}=12,\ a_{23}=\dfrac{5}{2}\).

  2. Exercise 2

    If a matrix has $\displaystyle 24$ elements, what are the possible orders it can have? What, if it has $\displaystyle 13$ elements?

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    NCERT’s answer
    $\displaystyle 1$ × $\displaystyle 24$, $\displaystyle 2$ × $\displaystyle 12$, $\displaystyle 3$ × $\displaystyle 8$, $\displaystyle 4$ × $\displaystyle 6$, $\displaystyle 6$ × $\displaystyle 4$, $\displaystyle 8$ × $\displaystyle 3$, $\displaystyle 12$ × $\displaystyle 2$, $\displaystyle 24$ × $\displaystyle 1$ ; $\displaystyle 1$ × $\displaystyle 13$ , $\displaystyle 13$ × $\displaystyle 1$
    A matrix of order \(\displaystyle m\times n\) has exactly \(\displaystyle mn\) elements, where \(\displaystyle m\) and \(\displaystyle n\) are positive integers. So finding the possible orders means finding all ORDERED pairs \(\displaystyle (m,n)\) of positive integers whose product is the given number — ordered, because \(\displaystyle 2\times 12\) and \(\displaystyle 12\times 2\) are different matrices.$\displaystyle 24$ elements. We need \(\displaystyle mn=24\). Listing the factor pairs of \(\displaystyle 24\) and counting each pair both ways: \[1\times 24,\quad 24\times 1,\quad 2\times 12,\quad 12\times 2,\quad 3\times 8,\quad 8\times 3,\quad 4\times 6,\quad 6\times 4.\] That is \(\displaystyle 8\) possible orders.$\displaystyle 13$ elements. We need \(\displaystyle mn=13\). Since \(\displaystyle 13\) is prime, its only positive factorisations are \(\displaystyle 1\times 13\) and \(\displaystyle 13\times 1\): \[1\times 13,\quad 13\times 1,\] so only \(\displaystyle 2\) possible orders — a single row of \(\displaystyle 13\) entries or a single column of \(\displaystyle 13\) entries.Final answer: with \(\displaystyle 24\) elements there are \(\displaystyle 8\) possible orders, namely \(\displaystyle 1\times 24,\ 24\times 1,\ 2\times 12,\ 12\times 2,\ 3\times 8,\ 8\times 3,\ 4\times 6,\ 6\times 4\); with \(\displaystyle 13\) elements only \(\displaystyle 2\), namely \(\displaystyle 1\times 13\) and \(\displaystyle 13\times 1\).

  3. Exercise 3

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

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    NCERT’s answer
    $\displaystyle 1$ × $\displaystyle 18$, $\displaystyle 2$ × $\displaystyle 9$, $\displaystyle 3$ × $\displaystyle 6$, $\displaystyle 6$ × $\displaystyle 3$, $\displaystyle 9$ × $\displaystyle 2$, $\displaystyle 18$ × $\displaystyle 1$; $\displaystyle 1$ × $\displaystyle 5$, $\displaystyle 5$ × $\displaystyle 1$
    As in the previous question, a matrix of order \(\displaystyle m\times n\) has \(\displaystyle mn\) elements, so the possible orders correspond to the ordered pairs \(\displaystyle (m,n)\) of positive integers with \(\displaystyle mn\) equal to the given count. Ordered pairs, not unordered: \(\displaystyle 3\times 6\) and \(\displaystyle 6\times 3\) are two different orders.$\displaystyle 18$ elements. Solve \(\displaystyle mn=18\) over positive integers: \[1\times 18,\quad 18\times 1,\quad 2\times 9,\quad 9\times 2,\quad 3\times 6,\quad 6\times 3.\] That gives \(\displaystyle 6\) possible orders.$\displaystyle 5$ elements. Solve \(\displaystyle mn=5\). As \(\displaystyle 5\) is prime, \[1\times 5,\quad 5\times 1,\] so \(\displaystyle 2\) possible orders.Final answer: \(\displaystyle 6\) possible orders for \(\displaystyle 18\) elements — \(\displaystyle 1\times 18,\ 18\times 1,\ 2\times 9,\ 9\times 2,\ 3\times 6,\ 6\times 3\); and \(\displaystyle 2\) possible orders for \(\displaystyle 5\) elements — \(\displaystyle 1\times 5\) and \(\displaystyle 5\times 1\).

  4. Exercise 4

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

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    NCERT’s answer
    (i)
    \(\displaystyle \begin{array}{rr}2 & \frac{9}{2} \\ & \frac{9}{2}\end{array} \quad \begin{aligned} & 8\end{aligned}\) (ii) \(\displaystyle \begin{array}{ll}1 & \frac{1}{2} \\ 2 & 1\end{array}\) (iii) \(\displaystyle \begin{array}{cc}\frac{9}{2} & \frac{25}{2} \\ 8 & 18\end{array}\)
    A \(\displaystyle 2\times 2\) matrix \(\displaystyle \mathrm{A}=[a_{ij}]\) has the four entries \[\mathrm{A}=\left[\begin{array}{rr}a_{11} & a_{12}\\ a_{21} & a_{22}\end{array}\right],\] so construct it by substituting \(\displaystyle (i,j)=(1,1),(1,2),(2,1),(2,2)\) into the given formula. Keep the order of the subscripts straight: in \(\displaystyle a_{12}\), \(\displaystyle i=1\) and \(\displaystyle j=2\).(i) \(\displaystyle a_{ij}=\dfrac{(i+j)^2}{2}\). \[a_{11}=\frac{(1+1)^2}{2}=\frac{4}{2}=2,\qquad a_{12}=\frac{(1+2)^2}{2}=\frac{9}{2},\] \[a_{21}=\frac{(2+1)^2}{2}=\frac{9}{2},\qquad a_{22}=\frac{(2+2)^2}{2}=\frac{16}{2}=8.\] \[\mathrm{A}=\left[\begin{array}{rr}2 & \dfrac{9}{2}\\[4pt] \dfrac{9}{2} & 8\end{array}\right].\] (Here \(\displaystyle i+j\) is symmetric in \(\displaystyle i,j\), which is why \(\displaystyle a_{12}=a_{21}\).)(ii) \(\displaystyle a_{ij}=\dfrac{i}{j}\). \[a_{11}=\frac{1}{1}=1,\qquad a_{12}=\frac{1}{2},\qquad a_{21}=\frac{2}{1}=2,\qquad a_{22}=\frac{2}{2}=1.\] \[\mathrm{A}=\left[\begin{array}{rr}1 & \dfrac{1}{2}\\[4pt] 2 & 1\end{array}\right].\] Note \(\displaystyle a_{12}\ne a_{21}\) here — the formula is not symmetric, so the row index must be used first.(iii) \(\displaystyle a_{ij}=\dfrac{(i+2j)^2}{2}\). \[a_{11}=\frac{(1+2)^2}{2}=\frac{9}{2},\qquad a_{12}=\frac{(1+4)^2}{2}=\frac{25}{2},\] \[a_{21}=\frac{(2+2)^2}{2}=\frac{16}{2}=8,\qquad a_{22}=\frac{(2+4)^2}{2}=\frac{36}{2}=18.\] \[\mathrm{A}=\left[\begin{array}{rr}\dfrac{9}{2} & \dfrac{25}{2}\\[4pt] 8 & 18\end{array}\right].\]

  5. Exercise 5

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

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    A \(\displaystyle 3\times 4\) matrix has rows \(\displaystyle i=1,2,3\) and columns \(\displaystyle j=1,2,3,4\), i.e. twelve entries \(\displaystyle a_{ij}\). Substitute each of the twelve pairs \(\displaystyle (i,j)\) into the given rule, taking \(\displaystyle i\) as the row and \(\displaystyle j\) as the column.(i) \(\displaystyle a_{ij}=\dfrac{1}{2}\left|-3i+j\right|\). The modulus must be evaluated BEFORE halving, and it makes every entry non-negative even though \(\displaystyle -3i+j\) is usually negative.Row \(\displaystyle i=1\): \(\displaystyle -3+j\) for \(\displaystyle j=1,2,3,4\) is \(\displaystyle -2,-1,0,1\), so \[a_{11}=\tfrac12(2)=1,\quad a_{12}=\tfrac12(1)=\tfrac12,\quad a_{13}=\tfrac12(0)=0,\quad a_{14}=\tfrac12(1)=\tfrac12.\] Row \(\displaystyle i=2\): \(\displaystyle -6+j\) is \(\displaystyle -5,-4,-3,-2\), so \[a_{21}=\tfrac52,\quad a_{22}=2,\quad a_{23}=\tfrac32,\quad a_{24}=1.\] Row \(\displaystyle i=3\): \(\displaystyle -9+j\) is \(\displaystyle -8,-7,-6,-5\), so \[a_{31}=4,\quad a_{32}=\tfrac72,\quad a_{33}=3,\quad a_{34}=\tfrac52.\] Hence \[\mathrm{A}=\left[\begin{array}{rrrr}1 & \dfrac{1}{2} & 0 & \dfrac{1}{2}\\[4pt] \dfrac{5}{2} & 2 & \dfrac{3}{2} & 1\\[4pt] 4 & \dfrac{7}{2} & 3 & \dfrac{5}{2}\end{array}\right].\](ii) \(\displaystyle a_{ij}=2i-j\). No modulus here, so negative entries are allowed and must be kept.Row \(\displaystyle i=1\): \(\displaystyle 2-j\) gives \(\displaystyle 1,0,-1,-2\). Row \(\displaystyle i=2\): \(\displaystyle 4-j\) gives \(\displaystyle 3,2,1,0\). Row \(\displaystyle i=3\): \(\displaystyle 6-j\) gives \(\displaystyle 5,4,3,2\). \[\mathrm{A}=\left[\begin{array}{rrrr}1 & 0 & -1 & -2\\ 3 & 2 & 1 & 0\\ 5 & 4 & 3 & 2\end{array}\right].\]

  6. Exercise 6

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

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    NCERT’s answer
    (i)
    \(\displaystyle x=1, \quad y=4, \quad z=3\) (ii) \(\displaystyle x=4, \quad y=2, \quad z=0 \quad\) or \(\displaystyle x=2, \quad y=4, z=0\) (iii) \(\displaystyle x=2, \quad y=4, \quad z=3\)
    Use the definition of equality of matrices: two matrices are equal if and only if they have the SAME ORDER and their corresponding elements are equal. So each equation below is really a system obtained by equating entries in matching positions.(i) Both matrices are \(\displaystyle 2\times 2\). Equating corresponding entries, \[4=y,\qquad 3=z,\qquad x=1,\qquad 5=5.\] The last is an identity and carries no information. \[\boxed{\,x=1,\ y=4,\ z=3\,}\](ii) Equating corresponding entries of the two \(\displaystyle 2\times 2\) matrices, \[x+y=6,\qquad 2=2,\qquad 5+z=5,\qquad xy=8.\] From \(\displaystyle 5+z=5\) we get \(\displaystyle z=0\).Now solve \(\displaystyle x+y=6\) together with \(\displaystyle xy=8\). Substituting \(\displaystyle y=6-x\) into \(\displaystyle xy=8\): \[x(6-x)=8\ \Longrightarrow\ x^{2}-6x+8=0\ \Longrightarrow\ (x-2)(x-4)=0,\] so \(\displaystyle x=2\) or \(\displaystyle x=4\), giving \(\displaystyle y=4\) or \(\displaystyle y=2\) respectively. Nothing in the equations distinguishes \(\displaystyle x\) from \(\displaystyle y\), so BOTH assignments are genuine solutions and both must be reported. \[\boxed{\,x=2,\ y=4,\ z=0\quad\text{or}\quad x=4,\ y=2,\ z=0\,}\](iii) Both matrices are \(\displaystyle 3\times 1\); equating corresponding entries gives the system \[x+y+z=9,\qquad x+z=5,\qquad y+z=7.\] Substituting \(\displaystyle x+z=5\) into the first equation: \(\displaystyle 5+y=9\), so \(\displaystyle y=4\). Then \(\displaystyle y+z=7\) gives \(\displaystyle z=7-4=3\), and \(\displaystyle x+z=5\) gives \(\displaystyle x=5-3=2\). Check: \(\displaystyle 2+4+3=9\). \(\displaystyle \checkmark\) \[\boxed{\,x=2,\ y=4,\ z=3\,}\]

  7. Exercise 7

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

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    NCERT’s answer
    \(\displaystyle a=1, b=2, c=3, d=4\)
    By the definition of equality of matrices, corresponding entries of the two \(\displaystyle 2\times 2\) matrices must be equal. Equating them position by position: \[a-b=-1,\qquad 2a+c=5,\qquad 2a-b=0,\qquad 3c+d=13.\]The two equations containing only \(\displaystyle a\) and \(\displaystyle b\) are solved first, since they form a closed pair. From \(\displaystyle 2a-b=0\), \[b=2a.\] Substitute into \(\displaystyle a-b=-1\): \[a-2a=-1\ \Longrightarrow\ -a=-1\ \Longrightarrow\ a=1,\] and hence \(\displaystyle b=2a=2\). (The sign here is where the slip usually happens: \(\displaystyle -a=-1\) gives \(\displaystyle a=+1\), not \(\displaystyle -1\).)Now use \(\displaystyle 2a+c=5\) with \(\displaystyle a=1\): \[2+c=5\ \Longrightarrow\ c=3.\] Finally use \(\displaystyle 3c+d=13\) with \(\displaystyle c=3\): \[9+d=13\ \Longrightarrow\ d=4.\]Check in the original matrix: \(\displaystyle a-b=1-2=-1\), \(\displaystyle 2a+c=2+3=5\), \(\displaystyle 2a-b=2-2=0\), \(\displaystyle 3c+d=9+4=13\). All four match. \(\displaystyle \checkmark\)Final answer: \(\displaystyle a=1,\ b=2,\ c=3,\ d=4\).

  8. Exercise 8

    \(\displaystyle \mathrm{A}=\left[a_{i j}\right]_{m \times n!}\) is a square matrix, if (A) \(\displaystyle m<n\) (B) \(\displaystyle m>n\) (C) \(\displaystyle m=n\) (D) None of these

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    NCERT’s answer
    C
    Definition: a matrix is called a SQUARE matrix when its number of rows equals its number of columns. For \(\displaystyle \mathrm{A}=[a_{ij}]_{m\times n}\) the number of rows is \(\displaystyle m\) and the number of columns is \(\displaystyle n\), so \(\displaystyle \mathrm{A}\) is square precisely when \(\displaystyle m=n\).Options (A) \(\displaystyle m<n\) and (B) \(\displaystyle m>n\) each describe a rectangular (non-square) matrix — e.g. a \(\displaystyle 2\times 3\) matrix is not square — so both fail, and (D) is therefore not needed.Final answer: (C) \(\displaystyle m=n\).

  9. Exercise 9

    Which of the given values of \(\displaystyle x\) and \(\displaystyle y\) make the following pair of matrices equal \(\displaystyle \left[\begin{array}{cc}3 x+7 & 5 \\ y+1 & 2-3 x\end{array}\right],\left[\begin{array}{cc}0 & y-2 \\ 8 & 4\end{array}\right]\) (A) \(\displaystyle x=\frac{-1}{3}, y=7\) (B) Not possible to find (C) \(\displaystyle y=7, \quad x=\frac{-2}{3}\) (D) \(\displaystyle x=\frac{-1}{3}, y=\frac{-2}{3}\)

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    NCERT’s answer
    B
    Two matrices are equal only if EVERY pair of corresponding entries is equal — it is not enough for one or two entries to match, and that is exactly what this question tests. Equating all four positions: \[3x+7=0,\qquad 5=y-2,\qquad y+1=8,\qquad 2-3x=4.\]Take the \(\displaystyle y\)-equations first: \[5=y-2\ \Longrightarrow\ y=7,\qquad y+1=8\ \Longrightarrow\ y=7.\] These agree, so \(\displaystyle y=7\) is consistent.Now the \(\displaystyle x\)-equations: \[3x+7=0\ \Longrightarrow\ x=-\frac{7}{3},\] \[2-3x=4\ \Longrightarrow\ -3x=2\ \Longrightarrow\ x=-\frac{2}{3}.\] These two demands on \(\displaystyle x\) contradict each other, since \(\displaystyle -\dfrac{7}{3}\ne-\dfrac{2}{3}\). No single value of \(\displaystyle x\) can satisfy both entries at once, so no pair \(\displaystyle (x,y)\) makes the matrices equal. In particular option (C), \(\displaystyle x=-\frac{2}{3},\ y=7\), satisfies the bottom-right entry but fails the top-left one, and option (A) does the reverse.Final answer: (B) Not possible to find.

  10. Exercise 10

    The number of all possible matrices of order \(\displaystyle 3 \times 3\) with each entry $\displaystyle 0$ or $\displaystyle 1$ is: (A) $\displaystyle 27$ (B) $\displaystyle 18$ (C) $\displaystyle 81$ (D) $\displaystyle 512$

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    NCERT’s answer
    D
    Use the multiplication principle of counting. A matrix of order \(\displaystyle 3\times 3\) has \[3\times 3=9\] entry positions, and the positions are filled independently of one another. Each position may be filled in \(\displaystyle 2\) ways (either \(\displaystyle 0\) or \(\displaystyle 1\)).Hence the total number of such matrices is \[\underbrace{2\times 2\times\cdots\times 2}_{9\ \text{factors}}=2^{9}=512.\](The common error is to compute \(\displaystyle 2\times 9=18\) or \(\displaystyle 3^{3}=27\); it is the number of POSITIONS that goes in the exponent and the number of CHOICES per position that goes in the base.)Final answer: (D) $\displaystyle 512$.