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NCERT Solutions · Class 12 Mathematics Determinants

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EXERCISE 4.3 1–5 (part 3 of 8)

  1. Write Minors and Cofactors of the elements of following determinants:

    Exercise 1

    (i)
    ∣2−403∣\displaystyle \left|\begin{array}{rr}2 & -4 \\ 0 & 3\end{array}\right|
    (ii)
    ∣acbd∣\displaystyle \left|\begin{array}{cc}a & c \\ b & d\end{array}\right|

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    NCERT’s answer
    (i)
    \(\displaystyle \mathrm{M}_{11}=3, \mathrm{M}_{12}=0, \mathrm{M}_{21}=-4, \mathrm{M}_{22}=2, \mathrm{~A}_{11}=3, \mathrm{~A}_{12}=0, \mathrm{~A}_{21}=4, \mathrm{~A}_{22}=2\) (ii) \(\displaystyle \mathrm{M}_{11}=d, \mathrm{M}_{12}=b, \mathrm{M}_{21}=c, \mathrm{M}_{22}=a\) \(\displaystyle \mathrm{A}_{11}=d, \mathrm{~A}_{12}=-b, \mathrm{~A}_{21}=-c, \mathrm{~A}_{22}=a\)
    Definition. The minor \(\displaystyle M_{ij}\) of the element \(\displaystyle a_{ij}\) is the determinant of what is left after deleting the \(\displaystyle i\)th row and the \(\displaystyle j\)th column, and the cofactor is \(\displaystyle A_{ij}=(-1)^{i+j}M_{ij}\). For a second order determinant, deleting one row and one column leaves a single entry, so each minor is just that entry; the whole work is in the sign \(\displaystyle (-1)^{i+j}\), which is \(\displaystyle +,-\) in the first row and \(\displaystyle -,+\) in the second.
    (i)
    For \(\displaystyle \left|\begin{array}{rr}2 & -4 \\ 0 & 3\end{array}\right|\):
    Minors: deleting row $\displaystyle 1$, column $\displaystyle 1$ leaves \(\displaystyle 3\); row $\displaystyle 1$, column $\displaystyle 2$ leaves \(\displaystyle 0\); row $\displaystyle 2$, column $\displaystyle 1$ leaves \(\displaystyle -4\); row $\displaystyle 2$, column $\displaystyle 2$ leaves \(\displaystyle 2\). So
    \[M_{11}=3,\qquad M_{12}=0,\qquad M_{21}=-4,\qquad M_{22}=2. \]
    Cofactors, attaching \(\displaystyle (-1)^{i+j}\):
    \[A_{11}=(-1)^{2}(3)=3,\qquad A_{12}=(-1)^{3}(0)=0, \]
    \[A_{21}=(-1)^{3}(-4)=4,\qquad A_{22}=(-1)^{4}(2)=2. \]
    Note that \(\displaystyle A_{21}=4\), not \(\displaystyle -4\): the sign is applied to the minor \(\displaystyle -4\), and the two minus signs cancel.
    (ii)
    For \(\displaystyle \left|\begin{array}{cc}a & c \\ b & d\end{array}\right|\):
    Minors:
    \[M_{11}=d,\qquad M_{12}=b,\qquad M_{21}=c,\qquad M_{22}=a. \]
    Cofactors:
    \[A_{11}=d,\qquad A_{12}=-b,\qquad A_{21}=-c,\qquad A_{22}=a. \]
    Check (each expansion must reproduce the determinant): in (i), \(\displaystyle a_{11}A_{11}+a_{12}A_{12}=2(3)+(-4)(0)=6\), and indeed \(\displaystyle \left|\begin{array}{rr}2 & -4 \\ 0 & 3\end{array}\right|=6\); in (ii), \(\displaystyle a\,A_{11}+c\,A_{12}=ad-bc\).

  2. Exercise 2

    (i)
    ∣100010001∣\displaystyle \left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|
    (ii)
    ∣10435−1012∣\displaystyle \left|\begin{array}{rrr}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|

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    NCERT’s answer
    (i)
    \(\displaystyle \mathrm{M}_{11}=1, \mathrm{M}_{12}=0, \mathrm{M}_{13}=0, \mathrm{M}_{21}=0, \mathrm{M}_{22}=1, \mathrm{M}_{23}=0, \mathrm{M}_{31}=0, \mathrm{M}_{32}=0, \mathrm{M}_{33}=1\), \(\displaystyle \mathrm{A}_{11}=1, \mathrm{~A}_{12}=0, \mathrm{~A}_{13}=0, \mathrm{~A}_{21}=0, \mathrm{~A}_{22}=1, \mathrm{~A}_{23}=0, \mathrm{~A}_{31}=0, \mathrm{~A}_{32}=0, \mathrm{~A}_{33}=1\) (ii) \(\displaystyle \mathrm{M}_{11}=11, \mathrm{M}_{12}=6, \mathrm{M}_{13}=3, \mathrm{M}_{21}=-4, \mathrm{M}_{22}=2, \mathrm{M}_{23}=1, \mathrm{M}_{31}=-20, \mathrm{M}_{32}=-13, \mathrm{M}_{33}=5\) \(\displaystyle \mathrm{A}_{11}=11, \mathrm{~A}_{12}=-6, \mathrm{~A}_{13}=3, \mathrm{~A}_{21}=4, \mathrm{~A}_{22}=2, \mathrm{~A}_{23}=-1, \mathrm{~A}_{31}=-20, \mathrm{~A}_{32}=13, \mathrm{~A}_{33}=5\)
    Definition. \(\displaystyle M_{ij}\) is the \(\displaystyle 2\times 2\) determinant obtained by deleting the \(\displaystyle i\)th row and \(\displaystyle j\)th column, and \(\displaystyle A_{ij}=(-1)^{i+j}M_{ij}\). The sign board for a third order determinant is
    \[\left[\begin{array}{rrr}+ & - & + \\ - & + & - \\ + & - & +\end{array}\right]. \]
    (i)
    For \(\displaystyle \left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|\):
    \[M_{11}=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|=1,\quad M_{12}=\left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right|=0,\quad M_{13}=\left|\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right|=0, \]
    \[M_{21}=\left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right|=0,\quad M_{22}=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|=1,\quad M_{23}=\left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right|=0, \]
    \[M_{31}=\left|\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right|=0,\quad M_{32}=\left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right|=0,\quad M_{33}=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|=1. \]
    Applying the signs (every off-diagonal minor is \(\displaystyle 0\), so its sign is immaterial):
    \[A_{11}=1,\; A_{12}=0,\; A_{13}=0,\; A_{21}=0,\; A_{22}=1,\; A_{23}=0,\; A_{31}=0,\; A_{32}=0,\; A_{33}=1. \]
    (ii)
    For \(\displaystyle \left|\begin{array}{rrr}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|\):
    Minors:
    \[M_{11}=\left|\begin{array}{rr}5 & -1 \\ 1 & 2\end{array}\right|=10+1=11,\quad M_{12}=\left|\begin{array}{rr}3 & -1 \\ 0 & 2\end{array}\right|=6-0=6,\quad M_{13}=\left|\begin{array}{rr}3 & 5 \\ 0 & 1\end{array}\right|=3-0=3, \]
    \[M_{21}=\left|\begin{array}{rr}0 & 4 \\ 1 & 2\end{array}\right|=0-4=-4,\quad M_{22}=\left|\begin{array}{rr}1 & 4 \\ 0 & 2\end{array}\right|=2-0=2,\quad M_{23}=\left|\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}\right|=1, \]
    \[M_{31}=\left|\begin{array}{rr}0 & 4 \\ 5 & -1\end{array}\right|=0-20=-20,\quad M_{32}=\left|\begin{array}{rr}1 & 4 \\ 3 & -1\end{array}\right|=-1-12=-13,\quad M_{33}=\left|\begin{array}{rr}1 & 0 \\ 3 & 5\end{array}\right|=5-0=5. \]
    Cofactors \(\displaystyle A_{ij}=(-1)^{i+j}M_{ij}\):
    \[A_{11}=11,\qquad A_{12}=-6,\qquad A_{13}=3, \]
    \[A_{21}=-(-4)=4,\qquad A_{22}=2,\qquad A_{23}=-1, \]
    \[A_{31}=-20,\qquad A_{32}=-(-13)=13,\qquad A_{33}=5. \]
    The places students slip are \(\displaystyle A_{21}\) and \(\displaystyle A_{32}\), where the minor is already negative and the sign change makes the cofactor positive.
    Check by expanding along the first row: \(\displaystyle 1(11)+0(-6)+4(3)=23\), and expanding along the second column gives \(\displaystyle 0(-6)+5(2)+1(13)=23\) as well.

  3. Exercise 3

    Using Cofactors of elements of second row, evaluate Δ=∣538201123∣\displaystyle \Delta=\left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|.

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    NCERT’s answer
    $\displaystyle 7$
    Rule used: expansion of a determinant along a row, \(\displaystyle \Delta=a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23}\), where \(\displaystyle A_{ij}=(-1)^{i+j}M_{ij}\). Along the second row the signs are \(\displaystyle -,+,-\).Here \(\displaystyle \Delta=\left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|\), so \(\displaystyle a_{21}=2,\ a_{22}=0,\ a_{23}=1\).Cofactors of the second row: \[A_{21}=(-1)^{2+1}\left|\begin{array}{ll}3 & 8 \\ 2 & 3\end{array}\right|=-(9-16)=7, \] \[A_{22}=(-1)^{2+2}\left|\begin{array}{ll}5 & 8 \\ 1 & 3\end{array}\right|=15-8=7, \] \[A_{23}=(-1)^{2+3}\left|\begin{array}{ll}5 & 3 \\ 1 & 2\end{array}\right|=-(10-3)=-7. \] Each minor is formed from the rows and columns that remain, so \(\displaystyle M_{21}\) keeps rows $\displaystyle 1$ and $\displaystyle 3$ and columns $\displaystyle 2$ and $\displaystyle 3$, and so on.Hence \[\Delta=a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23}=2(7)+0(7)+1(-7)=14+0-7=7. \](Confirming along the first row: \(\displaystyle 5(0-2)-3(6-1)+8(4-0)=-10-15+32=7\).)\[\Delta=7. \]

  4. Exercise 4

    Using Cofactors of elements of third column, evaluate Δ=∣1xyz1yzx1zxy∣\displaystyle \Delta=\left|\begin{array}{lll}1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{array}\right|.

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    NCERT’s answer
    \(\displaystyle (x-y)(y-z)(z-x)\)
    Rule used: expansion along a column, \(\displaystyle \Delta=a_{13}A_{13}+a_{23}A_{23}+a_{33}A_{33}\), with \(\displaystyle A_{ij}=(-1)^{i+j}M_{ij}\). Down the third column the signs are \(\displaystyle +,-,+\).For \(\displaystyle \Delta=\left|\begin{array}{lll}1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{array}\right|\) the third column entries are \(\displaystyle a_{13}=yz,\ a_{23}=zx,\ a_{33}=xy\).Cofactors of the third column (delete column $\displaystyle 3$ and the relevant row): \[A_{13}=(-1)^{1+3}\left|\begin{array}{ll}1 & y \\ 1 & z\end{array}\right|=z-y, \] \[A_{23}=(-1)^{2+3}\left|\begin{array}{ll}1 & x \\ 1 & z\end{array}\right|=-(z-x)=x-z, \] \[A_{33}=(-1)^{3+3}\left|\begin{array}{ll}1 & x \\ 1 & y\end{array}\right|=y-x. \] The middle one is where the sign is lost most often: the minor is \(\displaystyle z-x\), and the factor \(\displaystyle (-1)^{5}\) turns it into \(\displaystyle x-z\).Therefore \[\Delta=yz(z-y)+zx(x-z)+xy(y-x) \] \[=yz^{2}-y^{2}z+x^{2}z-xz^{2}+xy^{2}-x^{2}y. \]Now factorise. Group the terms containing \(\displaystyle x^{2}\), then the rest: \[\Delta=x^{2}(z-y)-x(z^{2}-y^{2})+yz(z-y)=(z-y)\left[x^{2}-x(z+y)+yz\right], \] using \(\displaystyle z^{2}-y^{2}=(z-y)(z+y)\). The bracket factorises as \(\displaystyle x^{2}-x(y+z)+yz=(x-y)(x-z)\). Hence \[\Delta=(z-y)(x-y)(x-z)=(x-y)(y-z)(z-x). \]\[\Delta=(x-y)(y-z)(z-x). \] In particular \(\displaystyle \Delta=0\) whenever any two of \(\displaystyle x,y,z\) are equal, as it must be, since then two rows of the determinant coincide.

  5. Exercise 5

    If Δ=∣a11a12a13a21a22a23a31a32a33∣\displaystyle \Delta=\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right| and Aij\displaystyle A_{i j} is Cofactors of aij\displaystyle a_{i j}, then value of Δ\displaystyle \Delta is given by (A) a11 A31+a12 A32+a13 A33\displaystyle a_{11} \mathrm{~A}_{31}+a_{12} \mathrm{~A}_{32}+a_{13} \mathrm{~A}_{33} (B) a11 A11+a12 A21+a13 A31\displaystyle a_{11} \mathrm{~A}_{11}+a_{12} \mathrm{~A}_{21}+a_{13} \mathrm{~A}_{31} (C) a21 A11+a22 A12+a23 A13\displaystyle a_{21} \mathrm{~A}_{11}+a_{22} \mathrm{~A}_{12}+a_{23} \mathrm{~A}_{13} (D) a11 A11+a21 A21+a31 A31\displaystyle a_{11} \mathrm{~A}_{11}+a_{21} \mathrm{~A}_{21}+a_{31} \mathrm{~A}_{31}

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    NCERT’s answer
    (D)
    Rule being tested: a determinant equals the sum of the products of the elements of ANY ONE row (or ANY ONE column) with their OWN corresponding cofactors,
    \[\Delta=a_{i1}A_{i1}+a_{i2}A_{i2}+a_{i3}A_{i3}\quad\text{(any fixed row }i\text{)}, \]
    \[\Delta=a_{1j}A_{1j}+a_{2j}A_{2j}+a_{3j}A_{3j}\quad\text{(any fixed column }j\text{)}. \]
    If instead the elements of one row are multiplied by the cofactors of a DIFFERENT row, the sum is \(\displaystyle 0\), not \(\displaystyle \Delta\).
    Test each option against this rule:
    (A)
    \(\displaystyle a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{33}\): elements of row $\displaystyle 1$ against the cofactors of row $\displaystyle 3$ — a cross combination, so this equals \(\displaystyle 0\), not \(\displaystyle \Delta\).
    (B)
    \(\displaystyle a_{11}A_{11}+a_{12}A_{21}+a_{13}A_{31}\): the elements come from row $\displaystyle 1$ while the cofactors \(\displaystyle A_{11},A_{21},A_{31}\) come from column $\displaystyle 1$ — the indices do not match, so this is not an expansion at all.
    (C)
    \(\displaystyle a_{21}A_{11}+a_{22}A_{12}+a_{23}A_{13}\): elements of row $\displaystyle 2$ against the cofactors of row $\displaystyle 1$ — again a cross combination, equal to \(\displaystyle 0\).
    (D)
    \(\displaystyle a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}\): the elements \(\displaystyle a_{11},a_{21},a_{31}\) are exactly the first column, each paired with its own cofactor. This is the expansion of \(\displaystyle \Delta\) along the first column.
    Hence the correct option is (D).