Tag: NCERT CBSE Class 12th Maths Solutions

1. Write minors and cofactors of the elements of the following determinants:

(i) ${\displaystyle \mathrm{\Delta }=\mid \begin{array}{cc}2& 4\\ 0& 3\end{array}\mid }$

For a $2\times 2$ determinant the minor $M_{ij}$ of element $a_{ij}$ is the determinant left after deleting its row and column; for $2\times 2$ deleting a row & column leaves a $1\times 1$ number.

Elements and their minors:

• $a_{11}=2:M_{11}=3$. Cofactor $A_{11}=(-1{)}^{1+1}{M}_{11}=+3$

• $a_{12}=4:M_{12}=0$. Cofactor $A_{12}=(-1{)}^{1+2}{M}_{12}=-0=0$

• $a_{21}=0:M_{21}=4$. Cofactor $A_{21}=(-1{)}^{2+1}{M}_{21}=-4$

• $a_{22}=3:M_{22}=2$. Cofactor $A_{22}=(-1{)}^{2+2}{M}_{22}=+2$
  

(You can check: expansion along first row gives $\mathrm{\Delta }=2\cdot A_{11}+4\cdot A_{12}=2\cdot 3+4\cdot 0=6$, and direct determinant $2\cdot 3-4\cdot 0=6$)

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(ii) ${\displaystyle \mathrm{\Delta }=\mid \begin{array}{cc}a& c\\ b& d\end{array}\mid }$

Minors (each $1\times 1$ entry):

• $M_{11}=d,A_{11}=+d\mathrm{}$

• $M_{12}=b,A_{12}=(-1{)}^{1+2}b=-b\mathrm{}$

• $M_{21}=c,A_{21}=(-1{)}^{2+1}c=-c\mathrm{}$

• $M_{22}=a,A_{22}=+a\mathrm{}$

(So cofactors matrix is $\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$)

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2. Write minors and cofactors for the following $3\times 3$ determinants:

(i) $I_3=\mid \begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\mid$

We give minors $M_{ij}$ and cofactors $A_{ij}=(-1{)}^{i+j}{M}_{ij}$

Because $I_3$ is diagonal, minors are determinants of the $2\times 2$submatrices:

Row 1:

• $M_{11}=\mid \begin{array}{cc}1& 0\\ 0& 1\end{array}\mid =1,A_{11}=+1$

• $M_{12}=\mid \begin{array}{cc}0& 0\\ 0& 1\end{array}\mid =0,A_{12}=-0=0$

• $M_{13}=\mid \begin{array}{cc}0& 1\\ 0& 0\end{array}\mid =0,A_{13}=+0=0$
  

Row 2:

• $M_{21}=\mid \begin{array}{cc}0& 0\\ 0& 1\end{array}\mid =0,A_{21}=-0=0$

• $M_{22}=\mid \begin{array}{cc}1& 0\\ 0& 1\end{array}\mid =1,A_{22}=+1$

• $M_{23}=\mid \begin{array}{cc}1& 0\\ 0& 0\end{array}\mid =0,A_{23}=-0=0$

Row 3:

• $M_{31}=\mid \begin{array}{cc}0& 0\\ 1& 0\end{array}\mid =0,A_{31}=+0=0$
  

• $M_{32}=\mid \begin{array}{cc}1& 0\\ 0& 0\end{array}\mid =0,A_{32}=-0=0$

• $M_{33}=\mid \begin{array}{cc}1& 0\\ 0& 1\end{array}\mid =1,A_{33}=+1$

(So adj $I_3$ = transpose of cofactor matrix = identity again.)

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(ii) ${\displaystyle \mathrm{\Delta }=\mid \begin{array}{ccc}1& 0& 4\\ 3& 5& 1\\ 0& 1& 2\end{array}\mid }$

Compute minors (each is a $2\times 2$ determinant) and cofactors:

Row 1:

• $M_{11}=\mid \begin{array}{cc}5& 1\\ 1& 2\end{array}\mid =5\cdot 2-1\cdot 1=10-1=9,A_{11}=+9$

• $M_{12}=\mid \begin{array}{cc}3& 1\\ 0& 2\end{array}\mid =3\cdot 2-1\cdot 0=6,A_{12}=(-1{)}^{1+2}6=-6$

• $M_{13}=\mid \begin{array}{cc}3& 5\\ 0& 1\end{array}\mid =3\cdot 1-5\cdot 0=3,A_{13}=+3$

Row 2:

• $M_{21}=\mid \begin{array}{cc}0& 4\\ 1& 2\end{array}\mid =0\cdot 2-4\cdot 1=-4,A_{21}=(-1{)}^{2+1}(-4)=+4$

• $M_{22}=\mid \begin{array}{cc}1& 4\\ 0& 2\end{array}\mid =1\cdot 2-4\cdot 0=2,A_{22}=+2$

• $M_{23}=\mid \begin{array}{cc}1& 0\\ 0& 1\end{array}\mid =1\cdot 1-0\cdot 0=1,A_{23}=(-1{)}^{2+3}1=-1$

Row 3:

• $M_{31}=\mid \begin{array}{cc}0& 4\\ 5& 1\end{array}\mid =0\cdot 1-4\cdot 5=-20,A_{31}=(-1{)}^{3+1}(-20)=-20$
  (note sign: $(-1{)}^4=+1$, so $A_{31}=+(-20)=-20$
  

• $M_{32}=\mid \begin{array}{cc}1& 4\\ 3& 1\end{array}\mid =1\cdot 1-4\cdot 3=1-12=-11,A_{32}=(-1{)}^{3+2}(-11)=+11$
  

• $M_{33}=\mid \begin{array}{cc}1& 0\\ 3& 5\end{array}\mid =1\cdot 5-0\cdot 3=5,A_{33}=(-1{)}^{3+3}5=+5$

(You can verify determinant by expansion: $\mathrm{\Delta }=1\cdot A_{11}+0\cdot A_{12}+4\cdot A_{13}=1\cdot 9+0+4\cdot 3=9+12=21$

Direct check yields same.)

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3. Using cofactors of elements of the second row, evaluate

$$\mathrm{\Delta }=\mid \begin{array}{ccc}5& 3& 8\\ 2& 0& 1\\ 1& 2& 3\end{array}\mid \mathrm{}$$

We will expand along the second row: $\mathrm{\Delta }=a_{21}{A}_{21}+a_{22}{A}_{22}+a_{23}{A}_{23}$

Compute the cofactors (minors first):

• For $a_{21}=2$: $M_{21}=\mid \begin{array}{cc}3& 8\\ 2& 3\end{array}\mid =3\cdot 3-8\cdot 2=9-16=-7$
  $A_{21}=(-1{)}^{2+1}{M}_{21}=-(-7)=+7$

• For $a_{22}=0$: $M_{22}=\mid \begin{array}{cc}5& 8\\ 1& 3\end{array}\mid =5\cdot 3-8\cdot 1=15-8=7$
  $A_{22}=(-1{)}^{2+2}{M}_{22}=+7$

• For $a_{23}=1$: $M_{23}=\mid \begin{array}{cc}5& 3\\ 1& 2\end{array}\mid =5\cdot 2-3\cdot 1=10-3=7$
  $A_{23}=(-1{)}^{2+3}{M}_{23}=-7$

Now expansion:

$$\mathrm{\Delta }=2\cdot A_{21}+0\cdot A_{22}+1\cdot A_{23}=2\cdot 7+0+1\cdot (-7)=14-7=\boxed{{\displaystyle 7}}\mathrm{}$$

(You may check by any other expansion; result is $7$.)

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4. Using cofactors of elements of the third column, evaluate

$$\mathrm{\Delta }=\mid \begin{array}{ccc}1& 1& 1\\ x& y& z\\ y& z& x\end{array}\mid \mathrm{}$$

We expand along third column: $\mathrm{\Delta }=a_{13}{A}_{13}+a_{23}{A}_{23}+a_{33}{A}_{33}$
where $a_{13}=1,a_{23}=z,a_{33}=x$

Compute minors and cofactors:

• $A_{13}=(-1{)}^{1+3}{M}_{13}=(-1{)}^4{M}_{13}=+M_{13}$
  $M_{13}=\mid \begin{array}{cc}x& y\\ y& z\end{array}\mid =xz-y^2\mathrm{}$

• $A_{23}=(-1{)}^{2+3}{M}_{23}=(-1{)}^5(-M_{23})=-M_{23}\mathrm{<span\; class="katex"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="katex-html"\; aria-hidden="true"><span\; class="base"><span\; class="mord"><span\; class="msupsub"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist-s"></span></span></span></span></span></span></span></span>}$$\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"> </span>}$$\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">(Simpler: </span>}$$A_{23}=-M_{23}$)
  $M_{23}=\mid \begin{array}{cc}1& 1\\ y& z\end{array}\mid =1\cdot z-1\cdot y=z-y\mathrm{}$
  So $A_{23}=-(z-y)=y-z\mathrm{}$

• $A_{33}=(-1{)}^{3+3}{M}_{33}=+M_{33}\mathrm{}$
  $M_{33}=\mid \begin{array}{cc}1& 1\\ x& y\end{array}\mid =1\cdot y-1\cdot x=y-x\mathrm{}$
  So $A_{33}=y-x\mathrm{}$

Now expand:

$$\mathrm{\Delta }=1\cdot (xz-y^2)+z\cdot (y-z)+x\cdot (y-x)\mathrm{}$$

Simplify term-by-term:

$$\mathrm{\Delta }=xz-y^2+zy-z^2+xy-x^2\mathrm{.}$$

Group like terms:

$$\mathrm{\Delta }=-x^2+(xz+zy+xy)-(y^2+z^2)\mathrm{}$$

We can rewrite symmetric grouping if desired. But we can also notice a factorization — rearrange as

$$\mathrm{\Delta }=-(x^2+y^2+z^2)+(xy+yz+zx)\mathrm{.}$$

Thus

$$\boxed{{\displaystyle \mathrm{\Delta }=(xy+yz+zx)-(x^2+y^2+z^2)\mathrm{}}}$$

(That is the simplest closed form. You may also write $\mathrm{\Delta }=-\frac{1}{2}[(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$ — indeed expanding that gives the same value. So $\mathrm{\Delta }\le 0$ and equals zero exactly when $x=y=z$.)

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5. If $\mathrm{\Delta }=[a_{ij}]$ is the $3\times 3$ determinant and $A_{ij}$ are cofactors of $a_{ij}$, then which of the following equals $\mathrm{\Delta }$?

Options:

• (A) $a_{11}{A}_{31}+a_{12}{A}_{32}+a_{13}{A}_{33}$

• (B) $a_{11}{A}_{11}+a_{12}{A}_{21}+a_{13}{A}_{31}$

• (C) $a_{21}{A}_{11}+a_{22}{A}_{12}+a_{23}{A}_{13}$

• (D) $a_{11}{A}_{11}+a_{21}{A}_{21}+a_{31}{A}_{31}$

Solution / reasoning.

Standard properties of cofactors / expansions:

• Determinant expansion along the first row gives
  $\mathrm{\Delta }=a_{11}{A}_{11}+a_{12}{A}_{12}+a_{13}{A}_{13}\mathrm{}$

• Expansion along the first column gives
  $\mathrm{\Delta }=a_{11}{A}_{11}+a_{21}{A}_{21}+a_{31}{A}_{31}\mathrm{}$

Option (D) exactly matches the expansion along the first column, so it equals $\mathrm{\Delta }$.
Options (A), (B), (C) are mixed-index combinations that do not in general equal $\mathrm{\Delta }$ (they give either other identities or zero). For example, the sum of elements of one row multiplied by cofactors of a different row equals $0$.

Therefore the correct choice is

$$\boxed{{\displaystyle (D)a_{11}{A}_{11}+a_{21}{A}_{21}+a_{31}{A}_{31}\mathrm{}}}$$