# Standard Determinants

There are certain standard determinants whose results are given by direct formulas. The standard results of a few types of determinants are given below which will help to solve questions more efficiently.

All Topics in Determinants

## Expressions for Standard Determinants

(i) $\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|=\left( a-b \right)\left( b-c \right)\left( c-a \right)$

(ii) $\left| \begin{matrix} a & b & c \\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ bc & ca & ab \\ \end{matrix} \right|=\left| \begin{matrix} 1 & 1 & 1 \\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ {{a}^{3}} & {{b}^{3}} & {{c}^{3}} \\ \end{matrix} \right|=\left( a-b \right)\left( b-c \right)\left( c-a \right)\left( ab+bc+ca \right)$

(iii) $\left| \begin{matrix} a & bc & abc \\ b & ca & abc \\ c & ab & abc \\ \end{matrix} \right|=\left| \begin{matrix} a & {{a}^{2}} & {{a}^{3}} \\ b & {{b}^{2}} & {{b}^{3}} \\ c & {{c}^{2}} & {{c}^{3}} \\ \end{matrix} \right|=abc\left( a-b \right)\left( b-c \right)\left( c-a \right);$

(iv) $\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ {{a}^{3}} & {{b}^{3}} & {{c}^{3}} \\ \end{matrix} \right|=\left( a-b \right)\left( b-c \right)\left( c-a \right)\left( a+b+c \right)$

(v) $\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|=-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}+3abc$

### Practice Problems From Determinant Standard Results

Illustration: Evaluate the determinant $\Delta =\left| \begin{matrix} \sqrt{p}+\sqrt{q} & 2\sqrt{r} & \sqrt{r} \\ \sqrt{qr}+\sqrt{2p} & r & \sqrt{2r} \\ q+\sqrt{pr} & \sqrt{qr} & r \\ \end{matrix} \right|,$

where p, q and r are positive real numbers.

Solution:

Taking $\sqrt{r}$ common from C2 and C3 of the given determinant using scalar multiple property and then expanding it using the invariance property we can evaluate the given problem.

We get $\Delta =r\left| \begin{matrix} \sqrt{p}+\sqrt{q} & 2 & 1 \\ \sqrt{qr}+\sqrt{2p} & \sqrt{r} & \sqrt{2} \\ q+\sqrt{pr} & \sqrt{q} & \sqrt{r} \\ \end{matrix} \right|$

Applying ${{C}_{1}}\to {{C}_{1}}-\sqrt{q}{{C}_{2}}-\sqrt{p}{{C}_{3}}$

We get D $=r\left| \begin{matrix} -\sqrt{q} & 2 & 1 \\ 0 & \sqrt{r} & \sqrt{2} \\ 0 & \sqrt{q} & \sqrt{r} \\ \end{matrix} \right|=-r\sqrt{q}\left( r-\sqrt{2}q \right)=r\left( q\sqrt{2}-r\sqrt{q} \right).$

Illustration: Let a, b, c be positive and not equal. Show that the value of the determinant $\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|$

is negative.

Solution:

By applying invariance and scalar multiple properties to the given determinant we can get the required result.

$D=\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|;\;\;then\;\;D=\left| \begin{matrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \\ \end{matrix} \right| \;\;\;\left[ {{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}} \right]$ $=\left( a+b+c \right)\left| \begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \\ \end{matrix} \right|$ [Taking (a+b+c) common from the first column] $=\left( a+b+c \right)\left| \begin{matrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \\ \end{matrix} \right|\;\; \left[ {{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\,\,and\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \right]$ $=\left( a+b+c+ \right)\left[ \left( c-b \right)\left( b-c \right)\left( a-b \right)\left( a-c \right) \right]=\left( a+b+c+ \right)\left[ bc+ca+ab-{{a}^{2}}-{{b}^{2}}-{{c}^{2}} \right]$ $=-\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-bc-ca-ab \right)=-\frac{1}{2}\left( a+b+c \right)\left( 2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2bc-2ca-2ab \right)$ $=-\frac{1}{2}\left( a+b+c \right)\left[ \left( {{a}^{2}}+{{b}^{2}}-2ab \right)+\left( {{b}^{2}}+{{c}^{2}}-2bc \right)+\left( {{c}^{2}}+{{a}^{2}}-2ac \right) \right]$ $=-\frac{1}{2}\left( a+b+c \right)\left[ {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}} \right]$ … (i)

a, b, c are positive $\Rightarrow a+b+c>0$

a, b, c are unequal $\Rightarrow {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}}>0$ … (ii)

From (i) and (ii), Δ <0.

Illustration: Show that $\Delta =\left| \begin{matrix} 1 & {{\cos }^{2}}\left( \alpha -\beta \right) & {{\cos }^{2}}\left( \alpha -\gamma \right) \\ {{\cos }^{2}}\left( \beta -\alpha \right) & 1 & {{\cos }^{2}}\left( \beta -\gamma \right) \\ {{\cos }^{2}}\left( \gamma -\alpha \right) & {{\cos }^{2}}\left( \gamma -\beta \right) & 1 \\ \end{matrix} \right|$ $=2{{\sin }^{2}}\left( \beta -\gamma \right){{\sin }^{2}}\left( \gamma -\alpha \right){{\sin }^{2}}\left( \alpha -\beta \right)$

Solution:

By Putting $\beta -\gamma =A,\gamma -\alpha =B,\alpha -\beta =C$ and then by using switching and invariance properties we can prove the above problem.

We can write Δ as, $\Delta =\left| \begin{matrix} 1 & {{\cos }^{2}}C & {{\cos }^{2}}B \\ {{\cos }^{2}}C & 1 & {{\cos }^{2}}A \\ {{\cos }^{2}}B & {{\cos }^{2}}A & 1 \\ \end{matrix} \right|$

(Note that A + B + C = 0).

Using ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}},{{C}_{1}}\to {{C}_{3}}-{{C}_{1}}$ we get

$\Delta =\left| \begin{matrix} 1 & -{{\sin }^{2}}C & -{{\sin }^{2}}B \\ {{\cos }^{2}}C & {{\sin }^{2}}C & {{\cos }^{2}}A-{{\cos }^{2}}C \\ {{\cos }^{2}}B & {{\cos }^{2}}A-{{\cos }^{2}}B & {{\sin }^{2}}B \\ \end{matrix} \right|\;\;\left| \begin{matrix} 1 & -{{\sin }^{2}}C & -{{\sin }^{2}}B \\ {{\cos }^{2}}C & {{\sin }^{2}}C & \sin B\sin \left( C-A \right) \\ {{\cos }^{2}}B & \sin C\sin \left( B-A \right) & {{\sin }^{2}}B \\ \end{matrix} \right|$ $={{\left( -1 \right)}^{2}}\left| \begin{matrix} 1 & {{\sin }^{2}}C & {{\sin }^{2}}B \\ {{\cos }^{2}}C & -{{\sin }^{2}}C & \sin B\sin \left( -A \right) \\ {{\cos }^{2}}B & \sin C\sin \left( B-A \right) & -{{\sin }^{2}}B \\ \end{matrix} \right|$

Since, $\left[ \,{{\cos }^{2}}A-{{\cos }^{2}}B=\sin \left( A+B \right)\sin \left( B-A \right),A+B=-C,C+A=-B \right]; \;\;=\sin C\sin B\left[ {{\Delta }_{1}} \right]$

Where ${{\Delta }_{1}}=\left| \begin{matrix} 1 & {{\sin }^{2}}C & \sin B \\ {{\cos }^{2}}C & -{{\sin }^{2}}C & \sin (C-A) \\ {{\cos }^{2}}B & \sin (B-A) & -\sin B \\ \end{matrix} \right|\; Using\; {{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\; and \;{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$

we get

${{\Delta }_{1}}=\left| \begin{matrix} 1 & \sin C & \sin B \\ -{{\sin }^{2}}C & -2{{\sin }^{2}}C & \sin (C-A)-sinB \\ -{{\sin }^{2}}B & \sin (B-A)-sinC & -2{{\sin }^{2}}B \\ \end{matrix} \right|$

But sin (C – A) – sin B = sin (C – A) + sin (C + A) = 2 sin C cos A and sin (B – A) – sin C = 2 sin B cos A

Therefore, ${{\Delta }_{1}}=\sin C\;\sin B\;{{\Delta }_{2}}\; where \;{{\Delta }_{2}}=\left| \begin{matrix} 1 & \sin C & \sin B \\ \sin C & 2 & -2\cos A \\ \sin B & -2\cos A & 2 \\ \end{matrix} \right|$

Applying ${{R}_{2}}\to {{R}_{2}}-\sin C\,{{R}_{1}} \;and \;{{R}_{3}}\to -\sin \,B\,{{R}_{1}}$ we get

${{\Delta }_{2}}=\left| \begin{matrix} 1 & \sin C & \sin B \\ 0 & 2-{{\sin }^{2}}C & -2\cos A-\sin B\sin C \\ 0 & -2\cos A-\sin B\sin C & 2-{{\sin }^{2}}B \\ \end{matrix} \right|=\left( 2-{{\sin }^{2}}B \right)\left( 2-\sin C \right)-{{\left( 2\cos A+\sin B\sin C \right)}^{2}}$ $=4-2{{\sin }^{2}}B-2{{\sin }^{2}}C+{{\sin }^{2}}B{{\sin }^{2}}C-\left[ 4{{\cos }^{2}}A+4\cos A\sin B\sin C+{{\sin }^{2}}B{{\sin }^{2}}C \right]$ $=4{{\sin }^{2}}A-2{{\sin }^{2}}B-2{{\sin }^{2}}C-4\cos A\sin B\sin C$ $=2{{\sin }^{2}}A-2\left[ {{\sin }^{2}}B+{{\sin }^{2}}C-{{\sin }^{2}}A+2\cos A\sin B\sin C \right]$

But A + B + C = 0 implies; ${{\sin }^{2}}B+{{\sin }^{2}}C-{{\sin }^{2}}A=-2\cos A\sin B\sin C$ ${{\Delta }_{2}}=2{{\sin }^{2}}A;$ Hence, $D=\sin C\sin B{{\Delta }_{1}}={{\sin }^{2}}C{{\sin }^{2}}B{{\Delta }_{2}}$ $=2{{\sin }^{2}}A\;{{\sin }^{2}}B\;{{\sin }^{2}}C=2{{\sin }^{2}}\left( \alpha -\beta \right)\;{{\sin }^{2}}\left( \beta -\gamma \right)\;{{\sin }^{2}}\left( \gamma -\alpha \right).$

Illustration: Prove that the following determinant vanishes if any two of x; y; z are equal

$\Delta =\left| \begin{matrix} \sin x & \sin y & \sin z \\ \cos x & \cos y & \cos z \\ {{\cos }^{3}}x & {{\cos }^{3}}y & {{\cos }^{3}}z \\ \end{matrix} \right|$

Solution:Taking cos x, cos y, and cos z common from first, second and third column using scalar multiple and then using the invariance property we can prove the given statement.

Here, $\Delta =\cos x\cos y\cos z\left| \begin{matrix} \tan x & \tan y & \tan z \\ 1 & 1 & 1 \\ {{\cos }^{2}}x & {{\cos }^{2}}y & {{\cos }^{2}}z \\ \end{matrix} \right|$ $=\cos x\cos y\cos z\left| \begin{matrix} \tan x & \tan y-\tan x & \tan z-\tan y \\ 1 & 0 & 0 \\ {{\cos }^{2}}x & {{\cos }^{2}}y-{{\cos }^{2}}x & {{\cos }^{2}}z-{{\cos }^{2}}y \\ \end{matrix} \right|\left( {{C}_{3}}\to {{C}_{3}}-{{C}_{2}},{{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \right)$

Expanding alongR2, $\Delta =-\cos x\cos y\cos z\left| \begin{matrix} \tan y-\tan x & \tan z-\tan y \\ {{\cos }^{2}}y-{{\cos }^{2}}x & {{\cos }^{2}}z-{{\cos }^{2}}y \\ \end{matrix} \right|$ $=-\cos x\cos y\cos z\left| \begin{matrix} \frac{\sin \left( y-x \right)}{\cos x\cos y} & \frac{\sin \left( z-y \right)}{\cos y.\cos z} \\ {{\sin }^{2}}x-{{\sin }^{2}}y & {{\sin }^{2}}y-{{\sin }^{2}}z \\ \end{matrix} \right|=\left| \begin{matrix} \cos z.\sin \left( x-y \right) & \cos x.\sin \left( y-z \right) \\ \sin \left( x+y \right).\sin \left( x-y \right) & \sin \left( y+z \right).\sin \left( y-z \right) \\ \end{matrix} \right|$

… (i)

$=\sin \left( x-y \right)\sin \left( y-z \right)\left| \begin{matrix} \cos z & \cos x \\ \sin \left( x+y \right) & \sin \left( y+z \right) \\ \end{matrix} \right|=\sin \left( x-y \right)\sin \left( y-z \right)\left[ \sin \left( y+z \right)\cos z-\sin \left( x+y \right)\cos \,x) \right]$ $=\frac{1}{2}\sin \left( x-y \right)\sin \left( y-z \right)\left[ \left\{ \sin \left( y+2z \right)+\sin y \right\}-\left\{ \sin \left( y+2x \right)+\sin y \right\} \right]$ $=\frac{1}{2}\sin \left( x-y \right)\sin \left( y-z \right)\left[ \sin \left( y+2x \right) \right]=\frac{1}{2}\sin \left( x-y \right)\sin \left( y-z \right)2\cos \left( x+y+z \right)\sin \left( z-x \right)$ $=\sin \left( x-y \right)\sin \left( y-z \right)\sin \left( z-x \right)\cos \left( x+y+z \right)$

Clearly, Δ is zero when any two of x, y, z are equal or $x+y+z=\frac{\pi }{2}.$

Hence proved.