Standard Determinant

A determinant is defined as a quantity which is obtained by adding the products of all elements in a square matrix. To find the determinant, a particular rule is followed. In this lesson, the concept of determinants is explained in detail along with solved examples, formulas, determinant types, and practice questions.

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) 1aa21bb21cc2=(ab)(bc)(ca)\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) abca2b2c2bccaab=111a2b2c2a3b3c3=(ab)(bc)(ca)(ab+bc+ca)\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) abcabcbcaabccababc=aa2a3bb2b3cc2c3=abc(ab)(bc)(ca);\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) 111abca3b3c3=(ab)(bc)(ca)(a+b+c)\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) abcbcacab=a3b3c3+3abc\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|=-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}+3abc

(vi) Determinant of order 3×3=a1b1c1a2b2c2a3b3c3=a1b2c2b3c3b1a2c2a3c3+c1a2b2b3b33\times 3=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|={{a}_{1}}\left| \begin{matrix} {{b}_{2}} & {{c}_{2}} \\ {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|-{{b}_{1}}\left| \begin{matrix} {{a}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{c}_{3}} \\ \end{matrix} \right|+{{c}_{1}}\left| \begin{matrix} {{a}_{2}} & {{b}_{2}} \\ {{b}_{3}} & {{b}_{3}} \\ \end{matrix} \right|

(vii) In the determinant D = a11a12a13a21a22a23a31a32a33,\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|,

minor of a12 is denoted as M12=a21a23a31a33{{M}_{12}}=\left| \begin{matrix} {{a}_{21}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{33}} \\ \end{matrix} \right|

and so on.

(viii) Cofactor of an element aij=Cij=(1)i+jMij{{a}_{i\,j}}={{C}_{i\,j}}={{\left( -1 \right)}^{i+j}}{{M}_{i\,j}}

Evaluation of the Determinant using SARRUS Diagram

If A=[a11a12a13a21a22a23a31a32a33]A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right]

is a square matrix of order 3, the below diagram is a Sarrus Diagram obtained by adjoining the first two columns on the right and draw dark and dotted lines as shown.

The value of the determinant is (a11a22a33+a12a23a31+a13a21a32)(a13a22a31+a11a23a32+a12a21a33).\left( {{a}_{11}}{{a}_{22}}{{a}_{33}}+{{a}_{12}}{{a}_{23}}{{a}_{31}}+{{a}_{13}}{{a}_{21}}{{a}_{32}} \right)-\left( {{a}_{13}}{{a}_{22}}{{a}_{31}}+{{a}_{11}}{{a}_{23}}{{a}_{32}}+{{a}_{12}}{{a}_{21}}{{a}_{33}} \right).

Solving Determinants

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Solved Problems on Determinant

Illustration 1: Evaluate the determinant Δ=p+q2rrqr+2pr2rq+prqrr,\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 r\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 Δ=rp+q21qr+2pr2q+prqr\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 C1C1qC2pC3{{C}_{1}}\to {{C}_{1}}-\sqrt{q}{{C}_{2}}-\sqrt{p}{{C}_{3}}

We get D =rq210r20qr=rq(r2q)=r(q2rq).=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 2: Let a, b, c be positive and not equal. Show that the value of the determinant abcbcacab\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=abcbcacab;    then    D=a+b+cbca+b+ccaa+b+cab      [C1C1+C2+C3]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] =(a+b+c)1bc1ca1ab=\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] =(a+b+c)1bc0cbac0abbc    [R2R2R1andR3R3R1]=\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] =(a+b+c+)[(cb)(bc)(ab)(ac)]=(a+b+c+)[bc+ca+aba2b2c2]=\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] =(a+b+c)(a2+b2+c2bccaab)=12(a+b+c)(2a2+2b2+2c22bc2ca2ab)=-\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) =12(a+b+c)[(a2+b22ab)+(b2+c22bc)+(c2+a22ac)]=-\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] =12(a+b+c)[(ab)2+(bc)2+(ca)2]=-\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 a+b+c>0\Rightarrow a+b+c>0

a, b, c are unequal (ab)2+(bc)2+(ca)2>0\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 3: Show that Δ=1cos2(αβ)cos2(αγ)cos2(βα)1cos2(βγ)cos2(γα)cos2(γβ)1\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| =2sin2(βγ)sin2(γα)sin2(αβ)=2{{\sin }^{2}}\left( \beta -\gamma \right){{\sin }^{2}}\left( \gamma -\alpha \right){{\sin }^{2}}\left( \alpha -\beta \right)

Solution:

By Putting βγ=A,γα=B,αβ=C\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, Δ=1cos2Ccos2Bcos2C1cos2Acos2Bcos2A1\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 C2C2C1,C1C3C1{{C}_{2}}\to {{C}_{2}}-{{C}_{1}},{{C}_{1}}\to {{C}_{3}}-{{C}_{1}} we get

Δ=1sin2Csin2Bcos2Csin2Ccos2Acos2Ccos2Bcos2Acos2Bsin2B    1sin2Csin2Bcos2Csin2CsinBsin(CA)cos2BsinCsin(BA)sin2B\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| =(1)21sin2Csin2Bcos2Csin2CsinBsin(A)cos2BsinCsin(BA)sin2B={{\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, [cos2Acos2B=sin(A+B)sin(BA),A+B=C,C+A=B];    =sinCsinB[Δ1]\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 Δ1=1sin2CsinBcos2Csin2Csin(CA)cos2Bsin(BA)sinB  Using  R2R2R1  and  R3R3R1{{\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

Δ1=1sinCsinBsin2C2sin2Csin(CA)sinBsin2Bsin(BA)sinC2sin2B{{\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, Δ1=sinC  sinB  Δ2  where  Δ2=1sinCsinBsinC22cosAsinB2cosA2{{\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 R2R2sinCR1  and  R3sinBR1{{R}_{2}}\to {{R}_{2}}-\sin C\,{{R}_{1}} \;and \;{{R}_{3}}\to -\sin \,B\,{{R}_{1}} we get

Δ2=1sinCsinB02sin2C2cosAsinBsinC02cosAsinBsinC2sin2B=(2sin2B)(2sinC)(2cosA+sinBsinC)2{{\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}} =42sin2B2sin2C+sin2Bsin2C[4cos2A+4cosAsinBsinC+sin2Bsin2C]=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] =4sin2A2sin2B2sin2C4cosAsinBsinC=4{{\sin }^{2}}A-2{{\sin }^{2}}B-2{{\sin }^{2}}C-4\cos A\sin B\sin C =2sin2A2[sin2B+sin2Csin2A+2cosAsinBsinC]=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; sin2B+sin2Csin2A=2cosAsinBsinC{{\sin }^{2}}B+{{\sin }^{2}}C-{{\sin }^{2}}A=-2\cos A\sin B\sin C Δ2=2sin2A;{{\Delta }_{2}}=2{{\sin }^{2}}A; Hence, D=sinCsinBΔ1=sin2Csin2BΔ2D=\sin C\sin B{{\Delta }_{1}}={{\sin }^{2}}C{{\sin }^{2}}B{{\Delta }_{2}} =2sin2A  sin2B  sin2C=2sin2(αβ)  sin2(βγ)  sin2(γα).=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 4: Prove that the following determinant vanishes if any two of x; y; z are equal

Δ=sinxsinysinzcosxcosycoszcos3xcos3ycos3z\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, Δ=cosxcosycosztanxtanytanz111cos2xcos2ycos2z\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| =cosxcosycosztanxtanytanxtanztany100cos2xcos2ycos2xcos2zcos2y(C3C3C2,C2C2C1)=\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, Δ=cosxcosycosztanytanxtanztanycos2ycos2xcos2zcos2y\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| =cosxcosycoszsin(yx)cosxcosysin(zy)cosy.coszsin2xsin2ysin2ysin2z=cosz.sin(xy)cosx.sin(yz)sin(x+y).sin(xy)sin(y+z).sin(yz)=-\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(xy)sin(yz)coszcosxsin(x+y)sin(y+z)=sin(xy)sin(yz)[sin(y+z)coszsin(x+y)cosx)]=\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] =12sin(xy)sin(yz)[{sin(y+2z)+siny}{sin(y+2x)+siny}]=\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] =12sin(xy)sin(yz)[sin(y+2x)]=12sin(xy)sin(yz)2cos(x+y+z)sin(zx)=\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(xy)sin(yz)sin(zx)cos(x+y+z)=\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=π2.x+y+z=\frac{\pi }{2}.

Hence proved.

Illustration 5: Show that a2+x2abcxac+bxab+cxb2+x2bcaxacbxbc+axc2+x2=xcbcxabax2.\left| \begin{matrix} {{a}^{2}}+{{x}^{2}} & ab-cx & ac+bx \\ ab+cx & {{b}^{2}}+{{x}^{2}} & bc-ax \\ ac-bx & bc+ax & {{c}^{2}}+{{x}^{2}} \\ \end{matrix} \right|={{\left| \begin{matrix} x & c & -b \\ -c & x & a \\ b & -a & x \\ \end{matrix} \right|}^{2}}.

Solution:

By replacing all elements of L.H.S. to their respective cofactors and using determinant property we will obtain the required result.

Let D =xcbcxabax=\left| \begin{matrix} x & c & -b \\ -c & x & a \\ b & -a & x \\ \end{matrix} \right|

Co-factors of 1st row of D are x2+a2,ab+cx,acbx.{{x}^{2}}+{{a}^{2}},ab+cx,ac-bx. Co-factors of 2nd row of D areabcx,x2+b2,ax+bcab-cx,{{x}^{2}}+{{b}^{2}},ax+bc and co-factors of 3rd row of D are ac+bx,bcax,x2+c2ac+bx,bc-ax,{{x}^{2}}+{{c}^{2}}