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 students solve questions more efficiently.
All Topics in Determinants
- Introduction to Determinants
- Minors and Cofactors
- Properties of Determinants
- System of Linear Equations Using Determinants
- Differentiation and Integration of Determinants
- Standard Determinants
Expressions for Standard Determinants
- \(\begin{array}{l}\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|=\left( a-b \right)\left( b-c \right)\left( c-a \right)\end{array} \)
- \(\begin{array}{l}\left| \begin{matrix} a & b & c \\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ bc & ca & ab \\ \end{matrix} \right|=\left| \begin{matrix} 1 & 1 & 1 \\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ {{a}^{3}} & {{b}^{3}} & {{c}^{3}} \\ \end{matrix} \right|=\left( a-b \right)\left( b-c \right)\left( c-a \right)\left( ab+bc+ca \right)\end{array} \)
- \(\begin{array}{l}\left| \begin{matrix} a & bc & abc \\ b & ca & abc \\ c & ab & abc \\ \end{matrix} \right|=\left| \begin{matrix} a & {{a}^{2}} & {{a}^{3}} \\ b & {{b}^{2}} & {{b}^{3}} \\ c & {{c}^{2}} & {{c}^{3}} \\ \end{matrix} \right|=abc\left( a-b \right)\left( b-c \right)\left( c-a \right);\end{array} \)
- \(\begin{array}{l}\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ {{a}^{3}} & {{b}^{3}} & {{c}^{3}} \\ \end{matrix} \right|=\left( a-b \right)\left( b-c \right)\left( c-a \right)\left( a+b+c \right)\end{array} \)
- \(\begin{array}{l}\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|=-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}+3abc\end{array} \)
- Determinant of order 3 Γ 3 \(\begin{array}{l}=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|={{a}_{1}}\left| \begin{matrix} {{b}_{2}} & {{c}_{2}} \\ {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|-{{b}_{1}}\left| \begin{matrix} {{a}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{c}_{3}} \\ \end{matrix} \right|+{{c}_{1}}\left| \begin{matrix} {{a}_{2}} & {{b}_{2}} \\ {{b}_{3}} & {{b}_{3}} \\ \end{matrix} \right|\end{array} \)
- In the determinant D = \(\begin{array}{l}\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|,\end{array} \)minor of a12 is denoted as\(\begin{array}{l}{{M}_{12}}=\left| \begin{matrix} {{a}_{21}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{33}} \\ \end{matrix} \right|\end{array} \)and so on.
- Cofactor of an element \(\begin{array}{l}{{a}_{i\,j}}={{C}_{i\,j}}={{\left( -1 \right)}^{i+j}}{{M}_{i\,j}}\end{array} \)
Evaluation of the Determinant Using the SARRUS Diagram
If
\(\begin{array}{l}A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right]\end{array} \)
is a square matrix of order 3, then the below diagram is a Sarrus Diagram, obtained by adjoining the first two columns on the right and drawing dark and dotted lines as shown.
The value of the determinant is
\(\begin{array}{l}\left( {{a}_{11}}{{a}_{22}}{{a}_{33}}+{{a}_{12}}{{a}_{23}}{{a}_{31}}+{{a}_{13}}{{a}_{21}}{{a}_{32}} \right)-\left( {{a}_{13}}{{a}_{22}}{{a}_{31}}+{{a}_{11}}{{a}_{23}}{{a}_{32}}+{{a}_{12}}{{a}_{21}}{{a}_{33}} \right).\end{array} \)
Recommended VideoΒ
Solved Problems on Determinant
Illustration 1: Evaluate the determinant
\(\begin{array}{l}\Delta =\left| \begin{matrix} \sqrt{p}+\sqrt{q} & 2\sqrt{r} & \sqrt{r} \\ \sqrt{qr}+\sqrt{2p} & r & \sqrt{2r} \\ q+\sqrt{pr} & \sqrt{qr} & r \\ \end{matrix} \right|,\end{array} \)
where p, q and r are positive real numbers.
Solution:
Taking βr common from C2 and C3 of the given determinant using the scalar multiple property and then expanding it using the invariance property, we can evaluate the given problem.
We get
\(\begin{array}{l}\Delta =r\left| \begin{matrix} \sqrt{p}+\sqrt{q} & 2 & 1 \\ \sqrt{qr}+\sqrt{2p} & \sqrt{r} & \sqrt{2} \\ q+\sqrt{pr} & \sqrt{q} & \sqrt{r} \\ \end{matrix} \right|\end{array} \)
\(\begin{array}{l}\text{Applying}\ {{C}_{1}}\to {{C}_{1}}-\sqrt{q}{{C}_{2}}-\sqrt{p}{{C}_{3}},\ \text{we get};\end{array} \)
\(\begin{array}{l}\Delta=r\left| \begin{matrix} -\sqrt{q} & 2 & 1 \\ 0 & \sqrt{r} & \sqrt{2} \\ 0 & \sqrt{q} & \sqrt{r} \\ \end{matrix} \right|=-r\sqrt{q}\left( r-\sqrt{2}q \right).\end{array} \)
Illustration 2: Let a, b, and c be positive and not equal. Show that the value of the determinant
\(\begin{array}{l}\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|\ \text{is negative.}\end{array} \)
Solution:
By applying invariance and scalar multiple properties to the given determinant, we can get the required result.
\(\begin{array}{l}D=\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|;\;\;then\;\;D=\left| \begin{matrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \\ \end{matrix} \right| \;\;\;\left[ {{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}} \right]\end{array} \)
\(\begin{array}{l}=\left( a+b+c \right)\left| \begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \\ \end{matrix} \right|\end{array} \)
[Taking (a+b+c) common from the first column]
\(\begin{array}{l}=\left( a+b+c \right)\left| \begin{matrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \\ \end{matrix} \right|\;\; \left[ {{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\,\,and\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \right]\end{array} \)
\(\begin{array}{l}=\left( a+b+c+ \right)\left[ \left( c-b \right)\left( b-c \right)\left( a-b \right)\left( a-c \right) \right]\\=\left( a+b+c+ \right)\left[ bc+ca+ab-{{a}^{2}}-{{b}^{2}}-{{c}^{2}} \right]\end{array} \)
\(\begin{array}{l}=-\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-bc-ca-ab \right)\\=-\frac{1}{2}\left( a+b+c \right)\left( 2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2bc-2ca-2ab \right)\end{array} \)
\(\begin{array}{l}=-\frac{1}{2}\left( a+b+c \right)\left[ \left( {{a}^{2}}+{{b}^{2}}-2ab \right)+\left( {{b}^{2}}+{{c}^{2}}-2bc \right)+\left( {{c}^{2}}+{{a}^{2}}-2ac \right) \right]\end{array} \)
\(\begin{array}{l}=-\frac{1}{2}\left( a+b+c \right)\left[ {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}} \right]….(i)\end{array} \)
a, b, c are positive β a + b + c > 0
a, b, c are unequal β (a – b)2 + (b – c)2 + (c – a)2 > 0 ….(ii)
From (i) and (ii), Ξ <0.
Illustration 3: Show that
\(\begin{array}{l}\Delta =\left| \begin{matrix} 1 & {{\cos }^{2}}\left( \alpha -\beta \right) & {{\cos }^{2}}\left( \alpha -\gamma \right) \\ {{\cos }^{2}}\left( \beta -\alpha \right) & 1 & {{\cos }^{2}}\left( \beta -\gamma \right) \\ {{\cos }^{2}}\left( \gamma -\alpha \right) & {{\cos }^{2}}\left( \gamma -\beta \right) & 1 \\ \end{matrix} \right|\end{array} \)
\(\begin{array}{l}=2{{\sin }^{2}}\left( \beta -\gamma \right){{\sin }^{2}}\left( \gamma -\alpha \right){{\sin }^{2}}\left( \alpha -\beta \right)\end{array} \)
Solution:
By putting
\(\begin{array}{l}\beta -\gamma =A,\gamma -\alpha =B,\alpha -\beta =C\end{array} \)
and then by using switching and invariance properties, we can prove the above problem.
We can write Ξ as,
\(\begin{array}{l}\Delta =\left| \begin{matrix} 1 & {{\cos }^{2}}C & {{\cos }^{2}}B \\ {{\cos }^{2}}C & 1 & {{\cos }^{2}}A \\ {{\cos }^{2}}B & {{\cos }^{2}}A & 1 \\ \end{matrix} \right|\end{array} \)
(Note that A + B + C = 0).
Using C2 β C2 – C1, C1 β C3 – C1, we get
\(\begin{array}{l}\Delta =\left| \begin{matrix} 1 & -{{\sin }^{2}}C & -{{\sin }^{2}}B \\ {{\cos }^{2}}C & {{\sin }^{2}}C & {{\cos }^{2}}A-{{\cos }^{2}}C \\ {{\cos }^{2}}B & {{\cos }^{2}}A-{{\cos }^{2}}B & {{\sin }^{2}}B \\ \end{matrix} \right|\;\;\left| \begin{matrix} 1 & -{{\sin }^{2}}C & -{{\sin }^{2}}B \\ {{\cos }^{2}}C & {{\sin }^{2}}C & \sin B\sin \left( C-A \right) \\ {{\cos }^{2}}B & \sin C\sin \left( B-A \right) & {{\sin }^{2}}B \\ \end{matrix} \right|\end{array} \)
\(\begin{array}{l}={{\left( -1 \right)}^{2}}\left| \begin{matrix} 1 & {{\sin }^{2}}C & {{\sin }^{2}}B \\ {{\cos }^{2}}C & -{{\sin }^{2}}C & \sin B\sin \left( -A \right) \\ {{\cos }^{2}}B & \sin C\sin \left( B-A \right) & -{{\sin }^{2}}B \\ \end{matrix} \right|\end{array} \)
Since,
\(\begin{array}{l}\left[ \,{{\cos }^{2}}A-{{\cos }^{2}}B=\sin \left( A+B \right)\sin \left( B-A \right),A+B=-C,C+A=-B \right]; \;\;=\sin C\sin B\left[ {{\Delta }_{1}} \right]\end{array} \)
Where
\(\begin{array}{l}{{\Delta }_{1}}=\left| \begin{matrix} 1 & {{\sin }^{2}}C & \sin B \\ {{\cos }^{2}}C & -{{\sin }^{2}}C & \sin (C-A) \\ {{\cos }^{2}}B & \sin (B-A) & -\sin B \\ \end{matrix} \right|\; Using\; {{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\; and \;{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\end{array} \)
Using R2 β R2 – R1 and R3 β R3 – R1, we get
\(\begin{array}{l}{{\Delta }_{1}}=\left| \begin{matrix} 1 & \sin C & \sin B \\ -{{\sin }^{2}}C & -2{{\sin }^{2}}C & \sin (C-A)-sinB \\ -{{\sin }^{2}}B & \sin (B-A)-sinC & -2{{\sin }^{2}}B \\ \end{matrix} \right|\end{array} \)
But sin (C β A) β sin B = sin (C β A) + sin (C + A) = 2 sin C cos A and sin (B β A) β sin C = 2 sin B cos A
Therefore,
\(\begin{array}{l}{{\Delta }_{1}}=\sin C\;\sin B\;{{\Delta }_{2}}\; where \;{{\Delta }_{2}}=\left| \begin{matrix} 1 & \sin C & \sin B \\ \sin C & 2 & -2\cos A \\ \sin B & -2\cos A & 2 \\ \end{matrix} \right|\end{array} \)
ApplyingΒ R2 β R2 – sin C R1 and R3 β R3 – sin B R1, we get
\(\begin{array}{l}{{\Delta }_{2}}=\left| \begin{matrix} 1 & \sin C & \sin B \\ 0 & 2-{{\sin }^{2}}C & -2\cos A-\sin B\sin C \\ 0 & -2\cos A-\sin B\sin C & 2-{{\sin }^{2}}B \\ \end{matrix} \right|\\=\left( 2-{{\sin }^{2}}B \right)\left( 2-\sin C \right)-{{\left( 2\cos A+\sin B\sin C \right)}^{2}}\end{array} \)
\(\begin{array}{l}=4-2{{\sin }^{2}}B-2{{\sin }^{2}}C+{{\sin }^{2}}B{{\sin }^{2}}C-\left[ 4{{\cos }^{2}}A+4\cos A\sin B\sin C+{{\sin }^{2}}B{{\sin }^{2}}C \right]\end{array} \)
\(\begin{array}{l}=4{{\sin }^{2}}A-2{{\sin }^{2}}B-2{{\sin }^{2}}C-4\cos A\sin B\sin C\end{array} \)
\(\begin{array}{l}=2{{\sin }^{2}}A-2\left[ {{\sin }^{2}}B+{{\sin }^{2}}C-{{\sin }^{2}}A+2\cos A\sin B\sin C \right]\end{array} \)
But A + B + C = 0 implies;
\(\begin{array}{l}{{\sin }^{2}}B+{{\sin }^{2}}C-{{\sin }^{2}}A=-2\cos A\sin B\sin C\end{array} \)
\(\begin{array}{l}{{\Delta }_{2}}=2{{\sin }^{2}}A;\end{array} \)
Hence, \(\begin{array}{l}D=\sin C\sin B{{\Delta }_{1}}={{\sin }^{2}}C{{\sin }^{2}}B{{\Delta }_{2}}\end{array} \)
\(\begin{array}{l}=2{{\sin }^{2}}A\;{{\sin }^{2}}B\;{{\sin }^{2}}C=2{{\sin }^{2}}\left( \alpha -\beta \right)\;{{\sin }^{2}}\left( \beta -\gamma \right)\;{{\sin }^{2}}\left( \gamma -\alpha \right).\end{array} \)
Illustration 4: Prove that the following determinant vanishes if any two of x, y, z are equal
\(\begin{array}{l}\Delta =\left| \begin{matrix} \sin x & \sin y & \sin z \\ \cos x & \cos y & \cos z \\ {{\cos }^{3}}x & {{\cos }^{3}}y & {{\cos }^{3}}z \\ \end{matrix} \right|\end{array} \)
Solution:
Taking cos x, cos y, and cos z common from the first, second and third columns using scalar multiple and then using the invariance property, we can prove the given statement.
Here,
\(\begin{array}{l}\Delta =\cos x\cos y\cos z\left| \begin{matrix} \tan x & \tan y & \tan z \\ 1 & 1 & 1 \\ {{\cos }^{2}}x & {{\cos }^{2}}y & {{\cos }^{2}}z \\ \end{matrix} \right|\end{array} \)
\(\begin{array}{l}=\cos x\cos y\cos z\left| \begin{matrix} \tan x & \tan y-\tan x & \tan z-\tan y \\ 1 & 0 & 0 \\ {{\cos }^{2}}x & {{\cos }^{2}}y-{{\cos }^{2}}x & {{\cos }^{2}}z-{{\cos }^{2}}y \\ \end{matrix} \right|\left( {{C}_{3}}\to {{C}_{3}}-{{C}_{2}},{{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \right)\end{array} \)
Expanding along R2,
\(\begin{array}{l}\Delta =-\cos x\cos y\cos z\left| \begin{matrix} \tan y-\tan x & \tan z-\tan y \\ {{\cos }^{2}}y-{{\cos }^{2}}x & {{\cos }^{2}}z-{{\cos }^{2}}y \\ \end{matrix} \right|\end{array} \)
\(\begin{array}{l}=-\cos x\cos y\cos z\left| \begin{matrix} \frac{\sin \left( y-x \right)}{\cos x\cos y} & \frac{\sin \left( z-y \right)}{\cos y.\cos z} \\ {{\sin }^{2}}x-{{\sin }^{2}}y & {{\sin }^{2}}y-{{\sin }^{2}}z \\ \end{matrix} \right|\\=\left| \begin{matrix} \cos z.\sin \left( x-y \right) & \cos x.\sin \left( y-z \right) \\ \sin \left( x+y \right).\sin \left( x-y \right) & \sin \left( y+z \right).\sin \left( y-z \right) \\ \end{matrix} \right|….(i)\end{array} \)
\(\begin{array}{l}=\sin \left( x-y \right)\sin \left( y-z \right)\left| \begin{matrix} \cos z & \cos x \\ \sin \left( x+y \right) & \sin \left( y+z \right) \\ \end{matrix} \right|\\=\sin \left( x-y \right)\sin \left( y-z \right)\left[ \sin \left( y+z \right)\cos z-\sin \left( x+y \right)\cos \,x) \right]\end{array} \)
\(\begin{array}{l}=\frac{1}{2}\sin \left( x-y \right)\sin \left( y-z \right)\left[ \left\{ \sin \left( y+2z \right)+\sin y \right\}-\left\{ \sin \left( y+2x \right)+\sin y \right\} \right]\end{array} \)
\(\begin{array}{l}=\frac{1}{2}\sin \left( x-y \right)\sin \left( y-z \right)\left[ \sin \left( y+2x \right) \right]\\=\frac{1}{2}\sin \left( x-y \right)\sin \left( y-z \right)2\cos \left( x+y+z \right)\sin \left( z-x \right)\end{array} \)
\(\begin{array}{l}=\sin \left( x-y \right)\sin \left( y-z \right)\sin \left( z-x \right)\cos \left( x+y+z \right)\end{array} \)
Clearly, Ξ is zero when any two of x, y, z are equal or
\(\begin{array}{l}x+y+z=\frac{\pi }{2}.\end{array} \)
Hence proved.
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Frequently Asked Questions
Q1
When 2 rows or columns are interchanged, what happens to a determinant?
When 2 rows or columns are interchanged, the determinant changes its sign.
Q2
What is the value of the determinant, if all the elements of a row or column are zero?
If all the elements of a row or column are zero, then the determinant is equal to zero.
Q3
What are determinants used for?
Determinants are used to give formulas for the area or volume of certain geometric figures and also to find the inverse of a matrix.
Q4
Are determinants always positive?
No, determinants can be positive, negative or zero.
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