NCERT Solutions for Class 12 Maths Chapter 4: Determinants - Exercise 4.4

Exercise (4.4)

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

1. $\begin{bmatrix} 2 & -4  \\0 & 3 \\\end{bmatrix}$

\[\left| \begin{matrix}\text{2} & \text{-4}  \\\text{0} & \text{3}  \\\end{matrix} \right|\]

Ans: Given, \[\left| \begin{matrix}\text{2} & \text{-4}  \\ \text{0} & \text{3}  \\\end{matrix} \right|\]

Minor of an element is termed as the determinant obtained by removing the row and the column in which that element is present.

Minor of element \[{{\text{a}}_{ij}}\] is denoted by \[{{M}_{ij}}\],where

$i$ and $j$ denotes the row and the column of the determinant respectively.

\[\therefore {{\text{M}}_{11}}\text{=3}\]

\[{{\text{M}}_{12}}\text{=0}\]

\[{{\text{M}}_{21}}\text{=-4}\]

\[{{\text{M}}_{22}}\text{=2}\]

Cofactor of an element is termed as the determinant obtained by removing the row and the column in which that element is present preceded by a negative or a positive sign based on the position of the element.

Thus,

Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{\text{1+1}}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}={{\left( \text{-1} \right)}^{\text{2}}}\left( \text{3} \right)\]

\[\therefore {{\text{A}}_{11}}\text{=3}\]

Similarly,

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{1+2}}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( \text{0} \right)\]

\[\therefore {{\text{A}}_{12}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( \text{-4} \right)\]

\[\therefore {{\text{A}}_{21}}\text{=4}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{2+2}}}{{M}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{4}}}\left( \text{2} \right)\]

\[\therefore {{\text{A}}_{22}}\text{=2}\]

2. \[\left| \begin{matrix} \text{a} & \text{c}  \\\text{b} & \text{d}  \\\end{matrix} \right|\]

Ans: Given, \[\left| \begin{matrix}\text{a} & \text{c}  \\\text{b} & \text{d}  \\\end{matrix} \right|\]

Minor of element \[{{\text{a}}_{ij}}\] is denoted by \[{{M}_{ij}}\].

\[\therefore {{\text{M}}_{11}}\text{=d}\]

\[{{\text{M}}_{12}}\text{=b}\]

\[{{\text{M}}_{21}}\text{=c}\]

\[{{\text{M}}_{22}}\text{=a}\]

Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{\text{1+1}}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}={{\left( \text{-1} \right)}^{\text{2}}}\left( d \right)\]

\[\therefore {{\text{A}}_{11}}\text{=d}\]

Similarly,

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{1+2}}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( b \right)\]

\[\therefore {{\text{A}}_{12}}\text{=}-b\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( c \right)\]

\[\therefore {{\text{A}}_{21}}\text{=}-\text{c}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{2+2}}}{{M}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{4}}}\left( a \right)\]

\[\therefore {{\text{A}}_{22}}\text{=a}\]

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

1. \[\left| \begin{matrix}\text{1} & \text{0} & \text{0}  \\\text{0} & \text{1} &\text{0}  \\\text{0} & \text{0} & \text{1}  \\\end{matrix} \right|\]

Ans: Given determinant, 

\[\left| \begin{matrix}\text{1} & \text{0} & \text{0}  \\\text{0} & \text{1} & \text{0}  \\\text{0} & \text{0} & \text{1}  \\\end{matrix} \right|\].

Minor of element \[{{\text{a}}_{ij}}\] is denoted by \[{{M}_{ij}}\].

\[\therefore {{\text{M}}_{11}}\text{=}\left| \begin{matrix}\text{1} & \text{0}  \\ \text{0} & \text{1}  \\\end{matrix} \right|\text{=1}\]

\[\Rightarrow {{\text{M}}_{12}}\text{=}\left| \begin{matrix}\text{0} & \text{0}  \\\text{0} & \text{1}  \\\end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{13}}\text{=}\left| \begin{matrix} \text{0} & \text{1}  \\\text{0} & \text{0}  \\\end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{21}}\text{=}\left| \begin{matrix}\text{0} & \text{0}  \\\text{0} & \text{1}  \\\end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{22}}\text{=}\left| \begin{matrix}\text{1} & \text{0}  \\\text{0} & \text{1}  \\\end{matrix} \right|\text{=1}\]

\[\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix}\text{1} & \text{0}  \\\text{0} & \text{0}  \\\end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{31}}\text{=}\left| \begin{matrix}\text{0} & \text{0}  \\\text{1} & \text{0}  \\\end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{32}}\text{=}\left| \begin{matrix}\text{1} & \text{0}  \\ \text{0} & \text{0}  \\\end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{33}}\text{=}\left| \begin{matrix}\text{1} & \text{0}  \\\text{0} & \text{1}  \\\end{matrix} \right|\text{=1}\]

Cofactor of \[{{\text{a}}_{ij}}\] is \[{{A}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{M}_{ij}}\]. 

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}=1\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}=0\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}=0\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}=0\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}=1\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}=0\]

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}=0\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}=0\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}=1\]

2. \[\left| \begin{matrix}\text{1} & \text{0} & \text{4}  \\\text{3} & \text{5} & \text{-1}  \\\text{0} & \text{1} & \text{2}  \\\end{matrix} \right|\] 

Ans: Given determinant, \[\left| \begin{matrix}\text{1} & \text{0} & \text{4}  \\\text{3} & \text{5} & \text{-1}  \\\text{0} & \text{1} & \text{2}  \\\end{matrix} \right|\] 

Minor of element \[{{\text{a}}_{ij}}\] is denoted by \[{{M}_{ij}}\].

\[\Rightarrow {{\text{M}}_{11}}\text{=}\left| \begin{matrix}\text{5} & \text{-1}  \\\text{1} & \text{2}  \\\end{matrix} \right|\text{=10+1=11}\]

\[\Rightarrow {{\text{M}}_{12}}\text{=}\left| \begin{matrix}\text{3} & \text{-1}  \\\text{0} & \text{2}  \\\end{matrix} \right|\text{=6-0=6}\]

\[\Rightarrow {{\text{M}}_{13}}\text{=}\left| \begin{matrix}\text{3} & \text{5}  \\\text{0} & \text{1}  \\\end{matrix} \right|\text{=3-0=3}\]

\[\Rightarrow {{\text{M}}_{21}}\text{=}\left| \begin{matrix}\text{0} & \text{4}  \\\text{1} & \text{2}  \\\end{matrix} \right|\text{=0-4=-4}\]

\[\Rightarrow {{\text{M}}_{22}}\text{=}\left| \begin{matrix}\text{1} & \text{4}  \\\text{0} & \text{2}  \\\end{matrix} \right|\text{=2-0=2}\]

\[\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix}\text{1} & \text{0}  \\\text{0} & \text{1}  \\\end{matrix} \right|\text{=1-0=1}\]

\[\Rightarrow {{\text{M}}_{31}}\text{=}\left| \begin{matrix}\text{0} & \text{4}  \\\text{5} & \text{-1}  \\\end{matrix} \right|\text{=0-20=-20}\]

\[\Rightarrow {{\text{M}}_{32}}\text{=}\left| \begin{matrix}\text{1} & \text{4}  \\\text{3} & \text{-1}  \\\end{matrix} \right|\text{=-1-12=-13}\]

\[\Rightarrow {{\text{M}}_{33}}\text{=}\left| \begin{matrix}\text{1} & \text{0}  \\\text{3} & \text{5}  \\\end{matrix} \right|\text{=5-0=5}\]

Cofactor of \[{{\text{a}}_{ij}}\] is \[{{A}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{M}_{ij}}\]. 

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}=11\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}=-6\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}=3\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}=4\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}=2\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}=-1\]

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}=-20\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}=13\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}=5\]

3. Using Cofactors of elements of second row, evaluate \[\text{ }\!\!\Delta\] =  \[\left| \begin{matrix}\text{5} & \text{3} & \text{8}  \\ \text{2} & \text{0} &\text{1}  \\\text{1} & \text{2} & \text{3}  \\\end{matrix} \right|\]

Ans: Given determinant, \[\left| \begin{matrix}\text{5} & \text{3} & \text{8}  \\\text{2} & \text{0} & \text{1}  \\\text{1} & \text{2} & \text{3}  \\\end{matrix} \right|\]

Determining the minors and cofactors, we get:

\[\Rightarrow {{\text{M}}_{21}}\text{=}\left| \begin{matrix}\text{3} & \text{8}  \\\text{2} & \text{3}  \\\end{matrix} \right|=9-16=-7\]

\[\therefore {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{21}}\text{=7}\]

\[\Rightarrow {{\text{M}}_{22}}\text{=}\left| \begin{matrix}\text{5} & \text{8}  \\\text{1} & \text{3}  \\\end{matrix} \right|=15-8=7\]

\[\therefore {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{2+2}}}{{M}_{22}}\text{=7}\]

\[\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix}\text{5} & \text{3}  \\\text{1} & \text{2}  \\\end{matrix} \right|=10-3=7\]

\[\therefore {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{23}}=-7\]

Since, \[\text{ }\!\!\Delta\!\!\text{ }\] is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

\[\therefore\Delta={{a}_{21}}{{A}_{21}}+{{a}_{22}}{{A}_{22}}+{{a}_{23}}{{A}_{23}}\]

\[\Rightarrow \Delta \text{=2}\left( \text{7} \right)\text{+0}\left( \text{7} \right)\text{+1}\left( \text{-7} \right)\]

Hence, \[\Delta \text{=7}\].

4. Using Cofactors of elements of third column, evaluate \[\text{ }\!\!\Delta\] = \[\left| \begin{matrix}\text{1} & \text{x} & \text{yz}  \\ \text{1} & \text{y} & \text{zx}  \\\text{1} & \text{z} & \text{xy}  \\\end{matrix} \right|\]

Ans: Given determinant, 

\[\left| \begin{matrix}\text{1} & \text{x} & \text{yz}  \\\text{1} & \text{y} & \text{zx}  \\\text{1} & \text{z} & \text{xy}  \\\end{matrix} \right|\]

Determining the minors and cofactors, we get:

\[\Rightarrow {{\text{M}}_{13}}\text{=}\left| \begin{matrix}1 & y  \\1 & z  \\\end{matrix} \right|=z-y\]

\[\therefore {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{\text{1+3}}}{{M}_{13}}\text{=}\left( z-y \right)\]

\[\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix}1 & x  \\\text{1} & z  \\\end{matrix} \right|=z-x\]

\[\therefore {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{\text{2+3}}}{{M}_{23}}\text{=}\left( x-z \right)\]

\[\Rightarrow {{\text{M}}_{33}}\text{=}\left| \begin{matrix}1 & x  \\\text{1} & y  \\\end{matrix} \right|=y-x\]

\[\therefore {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{\text{3+3}}}{{M}_{33}}=\left( y-x \right)\]

Since, \[\text{ }\!\!\Delta\!\!\text{ }\] is equal to the sum of the product of the elements of the first row with their corresponding cofactors. 

\[\therefore \Delta ={{a}_{13}}{{A}_{13}}+{{a}_{23}}{{A}_{23}}+{{a}_{33}}{{A}_{33}}\]

\[\Rightarrow \Delta =yz\left( z-y \right)+zx\left( x-z \right)+xy\left( y-x \right)\]

\[\Rightarrow \Delta =y{{z}^{2}}-{{y}^{2}}z+{{x}^{2}}z-x{{z}^{2}}+x{{y}^{2}}-{{x}^{2}}y\]

\[\Rightarrow \Delta =\left( {{x}^{2}}z-{{y}^{2}}z \right)+\left( y{{z}^{2}}-x{{z}^{2}} \right)+\left( x{{y}^{2}}-{{x}^{2}}y \right)\]

\[\Rightarrow \Delta =\left( x-y \right)\left[ zx+zy-{{z}^{2}}-xy \right]\]

\[\Rightarrow \Delta =\left( x-y \right)\left[ z\left( x-z \right)+y\left( z-x \right) \right]\]

Thus, \[\text{ }\!\!\Delta\!\!\text{ =}\left( \text{x-y} \right)\left( \text{y-z} \right)\left( \text{z-x} \right)\].

5. If evaluate 

\[\mathbf{\text{ }\!\!\Delta\!\!\text{ =}\left|\begin{matrix}\text{a}_{11} & \text{a}_{12} & \text{a}_{13} \\\text{a}_{21} & \text{a}_{22} & \text{a}_{23} \\ \text{a}_{31} & \text{a}_{32} & \text{a}_{33} \\\end{matrix} \right|}\]

and \[{{\text{A}}_{ij}}\] is cofactors of \[{{\text{a}}_{ij}}\], then value of \[\text{ }\!\!\Delta\] is given by 

1. \[\mathbf{{{\text{a}}_{11}}{{A}_{31}}+{{a}_{12}}{{A}_{32}}+{{a}_{13}}{{A}_{33}}}\]

2. \[\mathbf{{{\text{a}}_{11}}{{A}_{11}}+{{a}_{12}}{{A}_{21}}+{{a}_{13}}{{A}_{31}}}\]

3. \[\mathbf{{{\text{a}}_{21}}{{A}_{11}}+{{a}_{22}}{{A}_{12}}+{{a}_{23}}{{A}_{13}}}\]

4. \[\mathbf{{{\text{a}}_{11}}{{A}_{11}}+{{a}_{21}}{{A}_{21}}+{{a}_{31}}{{A}_{31}}}\]

Ans: D) \[{{\text{a}}_{11}}{{A}_{11}}+{{a}_{21}}{{A}_{21}}+{{a}_{31}}{{A}_{31}}\]

NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

NCERT Solution Class 12 Maths of Chapter 4 All Exercises

Chapter 4 of Class 12th NCERT 

Chapter 4 of Class 12th NCERT covers the topic of determinants. In this study material, there is a specified emphasis on Exercise 4.4 of this chapter that includes the following topics:

1. Introduction

2. Determinants 

3. Matrix

4. 2 × 2 Matrix Calculation

5. 3 × 3 Matrix Calculation

6. Minor 

7. Cofactor of a Minor 

Introduction

The Exercise 4.4 Class 12 Maths NCERT Solutions cover the concept of Determinants. Determinants are studied as a topic of linear algebra. Studying determinant is paramount to understand linear equations- its properties and system. The invention of this concept made to simplify arriving at solutions when it came to relatively large sets of concurrent equations while using the computer. This chapter gives an insight into how to compute a given structure of linear equations with the help of determinants. One can quickly learn how to calculate the determinant of 2 × 2 and 3 × 3 matrices through a reading of this chapter. Class 12 Exercise 4.4 has several problems on how to determine a determinant for matrix square of 2 × 2 and 3 × 3, which has been thoroughly covered in the given solutions.

Determinant 

This chapter, in most simple words, explains a determinant as a scalar quantity or value that can be arrived at by simplifying or computing a matrix square. Here, a determinant is symbolized with the help of vertical lines on either side of a value. For example, the determinant of a matrix ‘M’ can be seen denoted as |M|. A thorough understanding of this chapter leads to understanding that the calculation of a determinant is rather easy. All that is needed is a matrix square i.e., an equal number of rows and columns. Then mere simple arithmetic is used to arrive at a determinant.

Matrix 

A Matrix in Mathematics can be defined as an arrangement of values or numbers in a row and column format where particular operations of multiplication and addition are well-defined. A matrix can be in 2 × 2 formats, 3 × 3 formats, or even 4 × 4 formats. Every value in a matrix has a distinct location. The Exercise 4.4 Class 12 Maths covers various questions that require simplification of the values in a matrix to calculate a determinant. 

 2 × 2 Matrix Calculation

Several questions in Class 12 Maths Ex 4.4. Solutions require computing a determinant of a 2 × 2 Matrix. If one notices correctly, a 2 × 2 Matrix looks like:

M= p q r s

where, 

|M| = determinant

The study material easily explains the calculation of 2 × 2 matrices using a simple arithmetic formula described below: 

and, 

|M| = ps – qr

The exercise involves various questions that require one to apply the above method to arrive at determinant values to give enough practice and confidence in solving these questions.  

3 × 3 Matrices Calculation

As a student, when you will proceed further with the Exercise 4.4 Class 12th Maths a student will encounter many questions that will require the calculation of 3 × 3 matrices. A 3 × 3 matrix will look somewhat like: 

M= p   q   r s   t   u v   w   x   

where, 

|M| = determinant

and the calculation of such determinant will require the application of the formula 

|M| = p (tx - uw) – q (sx – uv) + r ( sw – th )

The calculation of a 3 × 3 matrix can be tough, and hence, the solutions are given in the study material break the calculations into a step by step method to make it understandable and easier to comprehend. 

Minor 

Some questions in this Exercise 4.4 Maths Class 12 require one to write a minor of elements in a determinant. A minor in this topic of linear algebra can be understood as a determinant of a particular square matrix, which is formulated by the elimination or deletion of a single row and column from a large matrix square. It is necessary to know that a minor plays an essential role in this concept because, in every square matrix, every element has its minor.

Cofactor of a Minor 

In this chapter, you will find some solutions that relate to the calculation of the minor and cofactors of particular determinants. A cofactor is also known as the signed minor. For an element say ‘mij', the cofactor shall be denoted as ‘Mij’ where the said cofactor shall be defined with the help of the formula M = (-1)i+j B, where B is minor of the particular element 'mij'.

The calculation of minor and cofactor in this chapter has been covered in a very simple and thorough manner in the entire chapter. The Class 12 ex 4.4 is a very detailed material that includes not only the basic calculation of determinants but also gets into a complete step by step calculation of minors and cofactors of a determinant.  

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