How To Find Determinant Of 4x4 Matrix Shortcut

Determinant of a Matrix

Past Catalin David

Definition

The determinant of a square matrix A is the integer obtained through a range of methods using the elements of the matrix.

Notation

Let $ A = \begin{pmatrix} 1 & four & 2 \\ 5 & 3 & seven \\ 6 & 2 & 1 \finish{pmatrix}$

$det(A) = \left|A\right| = \brainstorm{vmatrix} 1 & 4 & 2 \\ 5 & 3 & 7 \\ half dozen & 2 & 1 \cease{vmatrix}$

Properties of the Determinant

If a matrix has a row or a column with all elements equal to 0 and then its determinant is 0.
Example 12
$\brainstorm{vmatrix} 1 & 4 & 2\\ 0 & 0 & 0\\ 3 & nine & 5 \end{vmatrix}= 0$ or $\begin{vmatrix} i & 4 & 0\\ 4 & 2 & 0\\ 3 & 9 & 0 \stop{vmatrix}=0$
If a matrix has two equal rows or 2 equal columns then its determinant is 0.
Example 13
$\begin{vmatrix} ane & four & 2\\ 1 & 4 & 2\\ 3 & 9 & v \end{vmatrix}= 0$ or $\begin{vmatrix} 1 & iv & 1\\ 4 & 2 & 4\\ 3 & ix & 3 \end{vmatrix}=0$
If a matrix has ii proportional rows or two proportional columns then its determinant is 0.
Instance 14
$\brainstorm{vmatrix} 1 & 4 & 2\\ ii & 8 & four\\ 3 & 9 & 5 \cease{vmatrix}= 0$ (the first two lines are proportional)
or
$\begin{vmatrix} 8 & four & vii\\ 4 & two & iii\\ 18 & ix & viii \end{vmatrix}=0$ (the first two columns are proportional)
If a line or a column is the sum or the divergence of other lines, respectively columns and so the determinant is 0.
Example 15
$\brainstorm{vmatrix} one & four & ii\\ 7 & 2 & 3\\ 8 & half dozen & 5 \finish{vmatrix}= 0$ $R_{1} +R_{2} =R_{iii}$ or

$ \brainstorm{vmatrix} 9 & 12 & 3\\ 1 & 8 & 7\\ 5 & vii & ii \terminate{vmatrix}=0$ $C_{one}+C_{3}=C_{2}$
In a determinant, nosotros can individually factor integers out of lines and columns.
Example 16
In determinant
$\begin{vmatrix} 3 & 9 & 12\\ 5 & one & viii \\ 7 & 4 & 2 \end{vmatrix}$, we factor 3 out of row ane $(R_{1})$ and we get:
$3 \cdot \begin{vmatrix} ane & three & 4\\ 5 & 1 & 8\\ 7 & 4 & 2 \end{vmatrix}$, so nosotros factor ii out of column 3 $(C_{three})$:
$6\cdot \begin{vmatrix} ane & 3 & two\\ 5 & ane & 4\\ 7 & 4 & 1 \end{vmatrix}$
In a determinant nosotros can add or decrease rows or columns to other rows, respectively columns and the value of the determinant remains the same.
Instance 17
$\begin{vmatrix} 1 & v\\ iii & viii \stop{vmatrix}$ $\xlongequal{R_{1}+R_{2}} \begin{vmatrix} 4 & thirteen\\ 3 & viii \end{vmatrix}$
Example 18
$\brainstorm{vmatrix} one & v\\ 3 & 8 \end{vmatrix}$ $\xlongequal{C_{1}+C_{2}} \begin{vmatrix} 6 & five\\ 11 & eight \cease{vmatrix}$
In a determinant we tin add or decrease multiples of lines or columns.
Example xix
$\begin{vmatrix} 1 & 5\\ 3 & 8 \cease{vmatrix}$ $\xlongequal{2R_{1}+3R_{2}} \begin{vmatrix} 11 & 34\\ 3 & 8 \finish{vmatrix}$

Instance xx
$\begin{vmatrix} i & 5\\ 3 & eight \end{vmatrix}$ $\xlongequal{5C_{i}-C_{two}} \begin{vmatrix} 0 & v\\ 7 & 8 \cease{vmatrix}$
The determinant of a matrix is equal to the determinant of its transpose.
The determinant of the product of two square matrices is equal to the product of the determinants of the given matrices.

Small-scale of a Matrix

The determinant obtained through the elimination of some rows and columns in a square matrix is chosen a minor of that matrix.

Example 21
$A=\brainstorm{pmatrix} 1 & four & 2 \\ 5 & 3 & 7 \\ six & two & ane \end{pmatrix}$

1 of the minors of the matrix A is $\brainstorm{vmatrix} 1 & 4\\ 5 & 3 \stop{vmatrix}$ (obtained through the elimination of row 3 and column iii from the matrix A)

Some other pocket-sized is $\begin{vmatrix} 1 & 2 \\ vi & 1 \end{vmatrix}$ (obtained through the elimination of row 2 and column 2 from the matrix A)

Example 22
$B=\begin{pmatrix} ii & 5 & ane & 3\\ 4 & one & 7 & 9\\ 6 & 8 & 3 & 2\\ vii & 8 & 1 & four \terminate{pmatrix} $

One of the minors of the matrix B is $ \begin{vmatrix} 1 & seven & ix\\ 8 & 3 & 2\\ eight & 1 & iv \stop{vmatrix}$ (obtained through the elimination of row 1 and column 1 from the matrix B)

Another small-scale is $\brainstorm{vmatrix} 1 & vii \\ 8 & iii \end{vmatrix}$ (obtained through the elimination of rows 1 and 4 and columns 1 and 4 from the matrix B)

Let $A= \begin{pmatrix} a_{i,1} & a_{1,two} & a_{ane,3} & . & . & a_{one,n}\\ a_{ii,ane} & a_{2,two} & a_{ii,3} & . & . & a_{2,due north}\\ a_{3,1} & a_{three,2} & a_{three,iii} & . & . & a_{3,n}\\ . & . & . & . & .& .\\ a_{n,1} & a_{northward,2} & a_{n,3} & . & . & a_{n,n} \end{pmatrix}$

We can acquaintance the minor $\Delta_{i,j}$ (obtained through the elimination of row i and cavalcade j) to any element $a_{i,j}$ of the matrix A.

Example 23
$ A = \begin{pmatrix} 4 & 7\\ 2 & 9 \finish{pmatrix}$

We accept to determine the small-scale associated to ii. Since this element is constitute on row 2, column 1, then 2 is $a_{ii,one}$.

Nosotros have to eliminate row ii and column 1 from the matrix A, resulting in

The minor of two is $\Delta_{2,ane} = seven$.

Example 24
$B=\begin{pmatrix} 1 & 4 & 2 \\ 5 & 3 & seven \\ six & 2 & ane \end{pmatrix}$

We accept to determine the small associated to 7. Since this element is found on row 2, column 3, then vii is $a_{ii,3}$.

Nosotros have to eliminate row ii and cavalcade 3 from the matrix B, resulting in

The minor of vii is $\Delta_{2,iii}= \begin{vmatrix} 1 & 4\\ 6 & 2 \terminate{vmatrix}$

Instance 25
$C=\begin{pmatrix} 2 & v & 1 & 3\\ 4 & i & 7 & 9\\ 6 & 8 & 3 & 2\\ 7 & 8 & one & 4 \end{pmatrix}$

We have to determine the minor associated to five. Since this element is found on row 1, column two, and then v is $a_{ane,two}$.

We have to eliminate row i and column 2 from matrix C, resulting in

The minor of 5 is $\Delta_{one,2}= \begin{vmatrix} 4 & 7 & 9\\ half-dozen & three & 2\\ seven & one & iv\\ \cease{vmatrix}$

The Cofactor of an Element of a Matrix

Permit $A=\brainstorm{pmatrix} a_{1,1} & a_{1,ii} & a_{one,3} & . & . & a_{i,n}\\ a_{two,1} & a_{2,ii} & a_{2,3} & . & . & a_{2,due north}\\ a_{iii,one} & a_{three,2} & a_{three,3} & . & . & a_{iii,north}\\ . & . & . & . & .& .\\ a_{n,1} & a_{north,2} & a_{n,3} & . & . & a_{n,n}\\ \end{pmatrix}$

The cofactor $(-one)^{i+j}\cdot\Delta_{i,j}$ corresponds to whatever chemical element $a_{i,j}$ in matrix A. For example, the cofactor $(-one)^{ii+5}\cdot\Delta_{2,5}=(-1)^{vii}\cdot\Delta_{2,5}= -\Delta_{2,5} $ corresponds to element $ a_{ii.5}$

The Club of a Determinant

The order of a determinant is equal to its number of rows and columns.

Example 26
$\begin{vmatrix} one & four\\ half-dozen & two\\ \end{vmatrix}$ (it has 2 lines and two columns, then its social club is two)

Example 27
$\begin{vmatrix} four & seven & 9\\ 6 & 3 & 2\\ 7 & 1 & 4\\ \cease{vmatrix}$ (it has 3 lines and 3 columns, and so its order is 3)

Calculating the Determinant of a Matrix

The determinant of a matrix is equal to the sum of the products of the elements of any ane row or column and their cofactors.

$\left| A\right| = \brainstorm{vmatrix} a_{1,1} & a_{1,2} & a_{one,iii} & . & . & a_{1,n}\\ a_{two,1} & a_{2,two} & a_{2,3} & . & . & a_{two,n}\\ a_{3,1} & a_{3,two} & a_{3,3} & . & . & a_{3,n}\\ . & . & . & . & .& .\\ a_{due north,i} & a_{n,2} & a_{north,iii} & . & . & a_{n,north}\\ \terminate{vmatrix}$

We tin summate the determinant using, for example, row i:

$\left| A\right| =a_{i,1}\cdot(-1)^{i+i}\cdot\Delta_{i,i}$ $+a_{i,2}\cdot(-1)^{i+two}\cdot\Delta_{i,2}+a_{i,iii}\cdot(-i)^{i+3}\cdot\Delta_{i,3}+...$ $+a_{i,due north}\cdot(-1)^{i+north}\cdot\Delta_{i,north}$

Alternatively, we can calculate the determinant using column j:

$\left| A\right| =a_{1,j}\cdot(-1)^{i+j}\cdot\Delta_{1,j}$ $+a_{2,j}\cdot(-1)^{2+j}\cdot\Delta_{ii,j}+a_{3,j}\cdot(-1)^{3+j}\cdot\Delta_{3,j}+...$ $+a_{north,j}\cdot(-1)^{northward+j}\cdot\Delta_{n,j}$

Computing a 2x2 Determinant

We apply row 1 to calculate the determinant.

$\left| A\right| = \brainstorm{vmatrix} a_{1,one} & a_{i,ii}\\ a_{2,1} & a_{2,2}\\ \cease{vmatrix} = a_{ane,1}\cdot(-i)^{1+ane}\cdot\Delta_{i,1}+a_{one.2}\cdot(-ane)^{1+ii}\cdot\Delta_{1,2}=$

$a_{ane,ane}\cdot(-ane)^{2}\cdot\Delta_{one,1}+a_{1.2}\cdot(-1)^{3}\cdot\Delta_{1,two}=a_{1,1}\cdot\Delta_{i,i}-a_{1.2}\cdot\Delta_{1,2}$

Withal, $ \Delta_{1,i}= a_{2,2} $ and $ \Delta_{1,ii}=a_{2,1}$

$ \left| A\right| =a_{i.1} \cdot a_{ii,2}- a_{1.2} \cdot a_{2,one}$

$\color{red}{ \brainstorm{vmatrix} a & b\\ c & d \stop{vmatrix} =a \cdot d - b \cdot c}$

Example 28
$\begin{vmatrix} 2 & 5\\ 3 & 8 \end{vmatrix} =2 \cdot eight - 3 \cdot 5 = 16 -15 =1$

Example 29
$\begin{vmatrix} -4 & seven\\ -2 & 9 \end{vmatrix} =-4 \cdot nine - 7 \cdot (-2) = -36 -(-14) =-36 + 14 = - 22$

Calculating a 3x3 Determinant

We utilise row 1 to calculate the determinant.

$ \left| A\correct| = \begin{vmatrix} a_{1,1} & a_{i,ii} & a_{one,iii}\\ a_{two,one} & a_{2,2} & a_{2,3}\\ a_{3,1} & a_{3,2} & a_{3,three} \end{vmatrix} =$ $a_{1,1}\cdot(-1)^{1+one}\cdot\Delta_{1,1}+a_{ane.ii}\cdot(-1)^{i+ii}\cdot\Delta_{1,2}$ $+a_{1.3}\cdot(-1)^{1+3}\cdot\Delta_{1,three}=$ $=a_{1,1}\cdot(-ane)^{ii}\cdot\Delta_{1,1}+a_{1.2}\cdot(-ane)^{3}\cdot\Delta_{1,2}$ $+a_{one.iii}\cdot(-1)^{iv}\cdot\Delta_{1,iii}=$ $a_{1,1}\cdot\Delta_{1,ane}-a_{1.2}\cdot\Delta_{one,two}+a_{ane.3}\cdot\Delta_{1,3}$

$\Delta_{ane,1}= \begin{vmatrix} a_{2,2} & a_{2,three}\\ a_{3,2} & a_{iii,3} \stop{vmatrix} = a_{ii,2}\cdot a_{3,3}-a_{2,3}\cdot a_{3,2}$

$\Delta_{one,ii}= \brainstorm{vmatrix} a_{2,1} & a_{2,3}\\ a_{iii,1} & a_{3,three} \end{vmatrix} = a_{2,one}\cdot a_{iii,iii}-a_{two,3}\cdot a_{3,1}$

$\Delta_{1,3}= \begin{vmatrix} a_{two,1} & a_{two,2}\\ a_{iii,one} & a_{3,ii} \end{vmatrix} = a_{2,1}\cdot a_{3,2}-a_{2,ii}\cdot a_{3,1}$

$\left| A\correct| =a_{ane,1}\cdot( a_{2,2}\cdot a_{3,3}-a_{2,iii}\cdot a_{3,two})-a_{1,2}\cdot(a_{2,1}\cdot a_{3,iii}-a_{2,3}\cdot a_{3,1})+$ $a_{1,3}\cdot(a_{2,ane}\cdot a_{3,ii}-a_{2,ii}\cdot a_{three,1})=$ $a_{i,one}\cdot a_{2,2}\cdot a_{3,3}-a_{1,one}\cdot a_{2,iii}\cdot a_{3,2}-a_{1,two}\cdot a_{ii.1}\cdot a_{3,iii}+a_{1,2}\cdot a_{2,3}\cdot a_{3,i}+$ $a_{one,3}\cdot a_{2,1}\cdot a_{3,ii}-a_{1,3}\cdot a_{2,2}\cdot a_{three,1}=$ $\color{ruddy}{a_{1,1}\cdot a_{2,2}\cdot a_{iii,iii}+a_{1,ii}\cdot a_{2,three}\cdot a_{3,ane}+a_{1,3}\cdot a_{2,one}\cdot a_{3,two}-}$ $\color{ruby}{(a_{one,ane}\cdot a_{2,3}\cdot a_{3,two}+a_{1,two}\cdot a_{2,ane}\cdot a_{3,3}+a_{1,3}\cdot a_{2,2}\cdot a_{3,1})}$

To faster attain the concluding relation we tin employ the following method.

Offset, we rewrite the first two rows nether the determinant, equally follows.

$\begin{vmatrix} \color{ruby-red}{a_{1,i}} & a_{one,2} & a_{one,three}\\ \color{red}{a_{ii,1}} & \color{red}{a_{ii,2}} & a_{2,3}\\ \color{scarlet}{a_{3,one}} & \color{carmine}{a_{3,2}} & \color{red}{a_{3,3}} \end{vmatrix}$
$\hspace{2mm}\begin{array}{ccc} a_{1,one} & \color{cherry-red}{a_{1,2}} & \color{ruby-red}{a_{ane,3}}\\ a_{2,1} & a_{two,ii} & \color{cherry-red}{a_{2,3}}\\ \terminate{array}$

We multiply the elements on each of the three cerise diagonals (the main diagonal and the ones underneath) and we add up the results:
$\color{red}{a_{1,one}\cdot a_{two,2}\cdot a_{three,three}+ a_{2,i}\cdot a_{3,2}\cdot a_{1,3}+a_{three,i}\cdot a_{1,2}\cdot a_{2,3}}$

$\begin{vmatrix} \colour{ruby-red}{a_{1,1}} & \colour{red}{a_{ane,ii}} & \color{blue}{a_{ane,iii}}\\ \color{red}{a_{2,1}} & \color{blueish}{a_{2,2}} & \colour{blueish}{a_{2,3}}\\ \colour{blue}{a_{iii,i}} & \color{blue}{a_{3,2}} & \color{bluish}{a_{3,three}} \end{vmatrix}$
$\hspace{2mm} \brainstorm{assortment}{ccc} \color{blueish}{a_{ane,1}} & \color{blue}{a_{ane,ii}} & \colour{red}{a_{one,3}}\\ \color{blueish}{a_{2,one}} & \color{crimson}{a_{two,two}} & \color{red}{a_{2,3}}\\ \end{array}$

We multiply the elements on each of the three blue diagonals (the secondary diagonal and the ones underneath) and we add upward the results:

$\color{blue}{a_{ane,3}\cdot a_{2,ii}\cdot a_{3,one}+ a_{2,iii}\cdot a_{3,two}\cdot a_{1,1}+a_{3,3}\cdot a_{i,2}\cdot a_{ii,1}}$

If we subtract the 2 relations we become the determinant'due south formula:

$\color{red}{a_{one,i}\cdot a_{2,2}\cdot a_{three,iii}+ a_{two,1}\cdot a_{3,ii}\cdot a_{1,3}+a_{3,one}\cdot a_{one,2}\cdot a_{ii,3}-}$ $\color{ruby-red}{(a_{i,iii}\cdot a_{2,two}\cdot a_{iii,1}+ a_{two,3}\cdot a_{three,2}\cdot a_{ane,i}+a_{3,3}\cdot a_{i,2}\cdot a_{2,one})}$

Example 30
$A=\brainstorm{pmatrix} 1 & 4 & 3 \\ 2 & 1 & 5\\ 3 & 2 & 1\\ \end{pmatrix}$

$\brainstorm{vmatrix} i & 4 & 3 \\ 2 & i & 5\\ 3 & 2 & 1\\ \end{vmatrix}$
$\hspace{2mm}\begin{array}{ccc} 1 & 4 & three\\ ii & one & 5\\ \cease{array}$

$ = i\cdot1\cdot1 + 2\cdot2\cdot3 + iii\cdot4\cdot5 -(iii\cdot1\cdot3 + 5\cdot2\cdot1 + 1\cdot4\cdot2) =$ $ 1 + 12 + threescore -(9 + x + eight)=73-27=46$

Case 31
$A=\brainstorm{pmatrix} three & 5 & one \\ ane & four & 2\\ 7 & one & ix\\ \terminate{pmatrix}$

$\begin{vmatrix} iii & v & 1 \\ 1 & 4 & ii\\ seven & i & ix\\ \end{vmatrix}$
$\hspace{2mm}\brainstorm{array}{ccc} 3 & five & 1\\ 1 & 4 & 2\\ \end{array} $

$= 3\cdot4\cdot9 + 1\cdot1\cdot1 + vii\cdot5\cdot2 -(1\cdot4\cdot7 + ii\cdot1\cdot3 + 9\cdot5\cdot1) =$ $ 108 + 1 + lxx -(28 + 6 + 45)=79-79=100$

There are determinants whose elements are letters. They can be calculated more hands using the backdrop of determinants. For example, we calculate the determinant of a matrix in which in that location are the aforementioned elements on whatsoever row or cavalcade, but reordered.

$\brainstorm{vmatrix} a & b & c\\ c & a & b\\ b & c & a \end{vmatrix}$ $ \xlongequal{C_{1}+C_{2}+C_{three}} \begin{vmatrix} a + b + c & b & c\\ c + a + b & a & b\\ b + c + a & c & a \end{vmatrix} = (a + b + c) \cdot \brainstorm{vmatrix} 1 & b & c\\ 1 & a & b\\ one & c & a \end{vmatrix}$

We summate the last determinant:

$\brainstorm{vmatrix} one & b & c\\ ane & a & b\\ one & c & a \cease{vmatrix}$
$\hspace{2mm}\begin{array}{ccc} 1 & b & c\\ 1 & a & b \end{array}$

$ = a^{two} + b^{2} + c^{two} -a\cdot c - b\cdot c - a\cdot b =$ $\frac{one}{2}\cdot(2a^{2} +2b^{ii}+2c^{2} -2a\cdot b -2a\cdot c-2b\cdot c) =$ $\frac{1}{2}\cdot(a^{ii}-2a\cdot b + b^{2}+ a^{2}-2a\cdot c +c^{2}+b^{two}-2b\cdot c + c^{2})=$ $\frac{1}{two}\cdot[(a-b)^{2}+(a-c)^{2}+(b-c)^{2}]$

In conclusion

$\begin{vmatrix} a & b & c\\ c & a & b\\ b & c & a \cease{vmatrix}=$ $\frac{1}{2}\cdot(a+b+c)\cdot[(a-b)^{ii}+(a-c)^{2}+(b-c)^{ii}]$

Instance 32
We summate the determinant of a Vandermonde matrix.
$\begin{vmatrix} 1 & i & 1\\ a & b & c\\ a^{2} & b^{2} & c^{two} \cease{vmatrix}$

Using the properties of determinants we modify row 1 in club to have two elements equal to 0. In this case, when we utilize the formula, there's no need to summate the cofactors of these elements because their product will be 0.

$\brainstorm{vmatrix} 1 & i & one\\ a & b & c\\ a^{two} & b^{ii} & c^{2}\\ \cease{vmatrix}$ $\xlongequal{C_{ane}- C_{iii}\\C_{2} -C_{3}} \begin{vmatrix} 0 & 0 & ane\\ a-c & b-c & c\\ a^{2}- c^{ii} & b^{two}-c^{two} & c^{2} \end{vmatrix}=$ $one\cdot(-i)^{1+3}\cdot \begin{vmatrix} a-c & b-c \\ a^{two}- c^{ii} & b^{ii}-c^{2} \end{vmatrix}= $

$\begin{vmatrix} a-c & b-c \\ (a-c)(a+c) & (b-c)(b+c) \cease{vmatrix}=$ $(a-c)(b-c)\brainstorm{vmatrix} 1 & 1\\ a+c & b+c \end{vmatrix}=$

$=(a-c)(b-c)[(b+c)-(a+c)]=$ $(a-c)(b-c)(b+c-a-c)=(a-c)(b-c)(b-a)$

Computing a 4x4 Determinant

In society to calculate 4x4 determinants, we utilize the general formula.

Earlier applying the formula using the properties of determinants:

We bank check if whatsoever of the atmospheric condition for the value of the determinant to exist 0 is met.
We check if we can factor out of whatever row or column.
We check if the determinant is a Vandermonde matrix or if information technology has the aforementioned elements, but reordered, on whatsoever row or column.

In whatsoever of these cases, we apply the corresponding methods for calculating 3x3 determinants. We modify a row or a column in order to fill information technology with 0, except for one element. The determinant volition exist equal to the production of that element and its cofactor. In this instance, the cofactor is a 3x3 determinant which is calculated with its specific formula.

Example 33
$\begin{vmatrix} 1 & 3 & nine & 2\\ 5 & eight & 4 & iii\\ 0 & 0 & 0 & 0\\ two & three & ane & 8 \end{vmatrix}$

We observe that all elements on row 3 are 0, and then the determinant is 0.

Case 34
$\begin{vmatrix} i & three & 1 & 2\\ 5 & 8 & 5 & 3\\ 0 & iv & 0 & 0\\ 2 & 3 & 2 & viii \end{vmatrix}$
We detect that $C_{1}$ and $C_{3}$ are equal, so the determinant is 0.

Case 35
$\begin{vmatrix} ane & 3 & 9 & 2\\ 5 & eight & four & three\\ 10 & 16 & 18 & 4\\ 2 & 3 & 1 & 8 \end{vmatrix}$
We notice that rows two and three are proportional, and so the determinant is 0.

Example 36
$\brainstorm{vmatrix} \colour{carmine}{four} & three & ii & ii\\ 0 & i & -iii & iii\\ 0 & -1 & iii & 3\\ 0 & three & i & ane \end{vmatrix}$

Since there is only one element different from 0 on column i, we use the general formula using this column. The cofactors corresponding to the elements which are 0 don't need to be calculated because the product of them and these elements will exist 0.

=
$=4(1\cdot3\cdot1 +(-1)\cdot1\cdot3+three\cdot(-3)\cdot3$ $-(3\cdot3\cdot3+3\cdot1\cdot1 +1\cdot(-3)\cdot(-1)))$ $=four(3-iii-27-(27+3+3))=four\cdot(-60)=-240$

Instance 37
$\begin{vmatrix} 4 & three & two & 2\\ 0 & one & 0 & -2\\ 1 & -i & 3 & 3\\ 2 & three & 1 & 1 \stop{vmatrix}$

To modify rows to accept more zeroes, we operate with columns and vice-versa. Nosotros pick a row or cavalcade containing the element i because we can obtain any number through multiplication.

We notice that there already two elements equal to 0 on row 2. We only brand 1 other 0 in guild to calculate only the cofactor of one.

$\begin{vmatrix} 4 & 3 & 2 & 2\\ 0 & 1 & 0 & -2\\ 1 & -1 & 3 & 3\\ 2 & three & 1 & 1 \stop{vmatrix} \xlongequal{C_{4}+2C_{2}}$ $\begin{vmatrix} iv & 3 & 2 & 8\\ 0 & \color{red}{1} & 0 & 0\\ 1 & -1 & 3 & i\\ ii & three & 1 & vii \end{vmatrix}=$ $=$

$= 1\cdot(-1)^{2+2}\cdot \begin{vmatrix} iv & 2 & 8\\ 1 & three & 1\\ 2 & 1 & vii \end{vmatrix}=$
$=iv\cdot3\cdot7 + 1\cdot1\cdot8 + two\cdot2\cdot1$ $-(viii\cdot3\cdot2 + one\cdot1\cdot4 + 7\cdot2\cdot1) =$ $ 84 + 8 + 4- 48-4-14=30$

Example 38
$\brainstorm{vmatrix} i & -2 & 3 & 2\\ ii & 3 & 1 & -1\\ 3 & 3 & 3 & 3\\ -1 & 4 & 2 & 1\\ \end{vmatrix}$

We can factor 3 out of row 3:
$iii\cdot \begin{vmatrix} ane & -2 & 3 & 2\\ 2 & iii & one & -1\\ 1 & 1 & i & 1\\ -1 & 4 & 2 & one\\ \stop{vmatrix}$

Since at that place are just elements equal to 1 on row three, nosotros can easily make zeroes.

$\begin{vmatrix} 1 & -2 & 3 & 2\\ ii & 3 & 1 & -1\\ 1 & i & 1 & 1\\ -1 & 4 & 2 & 1 \stop{vmatrix}$ $ \xlongequal{C_{i} - C_{4},C_{2}-C_{4},C_{three}-C_{iv}} \brainstorm{vmatrix} -1 & -iv & 1 & ii\\ 3 & iv & 2 & -1\\ 0 & 0 & 0 & \color{ruddy}{1}\\ -2 & three & 1 & 1 \end{vmatrix}$ $=1\cdot(-i)^{iii+iv}\cdot$ $=(-1)\cdot \begin{vmatrix} -i & -4 & 1\\ 3 & 4 & 2 \\ -2 & 3 & 1\\ \end{vmatrix}$
$=-((-one)\cdot 4\cdot one +3 \cdot 3\cdot1 + (-2)\cdot (-iv)\cdot ii$ $- (one\cdot 4\cdot (-two) + two\cdot 3\cdot (-1) + one\cdot (-4)\cdot3))$ $=-(-4 + 9 + 16 + viii + 6 + 12) =-47$

Example 39
$\begin{vmatrix} two & five & ane & 4\\ four & 1 & vi & 3\\ v & 3 & seven & 2\\ i & 0 & 2 & 4 \end{vmatrix}$

In this case, we can apply the last row (which contains 1) and nosotros can make zeroes on the beginning cavalcade.

$\begin{vmatrix} 2 & five & 1 & 4\\ 4 & one & 6 & three\\ v & 3 & 7 & two\\ 1 & 0 & ii & 4 \cease{vmatrix}$ $\xlongequal{R_{ane}-2R_{4},R_{two}-4R_{iv}, R_{3}-5R_{4}} \begin{vmatrix} 0 & v & -3 & -4\\ 0 & one & -2 & -13\\ 0 & 3 & -iii & -18\\ \colour{red}{one} & 0 & 2 & 4 \end{vmatrix}=$ $=i\cdot(-one)^{iv+1}\cdot \begin{vmatrix} 5 & -3 & -4\\ 1 & -2 & -13\\ 3 & -3 & -eighteen \end{vmatrix}=$ $(-1)\cdot \brainstorm{vmatrix} 5 & -3 & -4\\ i & -2 & -xiii\\ 3 & -three & -18 \cease{vmatrix}$

We factor -i out of column two and -1 out of column 3.
$ (-one)\cdot(-1)\cdot(-1)\cdot \begin{vmatrix} v & 3 & 4\\ 1 & 2 & thirteen\\ three & 3 & 18 \stop{vmatrix}=$ $(-1)\cdot \begin{vmatrix} 5 & three & iv\\ 1 & two & xiii\\ 3 & 3 & 18 \cease{vmatrix}=$ $-[5\cdot 2\cdot eighteen + i\cdot 3\cdot four+ three\cdot 3\cdot xiii - (4\cdot 2\cdot 3\cdot + 13\cdot 3\cdot v + 18\cdot 3\cdot 1)]=$ $-(180+12+117-24-195-54)=36$

Example 40
$\begin{vmatrix} 4 & 7 & 2 & 3\\ 1 & 3 & ane & 2\\ 2 & 5 & 3 & 4\\ 1 & four & 2 & 3 \cease{vmatrix}$

There is a ane on cavalcade iii, and so nosotros will make zeroes on row 2.

$\begin{vmatrix} 4 & vii & 2 & 3\\ i & 3 & ane & two\\ two & 5 & 3 & iv\\ one & 4 & 2 & 3 \end{vmatrix}$ $\xlongequal{C_{i}-C_{3}, C_{2}-3C_{iii},C_{4}-2C_{3}} \begin{vmatrix} ii & i & 2 & -1\\ 0 & 0 & \colour{red}{ane} & 0 \\ -ane & -4 & 3 & -2\\ -ane & -2 & ii & -1 \end{vmatrix}=$ $=ane\cdot(-1)^{2+v}\cdot \begin{vmatrix} ii & 1 & -i\\ -1 & -iv & -2\\ -1 & -2 & -ane \end{vmatrix}$

We factor -one out of row 2 and -one out of row 3.
$ (-ane)\cdot(-1)\cdot(-one)\cdot \begin{vmatrix} ii & one & -1\\ 1 & iv & 2\\ one & 2 & ane \end{vmatrix}=$ $(-1)\cdot \begin{vmatrix} two & i & -1\\ 1 & iv & 2\\ ane & 2 & i \end{vmatrix}=$ $-[2\cdot 4\cdot 1 + ane\cdot 2\cdot (-1)+ one\cdot 1\cdot 2 - ((-1)\cdot 4\cdot 1 + ii\cdot two\cdot 2 + one\cdot 1\cdot 1)]=$ $-(8-2+two+4-eight-1)=-3$

Case 41
$\begin{vmatrix} 2 & 1 & 3 & 4\\ one & 3 & 4 & 2\\ 3 & 4 & 2 & one\\ 4 & ii & 1 & 3\\ \end{vmatrix}$

Nosotros observe that any row or column has the same elements, but reordered. In this case, we add up all lines or all columns.

$\begin{vmatrix} 2 & one & 3 & 4\\ 1 & 3 & 4 & 2\\ 3 & 4 & two & 1\\ 4 & two & 1 & 3 \cease{vmatrix}$ $\xlongequal{L_{one}+L_{2}+L_{3}+L_{4}} \begin{vmatrix} x & 10 & 10 & 10\\ i & 3 & iv & 2\\ 3 & 4 & 2 & 1\\ 4 & 2 & 1 & three \end{vmatrix} =$ $10\cdot \begin{vmatrix} i & 1 & i & one\\ 1 & 3 & 4 & 2\\ 3 & iv & 2 & 1\\ 4 & 2 & i & 3 \stop{vmatrix}$ $\xlongequal{C_{1} - C_{4},C_{2}-C_{four},C_{three}-C_{4}}10\cdot \begin{vmatrix} 0 & 0 & 0 & \color{red}{1}\\ -1 & i & 2 & ii\\ 2 & iii & 1 & 1\\ one & -1 & -2 & 3 \cease{vmatrix}=$

$=x\cdot1\cdot(-1)^{1+4}$

$ = (-ten)\cdot \begin{vmatrix} -1 & 1 & 2\\ 2 & iii & 1\\ i & -1 & -two \end{vmatrix}=$ $(-10)\cdot((-1)\cdot 3\cdot (-2) +2 \cdot (-1)\cdot2 + 1\cdot 1\cdot 1$ $-(2\cdot three\cdot 1 + 1\cdot (-1)\cdot (-1) + (-two)\cdot1\cdot2))$ $= -10\cdot(6 -4 +ane -vi - one + iv) =0$