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TABLE OF CONTENTS
PROVING #1 .......................................................................................................3
PROVING #2 .......................................................................................................3
SAMPLE PROBLEM #1 ........................................................................................4
SAMPLE PROBLEM #2 ........................................................................................4
SAMPLE PROBLEM #3 ........................................................................................6
PROVING #1
Theorem 3. The determinant of an 𝒏 𝒙 𝒏 matrix is nonzero if and only if its rank is n, that
is to say, if and only if it's invertible.
Let 𝑨 be an 𝒏 𝒙 𝒏 matrix with full rank. This implies that the rows (or columns) of A are linearly
independent. Now, we know that a matrix is invertible if and only if its rows (or columns) are
linearly independent.
The determinant of a matrix can be expressed in terms of its rows or columns. Let's express it in
terms of rows. Consider the rows of A as vectors: 𝒓𝟏 ,𝒓𝟐 ,…,𝒓𝒏 .
The condition 𝒅𝒆𝒕(𝑨) ≠ 𝟎 means that the system of equations 𝑨𝒙 = 𝟎 has only the trivial solution
𝒙 = 𝟎. In other words:
𝑨𝒙 = 𝑪𝟏 𝒓𝟏 +𝑪𝟐 𝒓𝟐 +…+𝒄𝒏 𝒓𝒏 = 𝟎
where 𝐶1 ,𝐶2 ,…,𝐶𝑛 are scalars, and the only solution to this equation is 𝑪𝟏 = 𝑪𝟐 =…= 𝑪𝒏 = 𝟎 (the
trivial solution).
Now, putting A's determinant into consideration. If and only if the matrix is invertible, the
determinant is nonzero.
So, if 𝒅𝒆𝒕 (𝑨) ≠ 𝟎, it means that the system of equations formed by the rows of 𝑨 has only the
trivial solution, the rows are linearly independent. This implies full rank.
Conversely, if 𝒅𝒆𝒕(𝑨) = 𝟎, it means that the system of equations formed by the rows of 𝑨 has a
non-trivial solution, the rows are linearly dependent. This implies the matrix does not have full
rank.
Therefore, if 𝒅𝒆𝒕(𝑨) ≠ 𝟎, then A has full rank. 
PROVING #2
Theorem 22
If B is obtained from A by multiplying a row (column) of A by a real number k, then det(B) = k det(A).


For A:
det(A) = Σ±(𝑎1𝑗 1 𝑎2𝑗 2 𝑎3𝑗 3 … 𝑎𝑛𝑗 𝑛 )
For B:
det(B) = Σ±(𝑏1𝑗 1 𝑏2𝑗 2 𝑏3𝑗 3 … 𝑏𝑛𝑗 𝑛 )
Since det(B) = k det(A), we can say that:
det(B) = 𝑘[𝛴 ± (𝑎1𝑗 1 𝑎2𝑗 2 𝑎3𝑗 3 … 𝑎𝑛𝑗 𝑛 )
det(B) = k det(A) 
SAMPLE PROBLEM #1
Consider the following 3x3 matrix A:
5
6
7
|8
9 10|
11 12 13
Solution:
To determine the determinant of this matrix, we can use the formula for a 3x3 determinant:
det(A) = a(ei − fh) − b(di − fg) + c(dh − eg)
Where:



a, b, c are the elements in the first row of the matrix;
d, e, f are the elements in the second row of the matrix; and
g, h, i are the elements in the third row of the matrix.
In this case, the elements are:
 a = 5, b = 6, c = 7
 d = 8, e = 9, f = 10
 g = 11, h = 12, i = 13
Now, plug these values into the formula:
det(A) = 5(9*13 - 10*12) - 6(8*13 - 10*11) + 7(8*12 - 9*11)
det(A) = 5(117 - 120) - 6(104 - 110) + 7(96 - 99)
det(A) = 5(-3) - 6(-6) + 7(-3)
det(A) = -15 + 36 - 21
det(A) = 0
Therefore, the determinant of this 3x3 matrix B is 0.
SAMPLE PROBLEM #2
Solve the following system of 3 equations in 3 variables using Cramer’s rule:
x+y+z =2
2x + y + 3z = 9
x – 3y + z = 10
Solution:
The given system can be written in the matrix form AX = B where,
𝑥
1 1 1
2
A = [2 1 3], X = [𝑦], and B = [ 9 ]
𝑧
1 −3 1
10
We will compute the determinants.
1
D = det(A) = [2
1
1 1
1 3]
−3 1
= [(1)(1)(1) + (1)(3)(1) + (1)(2)(−3)] − [(1)(1)(1) + (1)(2)(1) + (1)(3)(−3)]
= (1 + 3 − 6) − (1 + 2 − 9)
= −2 + 6
=4
2
1 1
DX = [ 9
1 3]
10 −3 1
= [(2)(1)(1) + (1)(3)(10) + (1)(9)(−3)] − [(1)(1)(10) + (1)(9)(1) + (2)(3)(−3)]
= (2 + 30 − 27) − (10 + 9 − 18)
=5−1
=4
1 2
D y = [2 9
1 10
1
3]
1
= [(1)(9)(1) + (2)(3)(1) + (1)(2)(10)] − [(1)(9)(1) + (2)(2)(1) + (1)(3)(10)]
= (9 + 6 + 20) − (9 + 4 + 30)
= 35 − 43
= −8
1 1
2
D z = [2 1
9]
1 −3 10
= [(1)(1)(10) + (1)(9)(1) + (2)(2)(−3)] − [(2)(1)(1) + (1)(2)(10) + (1)(9)(−3)]
= (10 + 9 − 12) − (2 + 20 − 27)
=7+5
= 12
Now, we apply the formulas:
x=
𝐷𝑥
𝐷
=4=1
4
y=
𝐷𝑦
𝐷
= − 4 = −2
z=
𝐷𝑧
𝐷
=
8
12
4
=3
∴ The solution of the given system is x = 1, y = -2, and z = 3.
SAMPLE PROBLEM #3
The adjugate of 𝐴, denoted 𝑎𝑑𝑗(𝐴), is the transpose of this cofactor matrix; in symbols,
𝑎𝑑𝑗(𝐴) = [𝑐𝑖 𝑗(𝐴)]𝑇
Let 𝐴 = [𝑎𝑖𝑗] be an 𝑛 × 𝑛 matrix. The 𝑛 × 𝑛 matrix 𝑎𝑑𝑗(𝐴), called the adjoint of 𝐴, is the
matrix whose (𝑖, 𝑗)𝑡ℎ entry is the cofactor 𝐴𝑗𝑖 of 𝑎𝑗𝑖.
Thus,
𝐴11 𝐴12
𝐴21 𝐴22
adj(A) =[
⋮
⋮
𝐴1𝑛 𝐴2𝑛
Problem 1:
1
3 −2
A =[ 0
1
5]
−2 −6 7
1
3
Compute the adjugate of 𝐴 =[ 0
1
−2 −6
Step 1:
Compute det(A)
−2
5]
7
⋯ 𝐴𝑛1
⋯ 𝐴𝑛2
]
⋱
⋮
⋯ 𝐴𝑛𝑛
1
3
𝐴 =[ 0
1
−2 −6
−2
5]
7
1
3 −2 1 3
𝑑𝑒𝑡(𝐴) =[ 0
1
5] 0 1
−2 −6 7 −2−6
𝑑𝑒𝑡(𝐴) = [7 − 30 + 0] − [4 − 30 + 0]
𝑑𝑒𝑡(𝐴) = −23 + 26
𝑑𝑒𝑡(𝐴) = 3
Step 2:
1
3 −2
𝐴 =[ 0
1
5]
−2 −6 7
Compute 𝑎𝑑𝑗(𝐴)
1
𝐴11 =(−1)1+1 [
−6
𝐴11 = 37
𝐴21 =(−1)2+1 [
5
]
7
3 −2
]
−6 7
𝐴21 = −9
𝐴31 =(−1)3+1 [
𝐴31 = 17
0 5
]
−2 7
1 5
𝐴13 =(−1)1+3 [
]
−6 7
𝐴13 = 2
1 −2
]
−2 7
𝐴23 =(−1)2+3 [
𝐴12 =(−1)1+2 [
𝐴12 = −10
𝐴22 =(−1)2+2 [
𝐴22 = 3
3 −2
]
1 5
1
𝐴32 =(−1)3+2 [
0
𝐴32 = −5
1
3
]
−2 −6
𝐴23 = 0
−2
]
5
𝐴33 =(−1)3+3 [
𝐴33 = 1
Then, the adjugate of 𝐴 is the transpose of this cofactor matrix.
37 −10
𝑎𝑑𝑗(𝐴) = [−9
3
17 −5
2𝑇
0]
1
1 3
]
0 1
37 −9 17
𝑎𝑑𝑗(𝐴) = [−10 3 −5]
2
0
1
Then, (𝐴) × 𝑎𝑑𝑗(𝐴)
1
3 −2
𝐴 =[0
1
5]
−2 −6 7
37
𝑎𝑑𝑗(𝐴) = [−10
2
1
3 −2
37
(𝐴) × 𝑎𝑑𝑗(𝐴) = [ 0
]
×
[
1
5
−10
−2 −6 7
2
37 − 30 − 4
(𝐴) × 𝑎𝑑𝑗(𝐴) = [ 0 − 10 + 10
−74 + 60 + 14
−9 17
3 −5]
0
1
−9 17
3 −5]
0
1
−9 + 9 + 0
17 − 15 − 2
0+3+0
0−5+5 ]
18 − 18 + 0 −34 + 30 + 7
3 0
(𝐴) × 𝑎𝑑𝑗(𝐴) = [0 3
0 0
1 0
(𝐴) × 𝑎𝑑𝑗(𝐴) = 3 [0 1
0 0
(𝐴) × 𝑎𝑑𝑗(𝐴) = 𝐼𝑛
0
0]
3
0
0]
1