Determinants L-11 Determinant
→ \rightarrow → → \rightarrow → Determinant lecture 11 → \rightarrow → → \rightarrow →
Determinants L-11 System of linear equations
→ \rightarrow → → \rightarrow → System of linear equations → \rightarrow → → \rightarrow →
Determinants L-11 System of linear equations
Systems of linear equations we consider multiple equation with multiple variable, and we like to solve for the variables such that the obtained values satisfy all the equations simultaneously.
Determinant lecture 11 → \rightarrow → → \rightarrow → System of linear equations → \rightarrow → → \rightarrow →
Determinants L-11 Example 1
System of linear equations → \rightarrow → → \rightarrow → Example 1 → \rightarrow → → \rightarrow →
Determinants L-11 Example 1
System of linear equations → \rightarrow → → \rightarrow → Example 1 → \rightarrow → → \rightarrow →
Determinants L-11 Example 1
∴ 2 x + 3 y = − 10 + 15 = 5 \therefore 2 x+3 y=-10+15=5 ∴ 2 x + 3 y = − 10 + 15 = 5
4 ⋅ ( − 5 ) + 3 ( 5 ) = − 20 + 30 = 10 \cdot(-5)+3(5)=-20+30=10 ⋅ ( − 5 ) + 3 ( 5 ) = − 20 + 30 = 10
x = 2 , y = 1 3 x=2, y=\frac{1}{3} x = 2 , y = 3 1
∴ 2 x + 3 y = 4 + 1 = 5 \therefore 2 x+3 y= 4+1=5 ∴ 2 x + 3 y = 4 + 1 = 5
4 ⋅ 2 + 6 ⋅ 1 3 = 8 + 2 = 10 4 \cdot 2+6 \cdot \frac{1}{3}=8+2=10 4 ⋅ 2 + 6 ⋅ 3 1 = 8 + 2 = 10
Example 1 → \rightarrow → → \rightarrow → Example 1 → \rightarrow → → \rightarrow →
Determinants L-11 Systems of equation
We apply matrix & matrix inverse based techniques:
Let the systems of equation be
a 11 x 1 + a 12 x 2 + ⋯ + a 1 n x n = b 1 a_{11} x_1+a_{12} x_2+\cdots+a_{1 n} x_n=b_1 a 11 x 1 + a 12 x 2 + ⋯ + a 1 n x n = b 1
a 21 x 1 + a 22 x 2 + ⋯ + a 2 n x n = b 2 a_{21} x_1+a_{22} x_2+\cdots+a_{2 n} x_n=b_2 a 21 x 1 + a 22 x 2 + ⋯ + a 2 n x n = b 2
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.
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a n 1 x 1 + a n 2 x 2 + ⋯ + a n n x n = b 1 a_{n 1} x_1+a_{n 2} x_2+\cdots+a_{n n} x_n=b_1 a n 1 x 1 + a n 2 x 2 + ⋯ + a nn x n = b 1 We are looking at x x x n n n x 1 ⋯ x n . x_1 \cdots x_n \text {. } x 1 ⋯ x n .
Example 1 → \rightarrow → → \rightarrow → Systems of equation → \rightarrow → → \rightarrow →
Determinants L-11 AX = B
o r [ a 11 ⋯ a 1 n a 21 ⋯ a 2 n ⋮ a n 1 ⋯ a n n ] ⋅ [ x 1 ⋮ x n ] = [ b 1 b 2 ⋮ b n ] or \left[\begin{array}{ccc}a_{11} & \cdots & a_{1 n} \\ a_{21} & \cdots & a_{2 n} \\ \vdots & & \\ a_{n 1} & \cdots & a_{n n}\end{array}\right] \cdot\left[\begin{array}{c}x_1 \\ \vdots \ x_n\end{array}\right]=\left[\begin{array}{c}b_1 \\ b_2 \ \vdots \\ b_n\end{array}\right] or a 11 a 21 ⋮ a n 1 ⋯ ⋯ ⋯ a 1 n a 2 n a nn ⋅ [ x 1 ⋮ x n ] = b 1 b 2 ⋮ b n
Example 1 → \rightarrow → → \rightarrow → AX = B → \rightarrow → → \rightarrow →
Determinants L-11 AX = B
Case.1 suppose A A A
We know we can compute
A − 1 A^{-1} A − 1
∴ \therefore ∴ A − 1 ( A x ) = A − 1 B A^{-1}(A x)=A^{-1} B A − 1 ( A x ) = A − 1 B
or x = A − 1 B \text { or } x=A^{-1} B or x = A − 1 B
Systems of equation → \rightarrow → → \rightarrow → AX = B → \rightarrow → → \rightarrow →
Determinants L-11 Example 2
consider 2 x+3 y=8
3 x-y = 1
[ 2 3 3 − 1 ] [ x y ] = [ 8 1 ] \left[\begin{array}{cc}2 & 3 \\ 3 & -1\end{array}\right]\left[\begin{array}{l}x \\ y
\end{array}\right]=\left[\begin{array}{l}8 \\ 1\end{array}\right] [ 2 3 3 − 1 ] [ x y ] = [ 8 1 ]
N o w ∣ A ∣ = ∣ 2 3 3 − 1 ∣ = − 2 − 9 = − 11 ≠ 0 Now |A|=\left|\begin{array}{cc}2 & 3 \\ 3 & -1\end{array}\right|=-2-9=-11\neq 0 N o w ∣ A ∣ = 2 3 3 − 1 = − 2 − 9 = − 11 = 0
∴ A − 1 e x i s t s . \therefore A^{-1} exists. ∴ A − 1 e x i s t s .
AX = B → \rightarrow → → \rightarrow → Example 2 → \rightarrow → → \rightarrow →
Determinants L-11 Example 2
A − 1 = ( a d j A ) ∣ A ∣ A^{-1}=\frac{(a d j A)}{|A|} A − 1 = ∣ A ∣ ( a d j A )
$ Now \left[\begin{array}{cc}2 & 3 \
∴ A − 1 = [ − 1 − 3 − 3 2 ] − 11 = ( 1 / 11 3 / 11 3 / 11 − 2 / 11 ) . \therefore A^{-1}=\frac{\left[\begin{array}{cc}-1 & -3 \\ -3 & 2\end{array}\right]}{-11}=\left(\begin{array}{cc}1 / 11 & 3 / 11 \\ 3 / 11 & -2 / 11\end{array}\right) . ∴ A − 1 = − 11 [ − 1 − 3 − 3 2 ] = ( 1/11 3/11 3/11 − 2/11 ) .
∴ A − 1 B = ( 1 / 11 3 / 11 3 / 11 − 2 / 11 ) ( 8 1 ) \therefore A^{-1} B=\left(\begin{array}{cc}1 / 11 & 3 / 11 \\ 3 / 11 & -2 / 11\end{array}\right)\left(\begin{array}{l}8 \\ 1\end{array}\right) ∴ A − 1 B = ( 1/11 3/11 3/11 − 2/11 ) ( 8 1 )
= ( 8 11 + 3 11 24 11 − 2 11 ) = ( 1 2 ) . =\left(\begin{array}{l}\frac{8}{11}+\frac{3}{11} \\ \frac{24}{11}-\frac{2}{11}\end{array}\right)=\left(\begin{array}{l}1 \\ 2\end{array}\right) \text {. } = ( 11 8 + 11 3 11 24 − 11 2 ) = ( 1 2 ) .
AX = B → \rightarrow → → \rightarrow → Example 2 → \rightarrow → → \rightarrow →
Determinants L-11 Example 3
Ex: The cost of 4kg of onion 3kg of wheat & 2kg of rice in Rs. 60 .
The cost of 2kg of onion , 4kg of wheat & 6 kg of rice in Rs 90.
The cost of 6kg of onion 2kg of wheat & 3kg of rice in Rs 70.
Find the individual cost.
Example 2 → \rightarrow → → \rightarrow → Example 3 → \rightarrow → → \rightarrow →
Determinants L-11 Example 3
We represent the system of equation as:
[ 4 3 2 2 n 6 6 2 3 ] [ x y z ] = [ 60 90 70 ] \left[\begin{array}{lll}4 & 3 & 2 \\ 2 & n & 6 \\ 6 & 2 & 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z \end{array}\right]=\left[\begin{array}{l}60 \\ 90 \\ 70\end{array}\right] 4 2 6 3 n 2 2 6 3 x y z = 60 90 70
Example 2 → \rightarrow → → \rightarrow → Example 3 → \rightarrow → → \rightarrow →
Determinants L-11 Example 3
∴ \therefore ∴ A − 1 = ( adj A ) ∣ A ∣ A^{-1}=\frac{(\operatorname{adj} A)}{|A|} A − 1 = ∣ A ∣ ( adj A ) A A A
A 11 = 4.3 − 6.2 = 0
A_{11}=4.3-6.2=0 A 11 = 4.3 − 6.2 = 0
A 12 = ( − 1 ) ( 2.3 − 6.6 ) = 30 A_{12}=(-1)(2.3-6.6)=30 A 12 = ( − 1 ) ( 2.3 − 6.6 ) = 30
A 13 = 2.2 − 6.4 = − 20 A_{13}=2.2-6.4=-20 A 13 = 2.2 − 6.4 = − 20
A 21 = ( − 1 ) ( 3.3 − 2.2 ) = − 5 A_{21}=(-1)(3.3-2.2)=-5 A 21 = ( − 1 ) ( 3.3 − 2.2 ) = − 5
A 22 = 4.3 − 6.2 = 0 A_{22}=4.3-6.2=0 A 22 = 4.3 − 6.2 = 0
A 23 = ( − 1 ) ( 4.2 − 6.3 ) = 10 A_{23}=(-1)(4.2-6.3)=10 A 23 = ( − 1 ) ( 4.2 − 6.3 ) = 10
A 31 = 3.6 − 4.2 = 10 A_{31}=3.6-4.2=10 A 31 = 3.6 − 4.2 = 10
A 32 = ( − 1 ) ( 4.6 − 2.2 ) = − 20 A_{32}=(-1)(4.6-2.2)=-20 A 32 = ( − 1 ) ( 4.6 − 2.2 ) = − 20
A 33 = 4.4 − 3.2 = + 10 A_{33}=4.4-3.2=+10 A 33 = 4.4 − 3.2 = + 10
Example 3 → \rightarrow → → \rightarrow → Example 3 → \rightarrow → → \rightarrow →
Determinants L-11 Example 3
( adj A ) = ( 0 − 5 10 30 0 − 20 − 20 10 10 ) (\operatorname{adj} A)=\left(\begin{array}{ccc}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{array}\right) ( adj A ) = 0 30 − 20 − 5 0 10 10 − 20 10
∴ \therefore ∴ A − 1 A^{-1} A − 1
Example 3 → \rightarrow → → \rightarrow → Example 3 → \rightarrow → → \rightarrow →
Determinants L-11 Example 3
∣ A ∣ = 4.0 + 3.30 + 2 ( − 20 ) |A|=4.0+3.30+2(-20) ∣ A ∣ = 4.0 + 3.30 + 2 ( − 20 )
= 0 + 90 − 40 = 50 =0+90-40=50 = 0 + 90 − 40 = 50
∴ A − 1 = ( 0 − 5 10 30 0 − 20 − 20 10 10 ) 50 \therefore A^{-1}=\frac{\left(\begin{array}{ccc}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{array}\right)}{50} ∴ A − 1 = 50 ( 0 30 − 20 − 5 0 10 10 − 20 10 )
= ( 0 − 1 10 1 5 3 / 5 0 − 2 / 5 − 2 / 5 1 5 1 / 5 ) ⋅ [ 60 90 70 ] =\left(\begin{array}{ccc}0 & -\frac{1}{10} & \frac{1}{5} \\ 3 / 5 & 0 & -2 / 5 \\ -2 / 5 & \frac{1}{5} & 1 / 5\end{array}\right) \cdot\left[\begin{array}{l}60 \\ 90 \\ 70\end{array}\right] = 0 3/5 − 2/5 − 10 1 0 5 1 5 1 − 2/5 1/5 ⋅ 60 90 70
∴ ( x y 3 ) = ( 5 8 8 ) \therefore\left(\begin{array}{l}x \\ y \\ 3\end{array}\right)=\left(\begin{array}{l}\ 5 \\ 8 \\ 8\end{array}\right) ∴ x y 3 = 5 8 8
Example 3 → \rightarrow → → \rightarrow → Example 3 → \rightarrow → → \rightarrow →
Determinants L-11 Example 3
Example 3 → \rightarrow → → \rightarrow → Example 3 → \rightarrow → → \rightarrow →
Determinants L-11 Example 4
The sum of 3 numbers is 6
If we multiply the 3 rd 3^{\text {rd }} 3 rd 2 nd 2^{\text {nd }} 2 nd 3 rd 3^{\text {rd }} 3 rd
Find the three numbers.
Example 3 → \rightarrow → → \rightarrow → Example 4 → \rightarrow → → \rightarrow →
Determinants L-11 Example 4
Three equations :
x + y + z = 6
x + y + 2 z = 7
3 x + y + z = 12
or [ 1 1 1 1 0 2 3 1 1 ] [ x y z ] = [ 6 7 12 ] \text { or }\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z \end{array}\right]=\left[\begin{array}{l}6 \\ 7 \\ 12\end{array}\right] or 1 1 3 1 0 1 1 2 1 x y z = 6 7 12
Determinant of A A A
= 1 ⋅ ( 0 − 2 ) − 1 ⋅ ( 1 − 6 ) + 1 ( 1 − 0 ) =1 \cdot(0-2)-1 \cdot(1-6)+1(1-0) = 1 ⋅ ( 0 − 2 ) − 1 ⋅ ( 1 − 6 ) + 1 ( 1 − 0 )
= − 2 + 5 + 1 = 4 ≠ 0 =-2+5+1=4 \neq 0 = − 2 + 5 + 1 = 4 = 0
Example 3 → \rightarrow → → \rightarrow → Example 4 → \rightarrow → → \rightarrow →
Determinants L-11 Example 4
∴ \therefore ∴ A − 1 A^{-1} A − 1
= ( adj A ) ∣ A ∣ =\frac{(\operatorname{adj} A)}{|A|} = ∣ A ∣ ( adj A )
Therefore as before we compute the cofactor.
A 11 = − 2 A 12 = 5 A 13 = 1 A 21 = 0 A 22 = − 2 A 23 = 2 A 31 = 2 A 32 = − 1 A 33 = − 1 \begin{array}{lll}A_{11}=-2 & A_{12}=5 & A_{13}=1 \\ A_{21}=0 & A_{22}=-2 & A_{23}=2 \\ A_{31}=2 & A_{32}=-1 & A_{33}=-1\end{array} A 11 = − 2 A 21 = 0 A 31 = 2 A 12 = 5 A 22 = − 2 A 32 = − 1 A 13 = 1 A 23 = 2 A 33 = − 1
Example 4 → \rightarrow → → \rightarrow → Example 4 → \rightarrow → → \rightarrow →
Determinants L-11 Example 4
∴ ( adj A ) = ( − 2 0 2 5 − 2 − 1 1 2 − 1 ) \therefore(\operatorname{adj} A)=\left(\begin{array}{ccc}-2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1\end{array}\right) ∴ ( adj A ) = − 2 5 1 0 − 2 2 2 − 1 − 1
∴ A − 1 = 1 4 ( − 2 0 2 5 − 2 − 1 1 2 − 1 ) \therefore A^{-1}=\frac{1}{4}\left(\begin{array}{ccc}-2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1\end{array}\right) ∴ A − 1 = 4 1 − 2 5 1 0 − 2 2 2 − 1 − 1
∴ \therefore ∴
= A − 1 B = 1 4 ( − 2 0 2 5 − 2 − 1 1 2 − 1 ) ( 6 7 12 ) =A^{-1} B=\frac{1}{4}\left(\begin{array}{ccc}-2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1\end{array}\right)\left(\begin{array}{l}6 \\ 7 \\ 12 \end{array}\right) = A − 1 B = 4 1 − 2 5 1 0 − 2 2 2 − 1 − 1 6 7 12
= 1 4 ( − 12 + 24 30 − 14 − 12 6 + 14 − 12 ) = 1 4 ( 12 4 8 ) = ( 3 1 2 ) = \frac{1}{4}\left(\begin{array}{cc}-12+24 \\ 30-14 & -12 \\ 6+14 & -12\end{array}\right)=\frac{1}{4}\left(\begin{array}{c}12 \\ 4 \\ 8 \end{array}\right)=\left(\begin{array}{l}3 \\ 1 \\ 2\end{array}\right) = 4 1 − 12 + 24 30 − 14 6 + 14 − 12 − 12 = 4 1 12 4 8 = 3 1 2
Example 4 → \rightarrow → → \rightarrow → Example 4 → \rightarrow → → \rightarrow →
Determinants L-11 Cramer rule
If the given system of equation are A n × n X n × 1 = B n × 1 \underset{\tiny n \times n}{A} \underset{~ ~ \tiny n \times 1}{X}=\underset{\tiny n \times 1}{B} n × n A n × 1 X = n × 1 B
such that A A A
B ≠ 0 n × 1 B \neq 0_{n \times 1} B = 0 n × 1 n × 1 n \times 1 n × 1
Example 4 → \rightarrow → → \rightarrow → Cramer rule → \rightarrow → → \rightarrow →
Determinants L-11 Cramer Rule
Let D 1 D_1 D 1 A A A B B B
i.e D 1 = ∣ b 1 a 12 a 13 b 2 a 22 a 23 b 3 a 32 a 33 ∣ \text { i.e } D_1=\left|\begin{array}{lll}b_1 & a_{12} & a_{13} \\ b_2 & a_{22} & a_{23} \\ b_3 & a_{32} & a_{33}\end{array}\right| i.e D 1 = b 1 b 2 b 3 a 12 a 22 a 32 a 13 a 23 a 33
Example 4 → \rightarrow → → \rightarrow → Cramer Rule → \rightarrow → → \rightarrow →
Determinants L-11 Cramer Rule
similarly Let D 2 D_2 D 2
similarly campute D 3 D_3 D 3
∣ a 11 a 12 b 1 a 21 a 22 b 2 a 31 a 32 b 3 ∣ \left|\begin{array}{lll}a_{11} & a_{12} & b_1 \\ a_{21} & a_{22} & b_2 \\ a_{31} & a_{32} & b_3\end{array}\right| a 11 a 21 a 31 a 12 a 22 a 32 b 1 b 2 b 3
Cramer rule → \rightarrow → → \rightarrow → Cramer Rule → \rightarrow → → \rightarrow →
Determinants L-11 Cramer Rule
Then x = D 1 D y = D 2 D & z = D 3 D x=\frac{D_1}{D} \quad y=\frac{D_2}{D} \quad \& z=\frac{D_3}{D} x = D D 1 y = D D 2 & z = D D 3
A=[ 1 1 1 1 0 2 3 1 1 ] [ x y z ] = [ 6 7 12 ] \left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}6 \\ 7 \\ 12\end{array}\right] 1 1 3 1 0 1 1 2 1 x y z = 6 7 12
( 3 1 2 ) \left(\begin{array}{l}3 \\ 1 \\ 2\end{array}\right) 3 1 2
Cramer Rule → \rightarrow → → \rightarrow → Cramer Rule → \rightarrow → → \rightarrow →
Determinants L-11 Solution by Cramer'Rule
= 6 ( 0 − 2 ) − 1 ( 7 − 24 ) + 1 ( 7 ) =6(0-2)-1(7-24)+1(7)
= 6 ( 0 − 2 ) − 1 ( 7 − 24 ) + 1 ( 7 )
= − 12 − 7 + 24 + 7 = 12 =-12-7+24+7
=12 = − 12 − 7 + 24 + 7 = 12
∴ \therefore ∴ = 12 4 = 3 =\frac{12}{4}=3 = 4 12 = 3
Cramer Rule → \rightarrow → → \rightarrow → Solution by Cramer'Rule → \rightarrow → → \rightarrow →
Determinants L-11 Solution by Cramer'Rule
Now D 2 = ∣ 1 6 1 1 7 2 3 12 1 ∣ D_2= \left|\begin{array}{ccc}1 & 6 & 1 \\ 1 & 7 & 2 \\ 3 & 12 & 1\end{array}\right| D 2 = 1 1 3 6 7 12 1 2 1
∴ D 2 = 1 ( 7 − 24 ) − 6 ( 1 − 6 ) \therefore D_2 =1(7-24)-6(1-6) ∴ D 2 = 1 ( 7 − 24 ) − 6 ( 1 − 6 )
= − 17 + 30 − 9 ( 12 − 21 ) = 4 =-17+30-9(12-21) =4 = − 17 + 30 − 9 ( 12 − 21 ) = 4
∴ \therefore ∴ y = 4 4 = 1 y=\frac{4}{4}=1 y = 4 4 = 1
Cramer Rule → \rightarrow → → \rightarrow → Solution by Cramer'Rule → \rightarrow → → \rightarrow → A A A
Determinants L-11 Solution by Cramer'Rule
= 1 ( 0 − 7 ) − 1 ( 12 − 21 ) =1(0-7)-1(12-21)
= 1 ( 0 − 7 ) − 1 ( 12 − 21 )
= − 7 + 9 + 6 = 8 =-7+9+6=8
= − 7 + 9 + 6 = 8
∴ z = 8 4 = 2. \therefore z=\frac{8}{4}=2 . ∴ z = 4 8 = 2.
Solution by Cramer'Rule → \rightarrow → → \rightarrow → Solution by Cramer'Rule → \rightarrow → A A A → \rightarrow →
Determinants L-11 If A A A
If A A A
∴ \therefore ∴
a) If |A| = 0 & (adj A) B = 0 then we will have multiple solution, for system of equation.
b) If |A| = 0 & (adj A) B ≠ 0 \neq 0 = 0
Solution by Cramer'Rule → \rightarrow → → \rightarrow → If A A A → \rightarrow → → \rightarrow →
Determinants L-11 Example 5
∴ \therefore ∴ ( ( ( A A A
= ( 6 − 3 − 4 2 ) ( 5 10 ) = \left(\begin{array}{cc}6 & -3 \\ -4 & 2\end{array}\right)\left(\begin{array}{c}5 \\ 10\end{array}\right) = ( 6 − 4 − 3 2 ) ( 5 10 )
= ( 30 − 30 − 20 + 20 ) = ( 0 0 ) = 0 = \left(\begin{array}{c}30-30 \\ -20+20\end{array}\right)=\left(\begin{array}{l}0 \\ 0\end{array}\right)=0 = ( 30 − 30 − 20 + 20 ) = ( 0 0 ) = 0
There are infinitely many solution.
Solution by Cramer'Rule → \rightarrow → A A A → \rightarrow → Example 5 → \rightarrow → → \rightarrow →
Determinants L-11 Example 6
2 x+3 y=5
4 x+6 y=15
∴ ∣ A ∣ = 0 \therefore|A|=0 ∴ ∣ A ∣ = 0
(adj A)( 5 15 ) = ( 6 − 3 − 4 2 ) ( 5 15 ) \left(\begin{array}{c}5 \\ 15\end{array}\right)=\left(\begin{array}{cc}6 & -3 \\ -4 & 2\end{array}\right)\left(\begin{array}{c}5 \\ 15 \end{array}\right) ( 5 15 ) = ( 6 − 4 − 3 2 ) ( 5 15 )
= 30 − 45 = 30 - 45 = 30 − 45
− 20 + 30 = ( − 15 10 ) -20 + 30 =\left(\begin{array}{c}-15 \\ 10\end{array}\right) − 20 + 30 = ( − 15 10 )
Not a zero matrix of size 2 × 1 2 \times 1 2 × 1
If A A A → \rightarrow → → \rightarrow → Example 6 → \rightarrow → → \rightarrow →
Determinants L-11 Thank you
Example 5 → \rightarrow → → \rightarrow → Thank you → \rightarrow → → \rightarrow →
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Determinants L-11 Determinant lecture 11 $\rightarrow$ $\rightarrow$ Determinant lecture 11 $\rightarrow$ System of linear equations $\rightarrow$ System of linear equations
