CS6015: Linear Algebra and Random Processes
Lecture 15: Formula for determinant, co-factors, Finding A−1, Cramer's rule for solving Ax=b, Determinant=Volume
Learning Objectives
(for today's lecture)
What is the formula for computing a determinant?
What are co-factors?
What is Cramer's rule for computing x = A−1b ?
What is the connection between determinant and volume?
A formula for determinant
(the 2 x 2 case)
acbd
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}\\
=a+0c0+bd
= \begin{vmatrix}
a + 0 & 0 + b\\
c & d
\end{vmatrix}\\
=ac0d+0cbd
= \begin{vmatrix}
a & 0\\
c & d
\end{vmatrix}+
\begin{vmatrix}
0 & b\\
c & d
\end{vmatrix}
=ac+000+d+0c+0b0+d
= \begin{vmatrix}
a & 0\\
c+0 & 0+d
\end{vmatrix}+
\begin{vmatrix}
0 & b\\
c+0 & 0+d
\end{vmatrix}
=ac00+a00d
= \begin{vmatrix}
a & 0\\
c & 0
\end{vmatrix}+
\begin{vmatrix}
a & 0\\
0 & d
\end{vmatrix}
+0cb0+00bd
+ \begin{vmatrix}
0 & b\\
c & 0
\end{vmatrix}+
\begin{vmatrix}
0 & b\\
0 & d
\end{vmatrix}
0
0
0
0
(by prop. 6)
(by prop. 6)
a00d+0cb0
\begin{vmatrix}
a & 0\\
0 & d
\end{vmatrix}+
\begin{vmatrix}
0 & b\\
c & 0
\end{vmatrix}
=a00d+(−1)b00c
=\begin{vmatrix}
a & 0\\
0 & d
\end{vmatrix}+
(-1)\begin{vmatrix}
b & 0\\
0 & c
\end{vmatrix}
=a⋅d1001−b⋅c1001
=a\cdot d\begin{vmatrix}
1 & 0\\
0 & 1
\end{vmatrix}-
b\cdot c\begin{vmatrix}
1 & 0\\
0 & 1
\end{vmatrix}
=1
=1
=1
=1
=a⋅d−b⋅c
=a\cdot d-
b\cdot c
(by prop. 2)
(by prop. 1)
(by prop. 1)
A formula for determinant
(the 3 x 3 case)
acbd
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}\\
=a00d+0cb0
=\begin{vmatrix}
a & 0\\
0 & d
\end{vmatrix}+
\begin{vmatrix}
0 & b\\
c & 0
\end{vmatrix}
Only one entry per row, per column 0's everywhere else
a11a21a31a12a22a32a13a23a33
\begin{vmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\\
\end{vmatrix}\\
=a11a21a310a22a320a23a33+0a21a31a12a22a320a23a33+0a21a310a22a32a13a23a33+…27 terms
=\begin{vmatrix}
a_{11} & 0 & 0\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\\
\end{vmatrix}+
\begin{vmatrix}
0 & a_{12} & 0\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\\
\end{vmatrix}+
\begin{vmatrix}
0 & 0 & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\\
\end{vmatrix}+\dots 27~terms
=ad1001+bc0110
=ad\begin{vmatrix}
1 & 0\\
0 & 1
\end{vmatrix}+
bc\begin{vmatrix}
0 & 1\\
1 & 0
\end{vmatrix}
All permutations of the 2x2 identity matrix
=a11000a22000a33+a110000a320a230+0a210a120000a33+00a31a12000a230+00a310a220a1300+0a21000a32a1300
=\begin{vmatrix}
a_{11} & 0 & 0\\
0 & a_{22} & 0\\
0 & 0 & a_{33}\\
\end{vmatrix}+
\begin{vmatrix}
a_{11} & 0 & 0\\
0 & 0 & a_{23}\\
0 & a_{32} & 0\\
\end{vmatrix}+
\begin{vmatrix}
0 & a_{12} & 0\\
a_{21} & 0 & 0\\
0 & 0 & a_{33}\\
\end{vmatrix}+
\begin{vmatrix}
0 & a_{12} & 0\\
0 & 0 & a_{23}\\
a_{31} & 0 & 0\\
\end{vmatrix}+
\begin{vmatrix}
0 & 0 & a_{13}\\
0 & a_{22} & 0\\
a_{31} & 0 & 0\\
\end{vmatrix}+
\begin{vmatrix}
0 & 0 & a_{13}\\
a_{21} & 0 & 0\\
0 & a_{32} & 0\\
\end{vmatrix}
(21 of these would have a column of 0s)
A formula for determinant
(the 3 x 3 case)
a11a21a31a12a22a32a13a23a33
\begin{vmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\\
\end{vmatrix}\\
=a11a22a33100010001+a11a23a32100001010+a12a23a31001100010
=a_{11}a_{22}a_{33}\begin{vmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{vmatrix} +
a_{11}a_{23}a_{32}\begin{vmatrix}
1 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0\\
\end{vmatrix} +
a_{12}a_{23}a_{31}\begin{vmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
1 & 0 & 0\\
\end{vmatrix}
+a12a21a33010100001+a13a21a32010001100+a13a22a31001010100
+
a_{12}a_{21}a_{33}\begin{vmatrix}
0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 1\\
\end{vmatrix}+
a_{13}a_{21}a_{32}\begin{vmatrix}
0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0\\
\end{vmatrix}+
a_{13}a_{22}a_{31}\begin{vmatrix}
0 & 0 & 1\\
0 & 1 & 0\\
1 & 0 & 0\\
\end{vmatrix}
All permutations of the 3x3 identity matrix
+1 or -1 depending on number of row exchanges (k): (−1)k
+1
+1
−1
-1
+1
+1
−1
-1
+1
+1
−1
-1
A formula for determinant
(the 3 x 3 case)
a11a21a31a12a22a32a13a23a33
\begin{vmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\\
\end{vmatrix}\\
=a11a22a33100010001+a11a23a32100001010+a12a23a31001100010
=a_{11}a_{22}a_{33}\begin{vmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{vmatrix} +
a_{11}a_{23}a_{32}\begin{vmatrix}
1 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0\\
\end{vmatrix} +
a_{12}a_{23}a_{31}\begin{vmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
1 & 0 & 0\\
\end{vmatrix}
+a12a21a33010100001+a13a21a32010001100+a13a22a31001010100
+
a_{12}a_{21}a_{33}\begin{vmatrix}
0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 1\\
\end{vmatrix}+
a_{13}a_{21}a_{32}\begin{vmatrix}
0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0\\
\end{vmatrix}+
a_{13}a_{22}a_{31}\begin{vmatrix}
0 & 0 & 1\\
0 & 1 & 0\\
1 & 0 & 0\\
\end{vmatrix}
+1
+1
−1
-1
+1
+1
−1
-1
+1
+1
−1
-1
P∈3! Permutations
P \in 3!~Permutations
=∑
=\sum
det(P)
det(P)
a1__a2__a3__
a_{1\_\_}a_{2\_\_}a_{3\_\_}
α
\alpha
β
\beta
γ
\gamma
{α,β,γ}=some permutation of {1,2,3}
\{\alpha, \beta, \gamma\} = some~permutation~of~\{1,2,3\}
A formula for determinant
(the n x n case)
a11a21a31⋯an1a12a22a32⋯an2a13a23a33⋯an3⋯⋯⋯⋯⋯a1na2na3n⋯ann
\begin{vmatrix}
a_{11} & a_{12} & a_{13}&\cdots&a_{1n}\\
a_{21} & a_{22} & a_{23}&\cdots&a_{2n}\\
a_{31} & a_{32} & a_{33}&\cdots&a_{3n}\\
\cdots & \cdots & \cdots&\cdots&\cdots\\
a_{n1} & a_{n2} & a_{n3}&\cdots&a_{nn}\\
\end{vmatrix}\\
P∈n! Permutations
P \in n!~Permutations
=∑
=\sum
det(P)
det(P)
a1αa2βa3γ…anω
a_{1\alpha}a_{2\beta}a_{3\gamma}\dots a_{n\omega}
{α,β,γ…,ω}=some permutation of {1,2,3,…,n}
\{\alpha, \beta, \gamma\, \dots, \omega\} = some~permutation~of~\{1,2,3, \dots, n\}
Co-factors
a11a21a31a12a22a32a13a23a33
\begin{vmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\\
\end{vmatrix}\\
=a11a22a33(+1)+a11a23a32(−1)+a12a23a31(+1)
=a_{11}a_{22}a_{33}(+1) +
a_{11}a_{23}a_{32}(-1) +
a_{12}a_{23}a_{31}(+1)
+a12a21a33(−1)+a13a21a32(+1)+a13a22a31(−1)
+a_{12}a_{21}a_{33}(-1) +
a_{13}a_{21}a_{32}(+1)+
a_{13}a_{22}a_{31}(-1)
=a11(a22a33−a23a32)
=a_{11}(a_{22}a_{33} - a_{23}a_{32})
+ a12(a23a31−a21a33)
+~a_{12}(a_{23}a_{31} - a_{21}a_{33})
+ a13(a21a32−a22a31)
+~a_{13}(a_{21}a_{32} - a_{22}a_{31})
a11a21a31a12a22a32a13a23a33
\begin{vmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\\
\end{vmatrix}\\
det of a n−1×n−1 matrix
det~of~a~n-1\times n-1~matrix
det of a n−1×n−1 matrix
det~of~a~n-1\times n-1~matrix
det of a n−1×n−1 matrix
det~of~a~n-1\times n-1~matrix
+
+
−
-
+
+
a11a21a31a12a22a32a13a23a33
\begin{vmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\\
\end{vmatrix}\\
a11a21a31a12a22a32a13a23a33
\begin{vmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\\
\end{vmatrix}\\
Co-factors
Co-factors
a11a21a31⋯an1a12a22a32⋯an2a13a23a33⋯an3⋯⋯⋯⋯⋯a1na2na3n⋯ann
\begin{vmatrix}
a_{11} & a_{12} & a_{13}&\cdots&a_{1n}\\
a_{21} & a_{22} & a_{23}&\cdots&a_{2n}\\
a_{31} & a_{32} & a_{33}&\cdots&a_{3n}\\
\cdots & \cdots & \cdots&\cdots&\cdots\\
a_{n1} & a_{n2} & a_{n3}&\cdots&a_{nn}\\
\end{vmatrix}\\
=a11C11+a12C12+⋯+a1nC1n
=a_{11}C_{11}+a_{12}C_{12}+\cdots+a_{1n}C_{1n}
Cij=(−1)i+j(det of a n−1×n−1 matrix)
C_{ij} = (-1)^{i+j} (det~of~a~n-1\times n-1~matrix)
(obtained after dropping i-th row and j-th column)
=a21C21+a22C22+⋯+a2nC2n
=a_{21}C_{21}+a_{22}C_{22}+\cdots+a_{2n}C_{2n}
(in the big formula you can put the brackets wherever you want)
a21(−a12a33+a13a32)
a_{21}(- a_{12}a_{33} + a_{13}a_{32} )
C21
C_{21}
Co-factors
a11a21a31⋯an1a12a22a32⋯an2a13a23a33⋯an3⋯⋯⋯⋯⋯a1na2na3n⋯ann
\begin{vmatrix}
a_{11} & a_{12} & a_{13}&\cdots&a_{1n}\\
a_{21} & a_{22} & a_{23}&\cdots&a_{2n}\\
a_{31} & a_{32} & a_{33}&\cdots&a_{3n}\\
\cdots & \cdots & \cdots&\cdots&\cdots\\
a_{n1} & a_{n2} & a_{n3}&\cdots&a_{nn}\\
\end{vmatrix}\\
=a11C11+a12C12+⋯+a1nC1n
=a_{11}C_{11}+a_{12}C_{12}+\cdots+a_{1n}C_{1n}
=a21C21+a22C22+⋯+a2nC2n
=a_{21}C_{21}+a_{22}C_{22}+\cdots+a_{2n}C_{2n}
a21(−a12a33+a13a32)
a_{21}(- a_{12}a_{33} + a_{13}a_{32} )
C21
C_{21}
=a31C31+a32C32+⋯+a3nC3n
=a_{31}C_{31}+a_{32}C_{32}+\cdots+a_{3n}C_{3n}
=an1Cn1+an2Cn2+⋯+annCnn
=a_{n1}C_{n1}+a_{n2}C_{n2}+\cdots+a_{nn}C_{nn}
…
\dots
Formula for A−1
(2 x 2 case)
(and the n x n case)
A=[acbd]
A = \begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\\
A−1=det(A)1[d−c−ba]
A^{-1} = \frac{1}{det(A)}\begin{bmatrix}
d & -b\\
-c & a
\end{bmatrix}\\
C=[d−b−ca]
C = \begin{bmatrix}
d & -c\\
-b & a
\end{bmatrix}\\
(matrix of co-factors)
C⊤
C^\top
A=a11a21a31⋯an1a12a22a32⋯an2a13a23a33⋯an3⋯⋯⋯⋯⋯a1na2na3n⋯ann
A=\begin{bmatrix}
a_{11} & a_{12} & a_{13}&\cdots&a_{1n}\\
a_{21} & a_{22} & a_{23}&\cdots&a_{2n}\\
a_{31} & a_{32} & a_{33}&\cdots&a_{3n}\\
\cdots & \cdots & \cdots&\cdots&\cdots\\
a_{n1} & a_{n2} & a_{n3}&\cdots&a_{nn}\\
\end{bmatrix}\\
A−1=det(A)1C⊤
A^{-1} = \frac{1}{det(A)}C^\top
Why is this formula correct?
To prove:
A−1=det(A)1C⊤
A^{-1} = \frac{1}{det(A)}C^\top
A(det(A)1C⊤)=I
A (\frac{1}{det(A)}C^\top) = I
i.e., AC⊤=det(A)I
i.e.,~AC^\top = det(A)I
a11a21a31⋯an1a12a22a32⋯an2a13a23a33⋯an3⋯⋯⋯⋯⋯a1na2na3n⋯ann
\begin{bmatrix}
a_{11} & a_{12} & a_{13}&\cdots&a_{1n}\\
a_{21} & a_{22} & a_{23}&\cdots&a_{2n}\\
a_{31} & a_{32} & a_{33}&\cdots&a_{3n}\\
\cdots & \cdots & \cdots&\cdots&\cdots\\
a_{n1} & a_{n2} & a_{n3}&\cdots&a_{nn}\\
\end{bmatrix}\\
C11C12C13⋯C1nC21C22C23⋯C2nC31C32C33⋯C3n⋯⋯⋯⋯⋯Cn1Cn2Cn3⋯Cnn
\begin{bmatrix}
C_{11} & C_{21} & C_{31}&\cdots&C_{n1}\\
C_{12} & C_{22} & C_{32}&\cdots&C_{n2}\\
C_{13} & C_{23} & C_{33}&\cdots&C_{n3}\\
\cdots & \cdots & \cdots&\cdots&\cdots\\
C_{1n} & C_{2n} & C_{3n}&\cdots&C_{nn}\\
\end{bmatrix}\\
(1,1) entry =
a11C11+a12C12+a13C13+⋯+a1nC1n
a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} + \cdots + a_{1n}C_{1n}
(2,2) entry =
=det(A)
=det(A)
a21C21+a22C22+a23C23+⋯+a2nC2n
a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23} + \cdots + a_{2n}C_{2n}
=det(A)
=det(A)
(i,i) entry =
ai1Ci1+ai2Ci2+ai3Ci3+⋯+ainCin
a_{i1}C_{i1} + a_{i2}C_{i2} + a_{i3}C_{i3} + \cdots + a_{in}C_{in}
=det(A)
=det(A)
All the diagonal entries will be det(A)
(exactly as we wanted)
Why is this formula correct?
To prove:
A−1=det(A)1C⊤
A^{-1} = \frac{1}{det(A)}C^\top
A(det(A)1C⊤)=I
A (\frac{1}{det(A)}C^\top) = I
i.e., AC⊤=det(A)I
i.e.,~AC^\top = det(A)I
a11a21a31⋯an1a12a22a32⋯an2a13a23a33⋯an3⋯⋯⋯⋯⋯a1na2na3n⋯ann
\begin{bmatrix}
a_{11} & a_{12} & a_{13}&\cdots&a_{1n}\\
a_{21} & a_{22} & a_{23}&\cdots&a_{2n}\\
a_{31} & a_{32} & a_{33}&\cdots&a_{3n}\\
\cdots & \cdots & \cdots&\cdots&\cdots\\
a_{n1} & a_{n2} & a_{n3}&\cdots&a_{nn}\\
\end{bmatrix}\\
C11C12C13⋯C1nC21C22C23⋯C2nC31C32C33⋯C3n⋯⋯⋯⋯⋯Cn1Cn2Cn3⋯Cnn
\begin{bmatrix}
C_{11} & C_{21} & C_{31}&\cdots&C_{n1}\\
C_{12} & C_{22} & C_{32}&\cdots&C_{n2}\\
C_{13} & C_{23} & C_{33}&\cdots&C_{n3}\\
\cdots & \cdots & \cdots&\cdots&\cdots\\
C_{1n} & C_{2n} & C_{3n}&\cdots&C_{nn}\\
\end{bmatrix}\\
(2,1) entry =
a21C11+a22C12+a23C13+⋯+a2nC1n
a_{21}C_{11} + a_{22}C_{12} + a_{23}C_{13} + \cdots + a_{2n}C_{1n}
=0
=0
What about the off-diagonal entries?
Why?
(not very obvious but we will see on the next slide)
Why is this formula correct?
To prove:
A−1=det(A)1C⊤
A^{-1} = \frac{1}{det(A)}C^\top
A(det(A)1C⊤)=I
A (\frac{1}{det(A)}C^\top) = I
i.e., AC⊤=det(A)I
i.e.,~AC^\top = det(A)I
A=a11a21a31⋯an1a12a22a32⋯an2a13a23a33⋯an3⋯⋯⋯⋯⋯a1na2na3n⋯ann
A=\begin{bmatrix}
a_{11} & a_{12} & a_{13}&\cdots&a_{1n}\\
a_{21} & a_{22} & a_{23}&\cdots&a_{2n}\\
a_{31} & a_{32} & a_{33}&\cdots&a_{3n}\\
\cdots & \cdots & \cdots&\cdots&\cdots\\
a_{n1} & a_{n2} & a_{n3}&\cdots&a_{nn}\\
\end{bmatrix}\\
B=a21a21a31⋯an1a22a22a32⋯an2a23a23a33⋯an3⋯⋯⋯⋯⋯a2na2na3n⋯ann
B=\begin{bmatrix}
a_{21} & a_{22} & a_{23}&\cdots&a_{2n}\\
a_{21} & a_{22} & a_{23}&\cdots&a_{2n}\\
a_{31} & a_{32} & a_{33}&\cdots&a_{3n}\\
\cdots & \cdots & \cdots&\cdots&\cdots\\
a_{n1} & a_{n2} & a_{n3}&\cdots&a_{nn}\\
\end{bmatrix}\\
The last n-1 rows of A and B are equal
Hence, the co-factors of the first row for the two matrices will be equal
det(B)=0
det(B) = 0
(two equal rows)
a similar argument can be made for all off-diagonal entries of AC⊤
C11C12C13⋯C1n
\begin{matrix}
C_{11} & C_{12} & C_{13} & \cdots & C_{1n}
\end{matrix}
C11C12C13⋯C1n
\begin{matrix}
C_{11} & C_{12} & C_{13} & \cdots & C_{1n}
\end{matrix}
Cramer's rule
(solving Ax=b )
x=A−1b
\mathbf{x} = A^{-1}\mathbf{b}
x=det(A)1C⊤b
\mathbf{x} = \frac{1}{det(A)}C^{\top}\mathbf{b}
C11C12C13⋯C1nC21C22C23⋯C2nC31C32C33⋯C3n⋯⋯⋯⋯⋯Cn1Cn2Cn3⋯Cnn
\begin{bmatrix}
C_{11} & C_{21} & C_{31}&\cdots&C_{n1}\\
C_{12} & C_{22} & C_{32}&\cdots&C_{n2}\\
C_{13} & C_{23} & C_{33}&\cdots&C_{n3}\\
\cdots & \cdots & \cdots&\cdots&\cdots\\
C_{1n} & C_{2n} & C_{3n}&\cdots&C_{nn}\\
\end{bmatrix}\\
b1b2b3⋯bn
\begin{bmatrix}
b_1\\
b_2\\
b_3\\
\cdots\\
b_n
\end{bmatrix}
x1x2x3⋯xn=det(A)1
\begin{bmatrix}
x_1\\
x_2\\
x_3\\
\cdots\\
x_n
\end{bmatrix}=\frac{1}{det(A)}
x1=det(A)1(b1C11+b2C21+b3C31+⋯+bnCn1)
x_1 = \frac{1}{det(A)}(b_1C_{11} + b_2C_{21} + b_3C_{31} + \cdots + b_nC_{n1})
this looks like a determinant of some matrix
det(A)=a11C11+a12C12+⋯+a1nC1n
det(A)=a_{11}C_{11}+a_{12}C_{12}+\cdots+a_{1n}C_{1n}
What is that matrix?
b1b2b3⋯bna12a22a32⋯an2a13a23a33⋯an3⋯⋯⋯⋯⋯a1na2na3n⋯ann
\begin{bmatrix}
b_{1} & a_{12} & a_{13}&\cdots&a_{1n}\\
b_{2} & a_{22} & a_{23}&\cdots&a_{2n}\\
b_{3} & a_{32} & a_{33}&\cdots&a_{3n}\\
\cdots & \cdots & \cdots&\cdots&\cdots\\
b_{n} & a_{n2} & a_{n3}&\cdots&a_{nn}\\
\end{bmatrix}\\
B1
B_1
x1=det(A)det(B1)
x_1 = \frac{det(B_1)}{det(A)}
The first column of A replaced by B
(transpose)
Cramer's rule
(solving Ax=b )
x=A−1b
\mathbf{x} = A^{-1}\mathbf{b}
x=det(A)1C⊤b
\mathbf{x} = \frac{1}{det(A)}C^{\top}\mathbf{b}
xi=det(A)det(Bi)
x_i = \frac{det(B_i)}{det(A)}
Bi is the matrix obtained by replacing the i-th column of A by b
Practically, not very useful as computing the determinant is expensive (n! terms)
(GE is cheaper and hence preferred)
Determinant = Volume
The determinant is equal to the volume of the parallelepiped formed by the columns of the matrix
Learning Objectives
(achieved)
What is the formula for computing a determinant?
What are co-factors?
What is Cramer's rule for computing the inverse?
What is the connection between determinant and volume?
CS6015: Linear Algebra and Random Processes Lecture 15: Formula for determinant, co-factors, Finding A − 1 , Cramer's rule for solving A x = b , Determinant=Volume
CS6015: Lecture 15
By Mitesh Khapra
CS6015: Lecture 15
Lecture 15: Formula for determinant, co-factors, Finding the inverse of A, Cramer's rule for solving Ax=b, Determinant=Volume
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