3.5 Matrix inversion
Motivation and definition ¶ Gauss-Jordan elimination works well for solving Ax = b , but the process
needs to be repeated for every new b . Is there an alternative if we need to
solve Ax = b for many different b ?
Matrix inverse
For a square (n x n ) matrix A , the inverse A − 1 \vv{A}^{-1} A − 1
A A − 1 = A − 1 A = I \vv{A} \vv{A}^{-1} = \vv{A}^{-1} \vv{A} = \vv{I} A A − 1 = A − 1 A = I where I is the n x n identity matrix.
A matrix is called nonsingular or invertible if it has an inverse, but
singular if it does not.
Invertible matrix theorem
A is invertible if and only if the determinant of A is nonzero.
(There are many more such conditions!)
If the inverse of A exists, it is unique and can be used to
solve Ax = b .
A x = b A − 1 A x = A − 1 b x = A − 1 b \begin{align}
\vv{A} \vv{x} &= \vv{b} \\
\vv{A}^{-1} \vv{A} \vv{x} &= \vv{A}^{-1} \vv{b} \\
\vv{x} &= \vv{A}^{-1} \vv{b}
\end{align} Ax A − 1 Ax x = b = A − 1 b = A − 1 b Finding the inverse of A is usually hard. There is a general definition
based on cofactors, as well as advanced numerical methods, that we will not
cover. Instead, we focus on two options: a formula for 2 x 2 matrices, and
use of Gauss-Jordan elimination for larger matrices.
Inverse of a 2 x 2 matrix ¶ For a 2 x 2 matrix,
A = [ a b c d ] , \vv{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, A = [ a c b d ] , the matrix inverse is
A − 1 = 1 ∣ A ∣ [ d − b − c a ] \vv{A}^{-1} = \frac{1}{|\vv{A}|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} A − 1 = ∣ A ∣ 1 [ d − c − b a ] (Flip a and d , change the signs of b and c .)
For example, to find the inverse of
A = [ 3 1 2 4 ] \vv{A} = \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix} A = [ 3 2 1 4 ] First, compute its determinant:
∣ A ∣ = 3 × 4 − 2 × 1 = 12 − 2 = 10 |\vv{A}| = 3 \times 4 - 2 \times 1 = 12 - 2 = 10 ∣ A ∣ = 3 × 4 − 2 × 1 = 12 − 2 = 10 Then, compute its inverse
A − 1 = 1 10 [ 4 − 1 − 2 3 ] = [ 0.4 − 0.1 − 0.2 0.3 ] \vv{A}^{-1} = \frac{1}{10} \begin{bmatrix} 4 & -1 \\ -2 & 3 \end{bmatrix}
= \begin{bmatrix} 0.4 & -0.1 \\ -0.2 & 0.3 \end{bmatrix} A − 1 = 10 1 [ 4 − 2 − 1 3 ] = [ 0.4 − 0.2 − 0.1 0.3 ] Inverses using Gauss-Jordan elimination ¶ For larger matrices, we can use Gauss–Jordan elimination to
solve A A − 1 = I \vv{A} \vv{A}^{-1} = \vv{I} A A − 1 = I Ax = b .
Check that ∣ A ∣ ≠ 0 |\vv{A}| \ne 0 ∣ A ∣ = 0 A is invertible).
Form the 2n x n augmented matrix [ A ∣ I ] [ \vv{A} \, | \, \vv{I} ] [ A ∣ I ]
Perform row operations to bring to [ I ∣ A − 1 ] [ \vv{I} \, | \, \vv{A}^{-1} ] [ I ∣ A − 1 ]
Example: Pump circuit ¶ We will resolve the problem we used Gauss-Jordan elimination
for, but now using the matrix inverse. As a reminder, the equations to solve
were
Q 1 − Q 2 + Q 3 = 0 Q 1 − Q 2 + Q 3 = 0 20 Q 1 + 10 Q 2 + 0 Q 3 = 80 0 Q 1 + 10 Q 2 + 25 Q 3 = 90 \begin{align}
Q_1 - Q_2 + Q_3 &= 0 \\
Q_1 - Q_2 + Q_3 &= 0 \\
20Q_1 + 10Q_2 + 0Q_3 &= 80 \\
0Q_1 + 10Q_2 + 25Q_3 &= 90
\end{align} Q 1 − Q 2 + Q 3 Q 1 − Q 2 + Q 3 20 Q 1 + 10 Q 2 + 0 Q 3 0 Q 1 + 10 Q 2 + 25 Q 3 = 0 = 0 = 80 = 90 We need a square matrix to calculate the inverse, and we know the second
equation is redundant of the first. We write this in matrix form Ax = b
with
A = [ 1 − 1 1 20 10 0 0 10 25 ] b = [ 0 80 90 ] \vv{A} = \begin{bmatrix} 1 & -1 & 1 \\ 20 & 10 & 0 \\ 0 & 10 & 25\end{bmatrix}
\qquad
\vv{b} = \begin{bmatrix} 0 \\ 80 \\ 90 \end{bmatrix} A = ⎣ ⎡ 1 20 0 − 1 10 10 1 0 25 ⎦ ⎤ b = ⎣ ⎡ 0 80 90 ⎦ ⎤ First, compute the determinant to make sure A is invertible
∣ A ∣ = − 10 ∣ 1 1 20 0 ∣ + 25 ∣ 1 − 1 20 10 ∣ = 10 ( 0 − 20 ) + 25 ( 10 + 20 ) = 950 \begin{align}
|\vv{A}| &= -10 \begin{vmatrix} 1 & 1 \\ 20 & 0 \end{vmatrix} +
25 \begin{vmatrix} 1 & -1 \\ 20 & 10\end{vmatrix} \\
&= 10 (0-20) + 25 (10+20) = 950
\end{align} ∣ A ∣ = − 10 ∣ ∣ 1 20 1 0 ∣ ∣ + 25 ∣ ∣ 1 20 − 1 10 ∣ ∣ = 10 ( 0 − 20 ) + 25 ( 10 + 20 ) = 950 The determinant is nonzero, so A − 1 \vv{A}^{-1} A − 1
[ 1 − 1 1 1 0 0 20 10 0 0 1 0 0 10 25 0 0 1 ] R 1 − 20 R 1 R 1 → [ 1 − 1 1 1 0 0 0 30 − 20 − 20 1 0 0 10 25 0 0 1 ] R 1 s w a p R 3 R 1 → [ 1 − 1 1 1 0 0 0 10 25 0 0 1 0 30 − 20 − 20 1 0 ] R 1 R 1 − 3 R 2 → [ 1 − 1 1 1 0 0 0 10 25 0 0 1 0 0 − 95 − 20 1 − 3 ] R 1 ÷ 10 ÷ − 95 → [ 1 − 1 1 1 0 0 0 1 2.5 0 0 0.1 0 0 1 . 211 − . 010 0.032 ] − R 3 − 2.5 R 3 R 3 → [ 1 − 1 0 0.789 0.010 − 0.032 0 1 0 − 0.528 0.025 0.020 0 0 1 0.211 − 0.010 0.032 ] + R 2 R 2 R 3 → [ 1 0 0 0.261 0.035 − 0.012 0 1 0 − 0.528 0.025 0.020 0 0 1 0.211 − 0.010 0.032 ] \begin{align}
&\begin{bmatrix}
1 & -1 & 1 & 1 & 0 & 0 \\
20 & 10 & 0 & 0 & 1 & 0 \\
0 & 10 & 25 & 0 & 0 & 1
\end{bmatrix}
\begin{matrix}
\vphantom{R_1} \\ -20 R_1 \\ \vphantom{R_1}
\end{matrix} \\
\to &\begin{bmatrix}
1 & -1 & 1 & 1 & 0 & 0 \\
0 & 30 & -20 & -20 & 1 & 0 \\
0 & 10 & 25 & 0 & 0 & 1
\end{bmatrix}
\begin{matrix}
\vphantom{R_1} \\ {\rm swap}\,R_3 \\ \vphantom{R_1}
\end{matrix} \\
\to &\begin{bmatrix}
1 & -1 & 1 & 1 & 0 & 0 \\
0 & 10 & 25 & 0 & 0 & 1 \\
0 & 30 & -20 & -20 & 1 & 0
\end{bmatrix}
\begin{matrix}
\vphantom{R_1} \\ \vphantom{R_1} \\ -3 R_2
\end{matrix} \\
\to &\begin{bmatrix}
1 & -1 & 1 & 1 & 0 & 0 \\
0 & 10 & 25 & 0 & 0 & 1 \\
0 & 0 & -95 & -20 & 1 & -3
\end{bmatrix}
\begin{matrix}
\vphantom{R_1} \\ \div 10 \\ \div -95
\end{matrix} \\
\to &\begin{bmatrix}
1 & -1 & 1 & 1 & 0 & 0 \\
0 & 1 & 2.5 & 0 & 0 & 0.1 \\
0 & 0 & 1 & .211 & -.010 & 0.032
\end{bmatrix}
\begin{matrix}
-R_3 \\ -2.5 R_3 \\ \vphantom{R_3}
\end{matrix} \\
\to &\begin{bmatrix}
1 & -1 & 0 & 0.789 & 0.010 & -0.032 \\
0 & 1 & 0 & -0.528 & 0.025 & 0.020 \\
0 & 0 & 1 & 0.211 & -0.010 & 0.032
\end{bmatrix}
\begin{matrix}
+R_2 \\ \vphantom{R_2} \\ \vphantom{R_3}
\end{matrix} \\
\to &\begin{bmatrix}
1 & 0 & 0 & 0.261 & 0.035 & -0.012 \\
0 & 1 & 0 & -0.528 & 0.025 & 0.020 \\
0 & 0 & 1 & 0.211 & -0.010 & 0.032
\end{bmatrix}
\end{align} → → → → → → ⎣ ⎡ 1 20 0 − 1 10 10 1 0 25 1 0 0 0 1 0 0 0 1 ⎦ ⎤ R 1 − 20 R 1 R 1 ⎣ ⎡ 1 0 0 − 1 30 10 1 − 20 25 1 − 20 0 0 1 0 0 0 1 ⎦ ⎤ R 1 swap R 3 R 1 ⎣ ⎡ 1 0 0 − 1 10 30 1 25 − 20 1 0 − 20 0 0 1 0 1 0 ⎦ ⎤ R 1 R 1 − 3 R 2 ⎣ ⎡ 1 0 0 − 1 10 0 1 25 − 95 1 0 − 20 0 0 1 0 1 − 3 ⎦ ⎤ R 1 ÷ 10 ÷ − 95 ⎣ ⎡ 1 0 0 − 1 1 0 1 2.5 1 1 0 .211 0 0 − .010 0 0.1 0.032 ⎦ ⎤ − R 3 − 2.5 R 3 R 3 ⎣ ⎡ 1 0 0 − 1 1 0 0 0 1 0.789 − 0.528 0.211 0.010 0.025 − 0.010 − 0.032 0.020 0.032 ⎦ ⎤ + R 2 R 2 R 3 ⎣ ⎡ 1 0 0 0 1 0 0 0 1 0.261 − 0.528 0.211 0.035 0.025 − 0.010 − 0.012 0.020 0.032 ⎦ ⎤ Solve:
Q = A − 1 b = [ 0.261 0.035 − 0.012 − 0.528 0.025 0.020 0.211 − 0.010 0.032 ] [ 0 80 90 ] = [ 1.72 3.80 2.08 ] \vv{Q} = \vv{A}^{-1}{b} = \begin{bmatrix}
0.261 & 0.035 & -0.012 \\
-0.528 & 0.025 & 0.020 \\
0.211 & -0.010 & 0.032
\end{bmatrix}
\begin{bmatrix}
0 \\
80 \\
90
\end{bmatrix}
=
\begin{bmatrix}
1.72 \\
3.80 \\
2.08
\end{bmatrix} Q = A − 1 b = ⎣ ⎡ 0.261 − 0.528 0.211 0.035 0.025 − 0.010 − 0.012 0.020 0.032 ⎦ ⎤ ⎣ ⎡ 0 80 90 ⎦ ⎤ = ⎣ ⎡ 1.72 3.80 2.08 ⎦ ⎤ This is close to the solution we found before, with some errors due to rounding.
Skill builder problems ¶ First, write in matrix form Ax = b with:
A = [ 5 − 2 − 1 4 ] b = [ 20.9 − 19.3 ] \vv{A} = \begin{bmatrix}
5 & -2 \\
-1 & 4
\end{bmatrix}
\qquad
\vv{b} = \begin{bmatrix}
20.9 \\
-19.3
\end{bmatrix} A = [ 5 − 1 − 2 4 ] b = [ 20.9 − 19.3 ] Then, evaluate ∣ A ∣ |\vv{A}| ∣ A ∣
∣ A ∣ = ( 5 ⋅ 4 ) − ( − 1 ⋅ − 2 ) = 18 |\vv{A}| = (5 \cdot 4) - (-1 \cdot -2) = 18 ∣ A ∣ = ( 5 ⋅ 4 ) − ( − 1 ⋅ − 2 ) = 18 ∣ A ∣ ≠ 0 |\vv{A}| \ne 0 ∣ A ∣ = 0
A − 1 = 1 18 [ 5 − 2 − 1 4 ] \vv{A}^{-1} = \frac{1}{18}
\begin{bmatrix}
5 & -2 \\
-1 & 4
\end{bmatrix} A − 1 = 18 1 [ 5 − 1 − 2 4 ] Last, solve for x :
x = A − 1 b = 1 18 [ 5 − 2 − 1 4 ] [ 20.9 − 19.3 ] = 1 18 [ 4 ⋅ 20.9 + 2 ⋅ − 19.3 1 ⋅ 20.9 + 5 ⋅ − 19.3 ] = [ 2.5 − 4.2 ] \begin{align}
\vv{x} = \vv{A}^{-1}\vv{b} &= \frac{1}{18}
\begin{bmatrix}
5 & -2 \\
-1 & 4
\end{bmatrix}
\begin{bmatrix}
20.9 \\
-19.3
\end{bmatrix} \\
&= \frac{1}{18}
\begin{bmatrix}
4 \cdot 20.9 + 2 \cdot -19.3 \\
1 \cdot 20.9 + 5 \cdot -19.3
\end{bmatrix} \\
&= \begin{bmatrix}
2.5 \\
-4.2
\end{bmatrix}
\end{align} x = A − 1 b = 18 1 [ 5 − 1 − 2 4 ] [ 20.9 − 19.3 ] = 18 1 [ 4 ⋅ 20.9 + 2 ⋅ − 19.3 1 ⋅ 20.9 + 5 ⋅ − 19.3 ] = [ 2.5 − 4.2 ] Therefore, x 1 = 2.5 x_1 = 2.5 x 1 = 2.5 x 2 = − 4.2 x_2 = -4.2 x 2 = − 4.2
First, write in matrix form Ax = b with:
A = [ 1 4 2 8 ] b = [ 8 17 ] \vv{A} = \begin{bmatrix}
1 & 4 \\
2 & 8
\end{bmatrix}
\qquad
\vv{b} = \begin{bmatrix}
8 \\
17
\end{bmatrix} A = [ 1 2 4 8 ] b = [ 8 17 ] Then, evalaute ∣ A ∣ |\vv{A}| ∣ A ∣
∣ A ∣ = ( 1 ⋅ 8 ) − ( 2 ⋅ 4 ) = 0 |\vv{A}| = (1 \cdot 8) - (2 \cdot 4) = 0 ∣ A ∣ = ( 1 ⋅ 8 ) − ( 2 ⋅ 4 ) = 0 A is singular because ∣ A ∣ = 0 |\vv{A}| = 0 ∣ A ∣ = 0
First, write in matrix form Ax = b with:
A = [ 0 1 1 0 4 6 1 1 1 ] b = [ − 2 − 12 2 ] \vv{A} = \begin{bmatrix}
0 & 1 & 1 \\
0 & 4 & 6 \\
1 & 1 & 1
\end{bmatrix}
\qquad
\vv{b} = \begin{bmatrix}
-2 \\
-12 \\
2
\end{bmatrix} A = ⎣ ⎡ 0 0 1 1 4 1 1 6 1 ⎦ ⎤ b = ⎣ ⎡ − 2 − 12 2 ⎦ ⎤ Then, evaluate ∣ A ∣ |\vv{A}| ∣ A ∣
∣ A ∣ = 0 ⋅ ∣ 4 6 1 1 ∣ − 1 ⋅ ∣ 0 6 1 1 ∣ + 1 ⋅ ∣ 0 4 1 1 ∣ \begin{align}
|\vv{A}| =
0 \cdot \begin{vmatrix}
4 & 6 \\
1 & 1
\end{vmatrix}
-1 \cdot
\begin{vmatrix}
0 & 6 \\
1 & 1
\end{vmatrix}
+ 1 \cdot
\begin{vmatrix}
0 & 4 \\
1 & 1
\end{vmatrix}
\end{align} ∣ A ∣ = 0 ⋅ ∣ ∣ 4 1 6 1 ∣ ∣ − 1 ⋅ ∣ ∣ 0 1 6 1 ∣ ∣ + 1 ⋅ ∣ ∣ 0 1 4 1 ∣ ∣ ∣ A ∣ = − 1 ⋅ ( 0 − 6 ) + 1 ⋅ ( 0 − 4 ) = 2 |\vv{A}| = -1 \cdot (0-6) + 1 \cdot (0-4) = 2 ∣ A ∣ = − 1 ⋅ ( 0 − 6 ) + 1 ⋅ ( 0 − 4 ) = 2 Since ∣ A ∣ ≠ 0 |\vv{A}| \ne 0 ∣ A ∣ = 0 A is invertible. Use Gauss-Jordan elimination
to find the inverse. Start with the augmented matrix [ A ∣ I ] [\vv{A} | \vv{I}] [ A ∣ I ]
[ 0 1 1 1 0 0 0 4 6 0 1 0 1 1 1 0 0 1 ] → [ 1 1 1 0 0 1 0 1 1 1 0 0 0 4 6 0 1 0 ] R 1 R 1 − 4 R 1 → [ 1 1 1 0 0 1 0 1 1 1 0 0 0 0 2 − 4 1 0 ] R 1 R 1 ÷ 2 → [ 1 1 1 0 0 1 0 1 1 1 0 0 0 0 1 − 2 0.5 0 ] − R 3 − R 3 R 1 → [ 1 1 0 2 − 0.5 1 0 1 0 3 − 0.5 0 0 0 1 − 2 0.5 0 ] − R 2 R 1 R 1 → [ 1 0 0 − 1 0 1 0 1 0 3 − 0.5 0 0 0 1 − 2 0.5 0 ] \begin{align}
\begin{bmatrix}
0 & 1 & 1 & 1 & 0 & 0 \\
0 & 4 & 6 & 0 & 1 & 0 \\
1 & 1 & 1 & 0 & 0 & 1
\end{bmatrix}
&\to \begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 1 \\
0 & 1 & 1 & 1 & 0 & 0 \\
0 & 4 & 6 & 0 & 1 & 0
\end{bmatrix}
\begin{matrix}
\vphantom{R_1} \\ \vphantom{R_1} \\ -4 R_1
\end{matrix} \\
&\to \begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 1 \\
0 & 1 & 1 & 1 & 0 & 0 \\
0 & 0 & 2 & -4 & 1 & 0
\end{bmatrix}
\begin{matrix}
\vphantom{R_1} \\ \vphantom{R_1} \\ \div 2
\end{matrix} \\
&\to \begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 1 \\
0 & 1 & 1 & 1 & 0 & 0 \\
0 & 0 & 1 & -2 & 0.5 & 0
\end{bmatrix}
\begin{matrix}
-R_3 \\ -R_3 \\ \vphantom{R_1}
\end{matrix} \\
&\to \begin{bmatrix}
1 & 1 & 0 & 2 & -0.5 & 1 \\
0 & 1 & 0 & 3 & -0.5 & 0 \\
0 & 0 & 1 & -2 & 0.5 & 0
\end{bmatrix}
\begin{matrix}
-R_2 \\ \vphantom{R_1} \\ \vphantom{R_1}
\end{matrix} \\
&\to \begin{bmatrix}
1 & 0 & 0 & -1 & 0 & 1 \\
0 & 1 & 0 & 3 & -0.5 & 0 \\
0 & 0 & 1 & -2 & 0.5 & 0
\end{bmatrix}
\end{align} ⎣ ⎡ 0 0 1 1 4 1 1 6 1 1 0 0 0 1 0 0 0 1 ⎦ ⎤ → ⎣ ⎡ 1 0 0 1 1 4 1 1 6 0 1 0 0 0 1 1 0 0 ⎦ ⎤ R 1 R 1 − 4 R 1 → ⎣ ⎡ 1 0 0 1 1 0 1 1 2 0 1 − 4 0 0 1 1 0 0 ⎦ ⎤ R 1 R 1 ÷ 2 → ⎣ ⎡ 1 0 0 1 1 0 1 1 1 0 1 − 2 0 0 0.5 1 0 0 ⎦ ⎤ − R 3 − R 3 R 1 → ⎣ ⎡ 1 0 0 1 1 0 0 0 1 2 3 − 2 − 0.5 − 0.5 0.5 1 0 0 ⎦ ⎤ − R 2 R 1 R 1 → ⎣ ⎡ 1 0 0 0 1 0 0 0 1 − 1 3 − 2 0 − 0.5 0.5 1 0 0 ⎦ ⎤ Hence,
A − 1 = [ − 1 0 1 3 − 0.5 0 − 2 0.5 0 ] \begin{align}
\vv{A}^{-1} = \begin{bmatrix}
-1 & 0 & 1 \\
3 & -0.5 & 0 \\
-2 & 0.5 & 0
\end{bmatrix}
\end{align} A − 1 = ⎣ ⎡ − 1 3 − 2 0 − 0.5 0.5 1 0 0 ⎦ ⎤ and
x = A − 1 b = [ − 1 0 1 3 − 0.5 0 − 2 0.5 0 ] [ − 2 − 12 2 ] = [ − 1 ⋅ − 2 + 1 ⋅ 2 3 ⋅ − 2 + − 0.5 ⋅ − 12 − 2 ⋅ − 2 + 0.5 ⋅ − 12 ] = [ 4 0 − 2 ] \begin{align}
\vv{x} = \vv{A}^{-1} \vv{b} &=
\begin{bmatrix}
-1 & 0 & 1 \\
3 & -0.5 & 0 \\
-2 & 0.5 & 0
\end{bmatrix}
\begin{bmatrix}
-2 \\
-12 \\
2
\end{bmatrix} \\
& = \begin{bmatrix}
-1 \cdot -2 + 1 \cdot 2 \\
3 \cdot -2 + -0.5 \cdot -12 \\
-2 \cdot -2 + 0.5 \cdot -12
\end{bmatrix} \\
&= \begin{bmatrix}
4 \\
0 \\
-2
\end{bmatrix}
\end{align} x = A − 1 b = ⎣ ⎡ − 1 3 − 2 0 − 0.5 0.5 1 0 0 ⎦ ⎤ ⎣ ⎡ − 2 − 12 2 ⎦ ⎤ = ⎣ ⎡ − 1 ⋅ − 2 + 1 ⋅ 2 3 ⋅ − 2 + − 0.5 ⋅ − 12 − 2 ⋅ − 2 + 0.5 ⋅ − 12 ⎦ ⎤ = ⎣ ⎡ 4 0 − 2 ⎦ ⎤ So, x 1 = 4 x_1 = 4 x 1 = 4 x 2 = 0 x_2 = 0 x 2 = 0 x 3 = − 2 x_3 = -2 x 3 = − 2
First, write in matrix form Ax = b with:
A = [ 0 4 4 3 − 11 − 2 6 − 17 1 ] b = [ 24 − 6 18 ] \vv{A} = \begin{bmatrix}
0 & 4 & 4 \\
3 & -11 & -2 \\
6 & -17 & 1
\end{bmatrix}
\qquad
\vv{b} = \begin{bmatrix}
24 \\
-6 \\
18
\end{bmatrix} A = ⎣ ⎡ 0 3 6 4 − 11 − 17 4 − 2 1 ⎦ ⎤ b = ⎣ ⎡ 24 − 6 18 ⎦ ⎤ Then, evaluate ∣ A ∣ |\vv{A}| ∣ A ∣
∣ A ∣ = 0 ⋅ ∣ − 11 − 2 − 17 1 ∣ − 4 ⋅ ∣ 3 − 2 6 1 ∣ + 4 ⋅ ∣ 3 − 11 6 − 17 ∣ \begin{align}
|\vv{A}| =
0 \cdot \begin{vmatrix}
-11 & -2 \\
-17 & 1
\end{vmatrix}
-4 \cdot
\begin{vmatrix}
3 & -2 \\
6 & 1
\end{vmatrix}
+ 4 \cdot
\begin{vmatrix}
3 & -11 \\
6 & -17
\end{vmatrix}
\end{align} ∣ A ∣ = 0 ⋅ ∣ ∣ − 11 − 17 − 2 1 ∣ ∣ − 4 ⋅ ∣ ∣ 3 6 − 2 1 ∣ ∣ + 4 ⋅ ∣ ∣ 3 6 − 11 − 17 ∣ ∣ ∣ A ∣ = − 4 ( 3 ⋅ 1 − 6 ⋅ − 2 ) + 4 ( 3 ⋅ − 17 − 6 ⋅ − 11 ) = − 4 ⋅ 15 + 4 ⋅ 15 = 0 \begin{align}
|\vv{A}| &= -4 (3 \cdot 1 - 6 \cdot -2) + 4 (3 \cdot -17 - 6 \cdot -11) \\
&= -4 \cdot 15 + 4 \cdot 15 \\
&= 0
\end{align} ∣ A ∣ = − 4 ( 3 ⋅ 1 − 6 ⋅ − 2 ) + 4 ( 3 ⋅ − 17 − 6 ⋅ − 11 ) = − 4 ⋅ 15 + 4 ⋅ 15 = 0 Since ∣ A ∣ = 0 |\vv{A}| = 0 ∣ A ∣ = 0 A is not invertible.
First, write in matrix form Ax = b with:
A = [ 2 − 1 3 − 4 2 − 6 ] b = [ − 1 2 ] \vv{A} = \begin{bmatrix}
2 & -1 & 3 \\
-4 & 2 & -6
\end{bmatrix}
\qquad
\vv{b} = \begin{bmatrix}
-1 \\
2
\end{bmatrix} A = [ 2 − 4 − 1 2 3 − 6 ] b = [ − 1 2 ] Since A is not square, A is not invertible.