a)
det A = ∣ 3 0 5 9 ∣ = 3 ( 9 ) − 0 ( 5 ) = 27 ≠ 0 \det A=\begin{vmatrix}
3 & 0 \\
5 & 9
\end{vmatrix}=3(9)-0(5)=27\not=0 det A = ∣ ∣ 3 5 0 9 ∣ ∣ = 3 ( 9 ) − 0 ( 5 ) = 27 = 0 = > A − 1 e x i s t s =>A^{-1}\ exists => A − 1 e x i s t s
A − 1 = 1 27 ( 9 0 − 5 3 ) = ( 1 / 3 0 − 5 / 27 1 / 9 ) A^{-1}=\dfrac{1}{27}\begin{pmatrix}
9 & 0 \\
-5 & 3
\end{pmatrix}=\begin{pmatrix}
1/3 & 0 \\
-5/27 & 1/9
\end{pmatrix} A − 1 = 27 1 ( 9 − 5 0 3 ) = ( 1/3 − 5/27 0 1/9 )
b)
det A = ∣ − 3 7 9 1 1 3 4 9 3 ∣ = − 3 ∣ 1 3 9 3 ∣ − 7 ∣ 1 3 4 3 ∣ + 9 ∣ 1 1 4 9 ∣ \det A=\begin{vmatrix}
-3 & 7 & 9 \\
1 & 1 & 3 \\
4 & 9 & 3 \\
\end{vmatrix}=-3\begin{vmatrix}
1 & 3 \\
9 & 3
\end{vmatrix}-7\begin{vmatrix}
1 & 3\\
4 & 3
\end{vmatrix}+9\begin{vmatrix}
1 & 1 \\
4 & 9
\end{vmatrix} det A = ∣ ∣ − 3 1 4 7 1 9 9 3 3 ∣ ∣ = − 3 ∣ ∣ 1 9 3 3 ∣ ∣ − 7 ∣ ∣ 1 4 3 3 ∣ ∣ + 9 ∣ ∣ 1 4 1 9 ∣ ∣
= − 3 ( 3 − 27 ) − 7 ( 3 − 12 ) + 9 ( 9 − 4 ) = 180 ≠ 0 =-3(3-27)-7(3-12)+9(9-4)=180\not=0 = − 3 ( 3 − 27 ) − 7 ( 3 − 12 ) + 9 ( 9 − 4 ) = 180 = 0 = > A − 1 e x i s t s =>A^{-1}\ exists => A − 1 e x i s t s
To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be the inverse matrix.
( − 3 7 9 1 0 0 1 1 3 0 1 0 4 9 3 0 0 1 ) \begin{pmatrix}
-3 & 7 & 9 & & 1 & 0 & 0 \\
1 & 1 & 3 & & 0 & 1 & 0 \\
4 & 9 & 3 & & 0 & 0 & 1 \\
\end{pmatrix} ⎝ ⎛ − 3 1 4 7 1 9 9 3 3 1 0 0 0 1 0 0 0 1 ⎠ ⎞ R 1 = − R 1 / 3 R_1=-R_1/3 R 1 = − R 1 /3
( 1 − 7 / 3 − 3 − 1 / 3 0 0 1 1 3 0 1 0 4 9 3 0 0 1 ) \begin{pmatrix}
1 & -7/3 & -3 & & -1/3 & 0 & 0 \\
1 & 1 & 3 & & 0 & 1 & 0 \\
4 & 9 & 3 & & 0 & 0 & 1 \\
\end{pmatrix} ⎝ ⎛ 1 1 4 − 7/3 1 9 − 3 3 3 − 1/3 0 0 0 1 0 0 0 1 ⎠ ⎞ R 2 = R 2 − R 1 R_2=R_2-R_1 R 2 = R 2 − R 1
( 1 − 7 / 3 − 3 − 1 / 3 0 0 0 10 / 3 6 1 / 3 1 0 4 9 3 0 0 1 ) \begin{pmatrix}
1 & -7/3 & -3 & & -1/3 & 0 & 0 \\
0 & 10/3 & 6 & & 1/3 & 1 & 0 \\
4 & 9 & 3 & & 0 & 0 & 1 \\
\end{pmatrix} ⎝ ⎛ 1 0 4 − 7/3 10/3 9 − 3 6 3 − 1/3 1/3 0 0 1 0 0 0 1 ⎠ ⎞ R 3 = R 3 − 4 R 1 R_3=R_3-4R_1 R 3 = R 3 − 4 R 1
( 1 − 7 / 3 − 3 − 1 / 3 0 0 0 10 / 3 6 1 / 3 1 0 0 55 / 3 15 4 / 3 0 1 ) \begin{pmatrix}
1 & -7/3 & -3 & & -1/3 & 0 & 0 \\
0 & 10/3 & 6 & & 1/3 & 1 & 0 \\
0 & 55/3 & 15 & & 4/3 & 0 & 1 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 − 7/3 10/3 55/3 − 3 6 15 − 1/3 1/3 4/3 0 1 0 0 0 1 ⎠ ⎞ R 2 = ( 3 / 10 ) R 2 R_2=(3/10)R_2 R 2 = ( 3/10 ) R 2
( 1 − 7 / 3 − 3 − 1 / 3 0 0 0 1 9 / 5 1 / 10 3 / 10 0 0 55 / 3 15 4 / 3 0 1 ) \begin{pmatrix}
1 & -7/3 & -3 & & -1/3 & 0 & 0 \\
0 & 1 & 9/5 & & 1/10 & 3/10 & 0 \\
0 & 55/3 & 15 & & 4/3 & 0 & 1 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 − 7/3 1 55/3 − 3 9/5 15 − 1/3 1/10 4/3 0 3/10 0 0 0 1 ⎠ ⎞ R 1 = R 1 + ( 7 / 3 ) R 2 R_1=R_1+(7/3)R_2 R 1 = R 1 + ( 7/3 ) R 2
( 1 0 6 / 5 − 1 / 10 7 / 10 0 0 1 9 / 5 1 / 10 3 / 10 0 0 55 / 3 15 4 / 3 0 1 ) \begin{pmatrix}
1 & 0 & 6/5 & & -1/10 & 7/10 & 0 \\
0 & 1 & 9/5 & & 1/10 & 3/10 & 0 \\
0 & 55/3 & 15 & & 4/3 & 0 & 1 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 55/3 6/5 9/5 15 − 1/10 1/10 4/3 7/10 3/10 0 0 0 1 ⎠ ⎞ R 3 = R 3 − ( 55 / 3 ) R 2 R_3=R_3-(55/3)R_2 R 3 = R 3 − ( 55/3 ) R 2
( 1 0 6 / 5 − 1 / 10 7 / 10 0 0 1 9 / 5 1 / 10 3 / 10 0 0 0 − 18 − 1 / 2 − 11 / 2 1 ) \begin{pmatrix}
1 & 0 & 6/5 & & -1/10 & 7/10 & 0 \\
0 & 1 & 9/5 & & 1/10 & 3/10 & 0 \\
0 & 0 & -18 & & -1/2 & -11/2 & 1 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 6/5 9/5 − 18 − 1/10 1/10 − 1/2 7/10 3/10 − 11/2 0 0 1 ⎠ ⎞ R 3 = − R 3 / 18 R_3=-R_3/18 R 3 = − R 3 /18
( 1 0 6 / 5 − 1 / 10 7 / 10 0 0 1 9 / 5 1 / 10 3 / 10 0 0 0 1 1 / 36 11 / 36 − 1 / 18 ) \begin{pmatrix}
1 & 0 & 6/5 & & -1/10 & 7/10 & 0 \\
0 & 1 & 9/5 & & 1/10 & 3/10 & 0 \\
0 & 0 & 1 & & 1/36 & 11/36 & -1/18 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 6/5 9/5 1 − 1/10 1/10 1/36 7/10 3/10 11/36 0 0 − 1/18 ⎠ ⎞ R 1 = R 1 − ( 6 / 5 ) R 3 R_1=R_1-(6/5)R_3 R 1 = R 1 − ( 6/5 ) R 3
( 1 0 0 − 2 / 15 1 / 3 1 / 15 0 1 9 / 5 1 / 10 3 / 10 0 0 0 1 1 / 36 11 / 36 − 1 / 18 ) \begin{pmatrix}
1 & 0 & 0 & & -2/15 & 1/3 & 1/15 \\
0 & 1 & 9/5 & & 1/10 & 3/10 & 0 \\
0 & 0 & 1 & & 1/36 & 11/36 & -1/18 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 0 9/5 1 − 2/15 1/10 1/36 1/3 3/10 11/36 1/15 0 − 1/18 ⎠ ⎞ R 2 = R 2 − ( 9 / 5 ) R 3 R_2=R_2-(9/5)R_3 R 2 = R 2 − ( 9/5 ) R 3
( 1 0 0 − 2 / 15 1 / 3 1 / 15 0 1 0 1 / 20 − 1 / 4 1 / 10 0 0 1 1 / 36 11 / 36 − 1 / 18 ) \begin{pmatrix}
1 & 0 & 0 & & -2/15 & 1/3 & 1/15 \\
0 & 1 & 0 & & 1/20 & -1/4 & 1/10 \\
0 & 0 & 1 & & 1/36 & 11/36 & -1/18 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 0 0 1 − 2/15 1/20 1/36 1/3 − 1/4 11/36 1/15 1/10 − 1/18 ⎠ ⎞ On the left is the identity matrix. On the right is the inverse matrix.
A − 1 = ( − 2 / 15 1 / 3 1 / 15 1 / 20 − 1 / 4 1 / 10 1 / 36 11 / 36 − 1 / 18 ) A^{-1}=\begin{pmatrix}
-2/15 & 1/3 & 1/15 \\
1/20 & -1/4 & 1/10 \\
1/36 & 11/36 & -1/18 \\
\end{pmatrix} A − 1 = ⎝ ⎛ − 2/15 1/20 1/36 1/3 − 1/4 11/36 1/15 1/10 − 1/18 ⎠ ⎞
Comments