The coefficient matrix is:
A = ( 2 1 5 0 ) A=\begin{pmatrix}
2 & 1 \\
5 & 0
\end{pmatrix} A = ( 2 5 1 0 ) The variable matrix is:
X = ( p q ) X=\begin{pmatrix}
p \\
q
\end{pmatrix} X = ( p q ) The constant matrix is:
B = ( 5 8 ) B=\begin{pmatrix}
5 \\
8
\end{pmatrix} B = ( 5 8 ) Thus, to solve a system A X = B , AX=B, A X = B , for X , X, X , multiply both sides by the inverse of A A A
A − 1 A X = A − 1 B A^{-1}AX=A^{-1}B A − 1 A X = A − 1 B and we shall obtain the solution:
X = A − 1 B X=A^{-1}B X = A − 1 B Provided the inverse A − 1 A^{-1} A − 1 exists, this formula will solve the system.
det A = ∣ A ∣ = ∣ 2 1 5 0 ∣ = 2 ( 0 ) − 1 ( 5 ) = − 5 ≠ 0 \det A=|A|=\begin{vmatrix}
2 & 1 \\
5 & 0
\end{vmatrix}=2(0)-1(5)=-5\not=0 det A = ∣ A ∣ = ∣ ∣ 2 5 1 0 ∣ ∣ = 2 ( 0 ) − 1 ( 5 ) = − 5 = 0 The inverse A − 1 A^{-1} A − 1 exists.
A − 1 = 1 − 5 ( 0 − 1 − 5 2 ) = ( 0 1 / 5 1 − 2 / 5 ) A^{-1}=\dfrac{1}{-5}\begin{pmatrix}
0 & -1 \\
-5 & 2
\end{pmatrix}=\begin{pmatrix}
0 & 1/5 \\
1 & -2/5
\end{pmatrix} A − 1 = − 5 1 ( 0 − 5 − 1 2 ) = ( 0 1 1/5 − 2/5 )
X = A − 1 B = ( 0 1 / 5 1 − 2 / 5 ) ( 5 8 ) X=A^{-1}B=\begin{pmatrix}
0 & 1/5 \\
1 & -2/5
\end{pmatrix}\begin{pmatrix}
5 \\
8
\end{pmatrix} X = A − 1 B = ( 0 1 1/5 − 2/5 ) ( 5 8 )
= ( 0 ( 5 ) + ( 1 / 5 ) ( 8 ) 1 ( 5 ) + ( − 2 / 5 ) ( 8 ) ) = ( 8 / 5 9 / 5 ) =\begin{pmatrix}
0(5)+(1/5)(8) \\
1(5)+(-2/5)(8)
\end{pmatrix}=\begin{pmatrix}
8/5 \\
9/5
\end{pmatrix} = ( 0 ( 5 ) + ( 1/5 ) ( 8 ) 1 ( 5 ) + ( − 2/5 ) ( 8 ) ) = ( 8/5 9/5 )
( p q ) = ( 8 / 5 9 / 5 ) \begin{pmatrix}
p \\
q
\end{pmatrix}=\begin{pmatrix}
8/5 \\
9/5
\end{pmatrix} ( p q ) = ( 8/5 9/5 )
p = 8 / 5 , q = 9 / 5 p=8/5, q=9/5 p = 8/5 , q = 9/5
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