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Inverse of a Matrix

Use product ...

Question

Use product $⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 & 2 0 & 2 & - 3 3 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦$ $⎡ ⎢ ⎣ \begin{matrix} - 2 & 0 & 1 9 & 2 & - 3 6 & 1 & - 2 \end{matrix} ⎤ ⎥ ⎦$
to solve the system of equation:
$x - y + 2 z = 1$
$2 y - 3 z = 1$
$3 x - 2 y + 4 z = 2$

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Solution

$⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 & 2 0 & 2 & - 3 3 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦ \times ⎡ ⎢ ⎣ \begin{matrix} - 2 & 0 & 1 9 & 2 & - 3 6 & 1 & - 2 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} - 2 - 9 + 12 & - 2 + 2 & 1 + 3 - 4 18 - 18 & 4 - 3 & - 6 + 6 - 6 - 18 + 24 & - 4 + 4 & 3 + 6 - 8 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = I_{3 \times 3} - - - - - - - (1)$
From the equations
$⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 & 2 0 & 2 & - 3 3 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦ \times ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 112 \end{matrix} ⎤ ⎥ ⎦$
or $A X = B$
Where $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 & 2 0 & 2 & - 3 3 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦, X = ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦, B = ⎡ ⎢ ⎣ \begin{matrix} 112 \end{matrix} ⎤ ⎥ ⎦$
or, $X = A^{- 1} B$
From the equation $(1) A^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} - 2 & 0 & 1 9 & 2 & - 3 6 & 1 & - 2 \end{matrix} ⎤ ⎥ ⎦$
or $X = ⎡ ⎢ ⎣ \begin{matrix} - 2 & 0 & 1 9 & 2 & - 3 6 & 1 & - 2 \end{matrix} ⎤ ⎥ ⎦ \times ⎡ ⎢ ⎣ \begin{matrix} 112 \end{matrix} ⎤ ⎥ ⎦$
or $X = ⎡ ⎢ ⎣ \begin{matrix} 053 \end{matrix} ⎤ ⎥ ⎦$
or $x = 0, y = 5, z = 3$

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⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 & 2 0 & 2 & - 3 3 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} - 2 & 0 & 1 9 & 2 & - 3 6 & 1 & - 2 \end{matrix} ⎤ ⎥ ⎦

Q. Solve the following system of linear equation by matrix method

x - y + 2 z = 1

2 y - 3 z = 1

3 x - 2 y + 4 z = 2

A = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}]

, find A⁻¹. Using A⁻¹, solve the system of linear equations
x − 2y = 10, 2x + y + 3z = 8, −2y + z = 7

(ii)

A = [\begin{matrix} 3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{matrix}]

, find A⁻¹ and hence solve the following system of equations:
3x − 4y + 2z = −1, 2x + 3y + 5z = 7, x + z = 2

(iii)

A = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}] and B = [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}]

, find AB. Hence, solve the system of equations:
x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

(iv) If

A = [\begin{matrix} 1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1 \end{matrix}]

, find A⁻¹. Using A⁻¹, solve the system of linear equations
x − 2y = 10, 2x − y − z = 8, −2y + z = 7

(v) Given

A = [\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}], B = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}]

, find BA and use this to solve the system of equations
y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17

(vi) If

A = (\begin{array}{rrr} 2 & 3 & 1 \\ 1 & 2 & 2 \\ - 3 & 1 & - 1 \end{array})

, find A^–1 and hence solve the system of equations 2x + y – 3z = 13, 3x + 2y + z = 4, x + 2y – z = 8.
(vii) Use product

[\begin{matrix} 1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{matrix}] [\begin{matrix} - 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2 \end{matrix}]

to solve the system of equations x + 3z = 9, −x + 2y − 2z = 4, 2x − 3y + 4z = −3.

A = ⎡ ⎢ ⎣ \begin{matrix} 2 & - 3 & 5 3 & 2 & - 4 1 & 1 & - 2 \end{matrix} ⎤ ⎥ ⎦

A^{- 1}

. Use it to solve the system of equations

2 x - 3 y + 5 z = 11

3 x + 2 y - 4 z = - 5

x + y - 2 z = - 3

Q. Solve the system of equations

x + 2 y + 3 z = 1 2 x + 3 y + 2 z = 2 3 x + 3 y + 4 z = 1

Find the value of

14 (x + y + z)

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