If A = $⎡ ⎢ ⎣ \begin{matrix} 2 & - 3 & 5 3 & 2 & - 4 1 & 1 & - 2 \end{matrix} ⎤ ⎥ ⎦$ , find $A^{- 1}$ . Use it to solve the system of equation 2x-3y+5z = 11, 3x+2y-4z = -5, x+y-2z =-3.

OR

Using elementary row transformations, find the inverse of the matrix

A = $⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 2 & 5 & 7 - 2 & - 4 & - 5 \end{matrix} ⎤ ⎥ ⎦$

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Solution

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

$∴ | A | = A = ⎡ ⎢ ⎣ \begin{matrix} 2 & - 3 & 5 3 & 2 & - 4 1 & 1 & - 2 \end{matrix} ⎤ ⎥ ⎦$ = 0+3(-2) +5(1)= -1 $\neq 0$

$∴ A^{- 1}$ exists.

Consider $C_{i j}$ be the cofactor of corresponding element $a_{i j}$ of matrix A.

$C_{11} = 0, C_{12} = 2, C_{13} = 1, C_{21} = - 1, C_{22} = - 9, C_{23} = - 5, C_{31} = 2, C_{32} = 2, C_{33} = 13, ∴ a d j . A = ⎡ ⎢ ⎣ \begin{matrix} 0 & - 1 & 2 2 & - 9 & 23 1 & - 5 & 13 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow A^{- 1} = \frac{a d j . A}{| A |} = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & - 2 - 2 & 9 & - 23 - 1 & 5 & - 13 \end{matrix} ⎤ ⎥ ⎦$

Now 2x-3y+5z = 11, 3x+2y-4z= -5, x+y - 2z= -3

Let A = $⎡ ⎢ ⎣ \begin{matrix} 2 & - 3 & 5 3 & 2 & - 4 1 & 1 & - 2 \end{matrix} ⎤ ⎥ ⎦, X = ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ and B = ⎡ ⎢ ⎣ \begin{matrix} 11 - 5 - 3 \end{matrix} ⎤ ⎥ ⎦$

As AX = B $\Rightarrow X = A^{- 1} B = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & - 2 - 2 & 9 & - 23 - 1 & 5 & - 13 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 11 - 5 - 3 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 123 \end{matrix} ⎤ ⎥ ⎦$

By equality of matrices , we get : x = 1, y = 2, z = 3.

OR Given that A = $⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 2 & 5 & 7 - 2 & - 4 & - 5 \end{matrix} ⎤ ⎥ ⎦$

Using elementary row operation we have : A = IA i.e., $⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 2 & 5 & 7 - 2 & - 4 & - 5 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ A$

By $R_{2} \to R_{2} - 2 R_{1}, R_{3} \to R_{3} + 2 R_{1}, ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 0 & 1 & 1 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 - 2 & 1 & 0 2 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ A By R_{1} \to R_{1} - 2 R_{2}, R_{2} \to R_{2} + R_{3}, ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 5 & - 2 & 0 - 4 & 1 & - 1 2 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ A By R_{1} \Rightarrow R_{1} - R_{3}, ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 2 & - 1 - 4 & 1 & - 1 2 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ A$

Using I = $A^{- 1}$ A, we get $A^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 2 & - 1 - 4 & 1 & - 1 2 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$

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Similar questions

If $A = ⎡ ⎢ ⎣ \begin{matrix} 2 & - 3 & 5 3 & 2 & - 4 1 & 1 & - 2 \end{matrix} ⎤ ⎥ ⎦$ , find $A^{- 1}$ . Using $A^{- 1}$ 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. If

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

, find

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. If

A =

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

, then find

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. Find the inverse of the matrix

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

by elementary row transformation. Hence solve the system of equations

x + 2 y + z = 4, - x + 2 z = 0, 2 x + y - 3 z = 0

Q. If

A =

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

, then find

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

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If A =⎡⎢⎣2−3532−411−2⎤⎥⎦, find A−1. Use it to solve the system of equation 2x-3y+5z = 11, 3x+2y-4z = -5, x+y-2z =-3. OR Using elementary row transformations, find the inverse of the matrix A =⎡⎢⎣123257−2−4−5⎤⎥⎦