Solved on Nov 06, 2023

Find the inverse $A^{-1}$ of the given 3x3 matrix $A = \begin{bmatrix} 1 & 1 & 3 \\ -2 & -1 & -6 \\ 0 & 3 & 1 \end{bmatrix}$ .

STEP 1

Assumptions1. The given matrix is a3x3 matrix $A = \begin{bmatrix}1 &1 &3 \\ - & -1 & -6 \\0 &3 &1 \end{bmatrix}$ . We are asked to find the inverse of matrix A, denoted as $A^{-1}$

STEP 2

To find the inverse of a matrix, we first need to create an augmented matrix by appending the identity matrix to the given matrix. The identity matrix is a square matrix with ones on the diagonal and zeros elsewhere.
$[A|] = \left[\begin{array}{ccc|ccc}1 &1 & &1 &0 &0 \\ -2 & -1 & -6 &0 &1 &0 \\0 & &1 &0 &0 &1 \end{array}\right]$

STEP 3

Next, we perform row operations to transform the left side of the augmented matrix into the identity matrix. The goal is to get a diagonal of ones and zeros everywhere else on the left side of the augmented matrix.
First, we swap row1 and row2 to get a1 in the top left corner.
$\left[\begin{array}{ccc|ccc} -2 & -1 & -6 &0 &1 &0 \\1 &1 &3 &1 &0 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]$

STEP 4

Multiply the first row by -1/2 to get a1 in the top left corner.
$\left[\begin{array}{ccc|ccc}1 &0. &3 &0 & -0. &0 \\1 &1 &3 &1 &0 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]$

STEP 5

Subtract the first row from the second row to get a0 in the second row, first column.
$\left[\begin{array}{ccc|ccc}1 &0.5 &3 &0 & -0.5 &0 \\0 &0.5 &0 &1 &0.5 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]$

STEP 6

Subtract6 times the second row from the first row to get a0 in the first row, second column.
$\left[\begin{array}{ccc|ccc}1 &0 &3 & -6 & -3 &0 \\0 &0.5 &0 &1 &0.5 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]$

STEP 7

Multiply the second row by2 to get a1 in the second row, second column.
$\left[\begin{array}{ccc|ccc}1 &0 &3 & -6 & -3 &0 \\0 &1 &0 &2 &1 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]$

STEP 8

Subtract3 times the second row from the third row to get a0 in the third row, second column.
$\left[\begin{array}{ccc|ccc}1 &0 &3 & -6 & -3 &0 \\0 &1 &0 &2 &1 &0 \\0 &0 &1 & -6 & -3 &1 \end{array}\right]$

STEP 9

Subtract3 times the third row from the first row to get a in the first row, third column.
$\left[\begin{array}{ccc|ccc} & & &12 &6 & -3 \\ & & &2 & & \\ & & & -6 & -3 & \end{array}\right]$

STEP 10

Now, the left side of the augmented matrix is the identity matrix. The right side of the augmented matrix is the inverse of the original matrix.
$A^{-} = \begin{bmatrix}12 &6 & -3 \\2 & &0 \\ -6 & -3 & \end{bmatrix}$ So, the inverse of matrix A is $A^{-} = \begin{bmatrix}12 &6 & -3 \\2 & &0 \\ -6 & -3 & \end{bmatrix}$ .

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Find the inverse $A^{-1}$ of the given 3x3 matrix $A = \begin{bmatrix} 1 & 1 & 3 \\ -2 & -1 & -6 \\ 0 & 3 & 1 \end{bmatrix}$ .

STEP 1

Assumptions1. The given matrix is a3x3 matrix $A = \begin{bmatrix}1 &1 &3 \\ - & -1 & -6 \\0 &3 &1 \end{bmatrix}$ . We are asked to find the inverse of matrix A, denoted as $A^{-1}$

STEP 2

STEP 3

STEP 4

Multiply the first row by -1/2 to get a1 in the top left corner.
$\left[\begin{array}{ccc|ccc}1 &0. &3 &0 & -0. &0 \\1 &1 &3 &1 &0 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]$

STEP 5

Subtract the first row from the second row to get a0 in the second row, first column.
$\left[\begin{array}{ccc|ccc}1 &0.5 &3 &0 & -0.5 &0 \\0 &0.5 &0 &1 &0.5 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]$

STEP 6

Subtract6 times the second row from the first row to get a0 in the first row, second column.
$\left[\begin{array}{ccc|ccc}1 &0 &3 & -6 & -3 &0 \\0 &0.5 &0 &1 &0.5 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]$

STEP 7

Multiply the second row by2 to get a1 in the second row, second column.
$\left[\begin{array}{ccc|ccc}1 &0 &3 & -6 & -3 &0 \\0 &1 &0 &2 &1 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]$

STEP 8

Subtract3 times the second row from the third row to get a0 in the third row, second column.
$\left[\begin{array}{ccc|ccc}1 &0 &3 & -6 & -3 &0 \\0 &1 &0 &2 &1 &0 \\0 &0 &1 & -6 & -3 &1 \end{array}\right]$

STEP 9

Subtract3 times the third row from the first row to get a in the first row, third column.
$\left[\begin{array}{ccc|ccc} & & &12 &6 & -3 \\ & & &2 & & \\ & & & -6 & -3 & \end{array}\right]$

STEP 10

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Find the inverse A−1A^{-1}A−1 of the given 3x3 matrix A=[113−2−1−6031]A = \begin{bmatrix} 1 & 1 & 3 \\ -2 & -1 & -6 \\ 0 & 3 & 1 \end{bmatrix}A=⎣⎡​1−20​1−13​3−61​⎦⎤​.

STEP 1

Assumptions1. The given matrix is a3x3 matrix A=[113−−1−6031]A = \begin{bmatrix}1 &1 &3 \\ - & -1 & -6 \\0 &3 &1 \end{bmatrix}A=⎣⎡​1−0​1−13​3−61​⎦⎤​ . We are asked to find the inverse of matrix A, denoted as A−1A^{-1}A−1

STEP 2

STEP 3

STEP 4

Multiply the first row by -1/2 to get a1 in the top left corner.[10.30−0.0113100031001]\left[\begin{array}{ccc|ccc}1 &0. &3 &0 & -0. &0 \\1 &1 &3 &1 &0 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]⎣⎡​110​0.13​331​010​−0.00​001​⎦⎤​

STEP 5

Subtract the first row from the second row to get a0 in the second row, first column.[10.530−0.5000.5010.50031001]\left[\begin{array}{ccc|ccc}1 &0.5 &3 &0 & -0.5 &0 \\0 &0.5 &0 &1 &0.5 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]⎣⎡​100​0.50.53​301​010​−0.50.50​001​⎦⎤​

STEP 6

Subtract6 times the second row from the first row to get a0 in the first row, second column.[103−6−3000.5010.50031001]\left[\begin{array}{ccc|ccc}1 &0 &3 & -6 & -3 &0 \\0 &0.5 &0 &1 &0.5 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]⎣⎡​100​00.53​301​−610​−30.50​001​⎦⎤​

STEP 7

Multiply the second row by2 to get a1 in the second row, second column.[103−6−30010210031001]\left[\begin{array}{ccc|ccc}1 &0 &3 & -6 & -3 &0 \\0 &1 &0 &2 &1 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]⎣⎡​100​013​301​−620​−310​001​⎦⎤​

STEP 8

Subtract3 times the second row from the third row to get a0 in the third row, second column.[103−6−30010210001−6−31]\left[\begin{array}{ccc|ccc}1 &0 &3 & -6 & -3 &0 \\0 &1 &0 &2 &1 &0 \\0 &0 &1 & -6 & -3 &1 \end{array}\right]⎣⎡​100​010​301​−62−6​−31−3​001​⎦⎤​

STEP 9

Subtract3 times the third row from the first row to get a in the first row, third column.[126−32−6−3]\left[\begin{array}{ccc|ccc} & & &12 &6 & -3 \\ & & &2 & & \\ & & & -6 & -3 & \end{array}\right]⎣⎡​​​​122−6​6−3​−3​⎦⎤​

STEP 10

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Find the inverse $A^{-1}$ of the given 3x3 matrix $A = \begin{bmatrix} 1 & 1 & 3 \\ -2 & -1 & -6 \\ 0 & 3 & 1 \end{bmatrix}$ .

Assumptions1. The given matrix is a3x3 matrix $A = \begin{bmatrix}1 &1 &3 \\ - & -1 & -6 \\0 &3 &1 \end{bmatrix}$ . We are asked to find the inverse of matrix A, denoted as $A^{-1}$

Multiply the first row by -1/2 to get a1 in the top left corner.
$\left[\begin{array}{ccc|ccc}1 &0. &3 &0 & -0. &0 \\1 &1 &3 &1 &0 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]$

Subtract the first row from the second row to get a0 in the second row, first column.
$\left[\begin{array}{ccc|ccc}1 &0.5 &3 &0 & -0.5 &0 \\0 &0.5 &0 &1 &0.5 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]$

Subtract6 times the second row from the first row to get a0 in the first row, second column.
$\left[\begin{array}{ccc|ccc}1 &0 &3 & -6 & -3 &0 \\0 &0.5 &0 &1 &0.5 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]$

Multiply the second row by2 to get a1 in the second row, second column.
$\left[\begin{array}{ccc|ccc}1 &0 &3 & -6 & -3 &0 \\0 &1 &0 &2 &1 &0 \\0 &3 &1 &0 &0 &1 \end{array}\right]$

Subtract3 times the second row from the third row to get a0 in the third row, second column.
$\left[\begin{array}{ccc|ccc}1 &0 &3 & -6 & -3 &0 \\0 &1 &0 &2 &1 &0 \\0 &0 &1 & -6 & -3 &1 \end{array}\right]$

Subtract3 times the third row from the first row to get a in the first row, third column.
$\left[\begin{array}{ccc|ccc} & & &12 &6 & -3 \\ & & &2 & & \\ & & & -6 & -3 & \end{array}\right]$