Solve ([0,1,4],[2,1,6],[3,2,8]) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\left[\begin{matrix} 0 & 1 & 4 \\2 & 1 & 6 \\3 & 2 & 8 \end{matrix}\right]$$

Calculate Determinant

$6$

Steps Using Sarrus' Rule

Steps Using Expansion by Minors

Steps Using Sarrus' Rule

Find the determinant of the matrix using the method of diagonals.

$$det(\left(\begin{matrix}0&1&4\\2&1&6\\3&2&8\end{matrix}\right))$$

Extend the original matrix by repeating the first two columns as the fourth and fifth columns.

$$\left(\begin{matrix}0&1&4&0&1\\2&1&6&2&1\\3&2&8&3&2\end{matrix}\right)$$

Starting at the upper left entry, multiply down along the diagonals, and add the resulting products.

$$6\times 3+4\times 2\times 2=34$$

Starting at the lower left entry, multiply up along the diagonals, and add the resulting products.

$$3\times 4+8\times 2=28$$

Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.

$$34-28$$

Subtract $28$ from $34$.

$$6$$

Steps Using Expansion by Minors

Find the determinant of the matrix using the method of expansion by minors (also known as expansion by cofactors).

$$det(\left(\begin{matrix}0&1&4\\2&1&6\\3&2&8\end{matrix}\right))$$

To expand by minors, multiply each element of the first row by its minor, which is the determinant of the $2\times 2$ matrix created by deleting the row and column containing that element, then multiply by the element's position sign.

$$-det(\left(\begin{matrix}2&6\\3&8\end{matrix}\right))+4det(\left(\begin{matrix}2&1\\3&2\end{matrix}\right))$$

For the $2\times 2$ matrix $\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$, the determinant is $ad-bc$.

$$-\left(2\times 8-3\times 6\right)+4\left(2\times 2-3\right)$$

Simplify.

$$-\left(-2\right)+4$$

Add the terms to obtain the final result.

$$6$$

Evaluate

$\left(\begin{matrix}0&1&4\\2&1&6\\3&2&8\end{matrix}\right)$

Invert Matrix

$\left(\begin{matrix}-\frac{2}{3}&0&\frac{1}{3}\\\frac{1}{3}&-2&\frac{4}{3}\\\frac{1}{6}&\frac{1}{2}&-\frac{1}{3}\end{matrix}\right)$

Calculate Trace

$9$

Transpose Matrix

$\left(\begin{matrix}0&2&3\\1&1&2\\4&6&8\end{matrix}\right)$

Matrix Size

$3,3$

Reduce Matrix

$\left(\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\right)$

Adjoint Matrix

$\left[\begin{array}{rrr}-4 & 0 & 2 \\ 2 & -12 & 8 \\ 1 & 3 & -2\end{array}\right]$

To find the adjoint of a matrix, first find the cofactor matrix of the given matrix. Then find the transpose of the cofactor matrix.

Cofactor of 0

$A_{11}=\left|\begin{array}{rr}1 & 6 \\ 2 & 8\end{array}\right|=-4$

Cofactor of 1

$A_{12}=-\left|\begin{array}{rr}2 & 6 \\ 3 & 8\end{array}\right|=2$

Cofactor of 4

$A_{13}=\left|\begin{array}{rr}2 & 1 \\ 3 & 2\end{array}\right|=1$

Cofactor of 2

$A_{21}=-\left|\begin{array}{rr}1 & 4 \\ 2 & 8\end{array}\right|=0$

Cofactor of 1

$A_{22}=\left|\begin{array}{rr}0 & 4 \\ 3 & 8\end{array}\right|=-12$

Cofactor of 6

$A_{23}=-\left|\begin{array}{rr}0 & 1 \\ 3 & 2\end{array}\right|=3$

Cofactor of 3

$A_{31}=\left|\begin{array}{rr}1 & 4 \\ 1 & 6\end{array}\right|=2$

Cofactor of 2

$A_{32}=-\left|\begin{array}{rr}0 & 4 \\ 2 & 6\end{array}\right|=8$

Cofactor of 8

$A_{33}=\left|\begin{array}{rr}0 & 1 \\ 2 & 1\end{array}\right|=-2$

The cofactor matrix of A is [Aij]

$\left[\begin{array}{rrr}-4 & 2 & 1 \\ 0 & -12 & 3 \\ 2 & 8 & -2\end{array}\right]$

The transpose of Aij

$\left[\begin{array}{rrr}-4 & 0 & 2 \\ 2 & -12 & 8 \\ 1 & 3 & -2\end{array}\right]$