Solve ([5,1,7],[1,2,0],[5,2,8]) | 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

$$\left[\begin{matrix} 5 & 1 & 7 \\1 & 2 & 0 \\5 & 2 & 8 \end{matrix}\right]$$

Calculate Determinant

$16$

Steps Using Sarrus' Rule

Steps Using Expansion by Minors

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&1&7\\1&2&0\\5&2&8\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&1&7&5&1\\1&2&0&1&2\\5&2&8&5&2\end{matrix}\right)$$

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

$$5\times 2\times 8+7\times 2=94$$

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

$$5\times 2\times 7+8=78$$

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

$$94-78$$

Subtract $78$ from $94$.

$$16$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&1&7\\1&2&0\\5&2&8\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}2&0\\2&8\end{matrix}\right))-det(\left(\begin{matrix}1&0\\5&8\end{matrix}\right))+7det(\left(\begin{matrix}1&2\\5&2\end{matrix}\right))$$

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

$$5\times 2\times 8-8+7\left(2-5\times 2\right)$$

Simplify.

$$5\times 16-8+7\left(-8\right)$$

Add the terms to obtain the final result.

$$16$$

Evaluate

$\left(\begin{matrix}5&1&7\\1&2&0\\5&2&8\end{matrix}\right)$

Invert Matrix

$\left(\begin{matrix}1&\frac{3}{8}&-\frac{7}{8}\\-\frac{1}{2}&\frac{5}{16}&\frac{7}{16}\\-\frac{1}{2}&-\frac{5}{16}&\frac{9}{16}\end{matrix}\right)$

Calculate Trace

$15$

Transpose Matrix

$\left(\begin{matrix}5&1&5\\1&2&2\\7&0&8\end{matrix}\right)$

Matrix Size

$3,3$

Reduce Matrix

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

Adjoint Matrix

$\left[\begin{array}{rrr}16 & 6 & -14 \\ -8 & 5 & 7 \\ -8 & -5 & 9\end{array}\right]$

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 5

$A_{11}=\left|\begin{array}{rr}2 & 0 \\ 2 & 8\end{array}\right|=16$

Cofactor of 1

$A_{12}=-\left|\begin{array}{rr}1 & 0 \\ 5 & 8\end{array}\right|=-8$

Cofactor of 7

$A_{13}=\left|\begin{array}{rr}1 & 2 \\ 5 & 2\end{array}\right|=-8$

Cofactor of 1

$A_{21}=-\left|\begin{array}{rr}1 & 7 \\ 2 & 8\end{array}\right|=6$

Cofactor of 2

$A_{22}=\left|\begin{array}{rr}5 & 7 \\ 5 & 8\end{array}\right|=5$

Cofactor of 0

$A_{23}=-\left|\begin{array}{rr}5 & 1 \\ 5 & 2\end{array}\right|=-5$

Cofactor of 5

$A_{31}=\left|\begin{array}{rr}1 & 7 \\ 2 & 0\end{array}\right|=-14$

Cofactor of 2

$A_{32}=-\left|\begin{array}{rr}5 & 7 \\ 1 & 0\end{array}\right|=7$

Cofactor of 8

$A_{33}=\left|\begin{array}{rr}5 & 1 \\ 1 & 2\end{array}\right|=9$

The cofactor matrix of A is [Aij]

$\left[\begin{array}{rrr}16 & -8 & -8 \\ 6 & 5 & -5 \\ -14 & 7 & 9\end{array}\right]$

The transpose of Aij

$\left[\begin{array}{rrr}16 & 6 & -14 \\ -8 & 5 & 7 \\ -8 & -5 & 9\end{array}\right]$