Solve ([-2,9,-3],[6,-6,7],[5,5,-1]) | 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

$$\left[\begin{matrix} -2 & 9 & -3 \\6 & -6 & 7 \\5 & 5 & -1 \end{matrix}\right]$$

Calculate Determinant

$247$

Steps Using Sarrus' Rule

Steps Using Expansion by Minors

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-2&9&-3\\6&-6&7\\5&5&-1\end{matrix}\right))$$

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

$$\left(\begin{matrix}-2&9&-3&-2&9\\6&-6&7&6&-6\\5&5&-1&5&5\end{matrix}\right)$$

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

$$-2\left(-6\right)\left(-1\right)+9\times 7\times 5-3\times 6\times 5=213$$

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

$$5\left(-6\right)\left(-3\right)+5\times 7\left(-2\right)-6\times 9=-34$$

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

$$213-\left(-34\right)$$

Subtract $-34$ from $213$.

$$247$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-2&9&-3\\6&-6&7\\5&5&-1\end{matrix}\right))$$

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.

$$-2det(\left(\begin{matrix}-6&7\\5&-1\end{matrix}\right))-9det(\left(\begin{matrix}6&7\\5&-1\end{matrix}\right))-3det(\left(\begin{matrix}6&-6\\5&5\end{matrix}\right))$$

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

$$-2\left(-6\left(-1\right)-5\times 7\right)-9\left(6\left(-1\right)-5\times 7\right)-3\left(6\times 5-5\left(-6\right)\right)$$

Simplify.

$$-2\left(-29\right)-9\left(-41\right)-3\times 60$$

Add the terms to obtain the final result.

$$247$$

Evaluate

$\left(\begin{matrix}-2&9&-3\\6&-6&7\\5&5&-1\end{matrix}\right)$

Invert Matrix

$\left(\begin{matrix}-\frac{29}{247}&-\frac{6}{247}&\frac{45}{247}\\\frac{41}{247}&\frac{17}{247}&-\frac{4}{247}\\\frac{60}{247}&\frac{55}{247}&-\frac{42}{247}\end{matrix}\right)$

Calculate Trace

$-9$

Transpose Matrix

$\left(\begin{matrix}-2&6&5\\9&-6&5\\-3&7&-1\end{matrix}\right)$

Matrix Size

$3,3$

Reduce Matrix

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

Adjoint Matrix

$\left[\begin{array}{rrr}-29 & -6 & 45 \\ 41 & 17 & -4 \\ 60 & 55 & -42\end{array}\right]$

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 -2

$A_{11}=\left|\begin{array}{rr}-6 & 7 \\ 5 & -1\end{array}\right|=-29$

Cofactor of 9

$A_{12}=-\left|\begin{array}{rr}6 & 7 \\ 5 & -1\end{array}\right|=41$

Cofactor of -3

$A_{13}=\left|\begin{array}{rr}6 & -6 \\ 5 & 5\end{array}\right|=60$

Cofactor of 6

$A_{21}=-\left|\begin{array}{rr}9 & -3 \\ 5 & -1\end{array}\right|=-6$

Cofactor of -6

$A_{22}=\left|\begin{array}{rr}-2 & -3 \\ 5 & -1\end{array}\right|=17$

Cofactor of 7

$A_{23}=-\left|\begin{array}{rr}-2 & 9 \\ 5 & 5\end{array}\right|=55$

Cofactor of 5

$A_{31}=\left|\begin{array}{rr}9 & -3 \\ -6 & 7\end{array}\right|=45$

Cofactor of 5

$A_{32}=-\left|\begin{array}{rr}-2 & -3 \\ 6 & 7\end{array}\right|=-4$

Cofactor of -1

$A_{33}=\left|\begin{array}{rr}-2 & 9 \\ 6 & -6\end{array}\right|=-42$

The cofactor matrix of A is [Aij]

$\left[\begin{array}{rrr}-29 & 41 & 60 \\ -6 & 17 & 55 \\ 45 & -4 & -42\end{array}\right]$

The transpose of Aij

$\left[\begin{array}{rrr}-29 & -6 & 45 \\ 41 & 17 & -4 \\ 60 & 55 & -42\end{array}\right]$