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

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

Calculate Determinant

$136$

Steps Using Sarrus' Rule

Steps Using Expansion by Minors

Steps Using Sarrus' Rule

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

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

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

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

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

$$7\times 2\times 9+3\times 2\times 5+2\left(-2\right)\times 3=144$$

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

$$5\times 2\times 2+3\times 2\times 7+9\left(-2\right)\times 3=8$$

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

$$144-8$$

Subtract $8$ from $144$.

$$136$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$7\times 12-3\left(-28\right)+2\left(-16\right)$$

Add the terms to obtain the final result.

$$136$$

Evaluate

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

Invert Matrix

$\left(\begin{matrix}\frac{3}{34}&-\frac{21}{136}&\frac{1}{68}\\\frac{7}{34}&\frac{53}{136}&-\frac{9}{68}\\-\frac{2}{17}&-\frac{3}{68}&\frac{5}{34}\end{matrix}\right)$

Calculate Trace

$18$

Transpose Matrix

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

Matrix Size

$3,3$

Reduce Matrix

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

Adjoint Matrix

$\left[\begin{array}{rrr}12 & -21 & 2 \\ 28 & 53 & -18 \\ -16 & -6 & 20\end{array}\right]$

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 7

$A_{11}=\left|\begin{array}{rr}2 & 2 \\ 3 & 9\end{array}\right|=12$

Cofactor of 3

$A_{12}=-\left|\begin{array}{rr}-2 & 2 \\ 5 & 9\end{array}\right|=28$

Cofactor of 2

$A_{13}=\left|\begin{array}{rr}-2 & 2 \\ 5 & 3\end{array}\right|=-16$

Cofactor of -2

$A_{21}=-\left|\begin{array}{rr}3 & 2 \\ 3 & 9\end{array}\right|=-21$

Cofactor of 2

$A_{22}=\left|\begin{array}{rr}7 & 2 \\ 5 & 9\end{array}\right|=53$

Cofactor of 2

$A_{23}=-\left|\begin{array}{rr}7 & 3 \\ 5 & 3\end{array}\right|=-6$

Cofactor of 5

$A_{31}=\left|\begin{array}{rr}3 & 2 \\ 2 & 2\end{array}\right|=2$

Cofactor of 3

$A_{32}=-\left|\begin{array}{rr}7 & 2 \\ -2 & 2\end{array}\right|=-18$

Cofactor of 9

$A_{33}=\left|\begin{array}{rr}7 & 3 \\ -2 & 2\end{array}\right|=20$

The cofactor matrix of A is [Aij]

$\left[\begin{array}{rrr}12 & 28 & -16 \\ -21 & 53 & -6 \\ 2 & -18 & 20\end{array}\right]$

The transpose of Aij

$\left[\begin{array}{rrr}12 & -21 & 2 \\ 28 & 53 & -18 \\ -16 & -6 & 20\end{array}\right]$