Solve ([4,1,2],[-4,10,1],[2,6,4]) | 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

$$\left[\begin{matrix} 4 & 1 & 2 \\-4 & 10 & 1 \\2 & 6 & 4 \end{matrix}\right]$$

Calculate Determinant

$66$

Steps Using Sarrus' Rule

Steps Using Expansion by Minors

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&1&2\\-4&10&1\\2&6&4\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&1&2&4&1\\-4&10&1&-4&10\\2&6&4&2&6\end{matrix}\right)$$

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

$$4\times 10\times 4+2+2\left(-4\right)\times 6=114$$

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

$$2\times 10\times 2+6\times 4+4\left(-4\right)=48$$

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

$$114-48$$

Subtract $48$ from $114$.

$$66$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}4&1&2\\-4&10&1\\2&6&4\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}10&1\\6&4\end{matrix}\right))-det(\left(\begin{matrix}-4&1\\2&4\end{matrix}\right))+2det(\left(\begin{matrix}-4&10\\2&6\end{matrix}\right))$$

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

$$4\left(10\times 4-6\right)-\left(-4\times 4-2\right)+2\left(-4\times 6-2\times 10\right)$$

Simplify.

$$4\times 34-\left(-18\right)+2\left(-44\right)$$

Add the terms to obtain the final result.

$$66$$

Evaluate

$\left(\begin{matrix}4&1&2\\-4&10&1\\2&6&4\end{matrix}\right)$

Invert Matrix

$\left(\begin{matrix}\frac{17}{33}&\frac{4}{33}&-\frac{19}{66}\\\frac{3}{11}&\frac{2}{11}&-\frac{2}{11}\\-\frac{2}{3}&-\frac{1}{3}&\frac{2}{3}\end{matrix}\right)$

Calculate Trace

$18$

Transpose Matrix

$\left(\begin{matrix}4&-4&2\\1&10&6\\2&1&4\end{matrix}\right)$

Matrix Size

$3,3$

Reduce Matrix

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

Adjoint Matrix

$\left[\begin{array}{rrr}34 & 8 & -19 \\ 18 & 12 & -12 \\ -44 & -22 & 44\end{array}\right]$

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 4

$A_{11}=\left|\begin{array}{rr}10 & 1 \\ 6 & 4\end{array}\right|=34$

Cofactor of 1

$A_{12}=-\left|\begin{array}{rr}-4 & 1 \\ 2 & 4\end{array}\right|=18$

Cofactor of 2

$A_{13}=\left|\begin{array}{rr}-4 & 10 \\ 2 & 6\end{array}\right|=-44$

Cofactor of -4

$A_{21}=-\left|\begin{array}{rr}1 & 2 \\ 6 & 4\end{array}\right|=8$

Cofactor of 10

$A_{22}=\left|\begin{array}{rr}4 & 2 \\ 2 & 4\end{array}\right|=12$

Cofactor of 1

$A_{23}=-\left|\begin{array}{rr}4 & 1 \\ 2 & 6\end{array}\right|=-22$

Cofactor of 2

$A_{31}=\left|\begin{array}{rr}1 & 2 \\ 10 & 1\end{array}\right|=-19$

Cofactor of 6

$A_{32}=-\left|\begin{array}{rr}4 & 2 \\ -4 & 1\end{array}\right|=-12$

Cofactor of 4

$A_{33}=\left|\begin{array}{rr}4 & 1 \\ -4 & 10\end{array}\right|=44$

The cofactor matrix of A is [Aij]

$\left[\begin{array}{rrr}34 & 18 & -44 \\ 8 & 12 & -22 \\ -19 & -12 & 44\end{array}\right]$

The transpose of Aij

$\left[\begin{array}{rrr}34 & 8 & -19 \\ 18 & 12 & -12 \\ -44 & -22 & 44\end{array}\right]$