Solve ([4,1,4],[3,10,-4],[5,6,2]) | 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

$$\left[\begin{matrix} 4 & 1 & 4 \\3 & 10 & -4 \\5 & 6 & 2 \end{matrix}\right]$$

Calculate Determinant

$22$

Steps Using Sarrus' Rule

Steps Using Expansion by Minors

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&1&4\\3&10&-4\\5&6&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&1&4&4&1\\3&10&-4&3&10\\5&6&2&5&6\end{matrix}\right)$$

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

$$4\times 10\times 2-4\times 5+4\times 3\times 6=132$$

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

$$5\times 10\times 4+6\left(-4\right)\times 4+2\times 3=110$$

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

$$132-110$$

Subtract $110$ from $132$.

$$22$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}4&1&4\\3&10&-4\\5&6&2\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}10&-4\\6&2\end{matrix}\right))-det(\left(\begin{matrix}3&-4\\5&2\end{matrix}\right))+4det(\left(\begin{matrix}3&10\\5&6\end{matrix}\right))$$

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

$$4\left(10\times 2-6\left(-4\right)\right)-\left(3\times 2-5\left(-4\right)\right)+4\left(3\times 6-5\times 10\right)$$

Simplify.

$$4\times 44-26+4\left(-32\right)$$

Add the terms to obtain the final result.

$$22$$

Evaluate

$\left(\begin{matrix}4&1&4\\3&10&-4\\5&6&2\end{matrix}\right)$

Invert Matrix

$\left(\begin{matrix}2&1&-2\\-\frac{13}{11}&-\frac{6}{11}&\frac{14}{11}\\-\frac{16}{11}&-\frac{19}{22}&\frac{37}{22}\end{matrix}\right)$

Calculate Trace

$16$

Transpose Matrix

$\left(\begin{matrix}4&3&5\\1&10&6\\4&-4&2\end{matrix}\right)$

Matrix Size

$3,3$

Reduce Matrix

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

Adjoint Matrix

$\left[\begin{array}{rrr}44 & 22 & -44 \\ -26 & -12 & 28 \\ -32 & -19 & 37\end{array}\right]$

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

$A_{11}=\left|\begin{array}{rr}10 & -4 \\ 6 & 2\end{array}\right|=44$

Cofactor of 1

$A_{12}=-\left|\begin{array}{rr}3 & -4 \\ 5 & 2\end{array}\right|=-26$

Cofactor of 4

$A_{13}=\left|\begin{array}{rr}3 & 10 \\ 5 & 6\end{array}\right|=-32$

Cofactor of 3

$A_{21}=-\left|\begin{array}{rr}1 & 4 \\ 6 & 2\end{array}\right|=22$

Cofactor of 10

$A_{22}=\left|\begin{array}{rr}4 & 4 \\ 5 & 2\end{array}\right|=-12$

Cofactor of -4

$A_{23}=-\left|\begin{array}{rr}4 & 1 \\ 5 & 6\end{array}\right|=-19$

Cofactor of 5

$A_{31}=\left|\begin{array}{rr}1 & 4 \\ 10 & -4\end{array}\right|=-44$

Cofactor of 6

$A_{32}=-\left|\begin{array}{rr}4 & 4 \\ 3 & -4\end{array}\right|=28$

Cofactor of 2

$A_{33}=\left|\begin{array}{rr}4 & 1 \\ 3 & 10\end{array}\right|=37$

The cofactor matrix of A is [Aij]

$\left[\begin{array}{rrr}44 & -26 & -32 \\ 22 & -12 & -19 \\ -44 & 28 & 37\end{array}\right]$

The transpose of Aij

$\left[\begin{array}{rrr}44 & 22 & -44 \\ -26 & -12 & 28 \\ -32 & -19 & 37\end{array}\right]$