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

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

Calculate Determinant

$-48$

Steps Using Sarrus' Rule

Steps Using Expansion by Minors

Steps Using Sarrus' Rule

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

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

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

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

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

$$4\left(-4\right)\times 4+4\times 5+2\times 3\times 2=-32$$

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

$$5\left(-4\right)\times 2+2\times 4+4\times 3\times 4=16$$

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

$$-32-16$$

Subtract $16$ from $-32$.

$$-48$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$4\left(-18\right)-4\times 7+2\times 26$$

Add the terms to obtain the final result.

$$-48$$

Evaluate

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

Invert Matrix

$\left(\begin{matrix}\frac{3}{8}&\frac{1}{4}&-\frac{1}{4}\\\frac{7}{48}&-\frac{1}{8}&-\frac{1}{24}\\-\frac{13}{24}&-\frac{1}{4}&\frac{7}{12}\end{matrix}\right)$

Calculate Trace

$4$

Transpose Matrix

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

Matrix Size

$3,3$

Reduce Matrix

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

Adjoint Matrix

$\left[\begin{array}{rrr}-18 & -12 & 12 \\ -7 & 6 & 2 \\ 26 & 12 & -28\end{array}\right]$

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

$A_{11}=\left|\begin{array}{rr}-4 & 1 \\ 2 & 4\end{array}\right|=-18$

Cofactor of 4

$A_{12}=-\left|\begin{array}{rr}3 & 1 \\ 5 & 4\end{array}\right|=-7$

Cofactor of 2

$A_{13}=\left|\begin{array}{rr}3 & -4 \\ 5 & 2\end{array}\right|=26$

Cofactor of 3

$A_{21}=-\left|\begin{array}{rr}4 & 2 \\ 2 & 4\end{array}\right|=-12$

Cofactor of -4

$A_{22}=\left|\begin{array}{rr}4 & 2 \\ 5 & 4\end{array}\right|=6$

Cofactor of 1

$A_{23}=-\left|\begin{array}{rr}4 & 4 \\ 5 & 2\end{array}\right|=12$

Cofactor of 5

$A_{31}=\left|\begin{array}{rr}4 & 2 \\ -4 & 1\end{array}\right|=12$

Cofactor of 2

$A_{32}=-\left|\begin{array}{rr}4 & 2 \\ 3 & 1\end{array}\right|=2$

Cofactor of 4

$A_{33}=\left|\begin{array}{rr}4 & 4 \\ 3 & -4\end{array}\right|=-28$

The cofactor matrix of A is [Aij]

$\left[\begin{array}{rrr}-18 & -7 & 26 \\ -12 & 6 & 12 \\ 12 & 2 & -28\end{array}\right]$

The transpose of Aij

$\left[\begin{array}{rrr}-18 & -12 & 12 \\ -7 & 6 & 2 \\ 26 & 12 & -28\end{array}\right]$