Solve ([0,-1,3],[2,1,9],[3,2,4]) | 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

$$\left[\begin{matrix} 0 & -1 & 3 \\2 & 1 & 9 \\3 & 2 & 4 \end{matrix}\right]$$

Calculate Determinant

$-16$

Steps Using Sarrus' Rule

Steps Using Expansion by Minors

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}0&-1&3\\2&1&9\\3&2&4\end{matrix}\right))$$

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

$$\left(\begin{matrix}0&-1&3&0&-1\\2&1&9&2&1\\3&2&4&3&2\end{matrix}\right)$$

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

$$-9\times 3+3\times 2\times 2=-15$$

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

$$3\times 3+4\times 2\left(-1\right)=1$$

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

$$-15-1$$

Subtract $1$ from $-15$.

$$-16$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}0&-1&3\\2&1&9\\3&2&4\end{matrix}\right))$$

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.

$$-\left(-det(\left(\begin{matrix}2&9\\3&4\end{matrix}\right))\right)+3det(\left(\begin{matrix}2&1\\3&2\end{matrix}\right))$$

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

$$-\left(-\left(2\times 4-3\times 9\right)\right)+3\left(2\times 2-3\right)$$

Simplify.

$$-\left(-\left(-19\right)\right)+3$$

Add the terms to obtain the final result.

$$-16$$

Evaluate

$\left(\begin{matrix}0&-1&3\\2&1&9\\3&2&4\end{matrix}\right)$

Invert Matrix

$\left(\begin{matrix}\frac{7}{8}&-\frac{5}{8}&\frac{3}{4}\\-\frac{19}{16}&\frac{9}{16}&-\frac{3}{8}\\-\frac{1}{16}&\frac{3}{16}&-\frac{1}{8}\end{matrix}\right)$

Calculate Trace

$5$

Transpose Matrix

$\left(\begin{matrix}0&2&3\\-1&1&2\\3&9&4\end{matrix}\right)$

Matrix Size

$3,3$

Reduce Matrix

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

Adjoint Matrix

$\left[\begin{array}{rrr}-14 & 10 & -12 \\ 19 & -9 & 6 \\ 1 & -3 & 2\end{array}\right]$

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 0

$A_{11}=\left|\begin{array}{rr}1 & 9 \\ 2 & 4\end{array}\right|=-14$

Cofactor of -1

$A_{12}=-\left|\begin{array}{rr}2 & 9 \\ 3 & 4\end{array}\right|=19$

Cofactor of 3

$A_{13}=\left|\begin{array}{rr}2 & 1 \\ 3 & 2\end{array}\right|=1$

Cofactor of 2

$A_{21}=-\left|\begin{array}{rr}-1 & 3 \\ 2 & 4\end{array}\right|=10$

Cofactor of 1

$A_{22}=\left|\begin{array}{rr}0 & 3 \\ 3 & 4\end{array}\right|=-9$

Cofactor of 9

$A_{23}=-\left|\begin{array}{rr}0 & -1 \\ 3 & 2\end{array}\right|=-3$

Cofactor of 3

$A_{31}=\left|\begin{array}{rr}-1 & 3 \\ 1 & 9\end{array}\right|=-12$

Cofactor of 2

$A_{32}=-\left|\begin{array}{rr}0 & 3 \\ 2 & 9\end{array}\right|=6$

Cofactor of 4

$A_{33}=\left|\begin{array}{rr}0 & -1 \\ 2 & 1\end{array}\right|=2$

The cofactor matrix of A is [Aij]

$\left[\begin{array}{rrr}-14 & 19 & 1 \\ 10 & -9 & -3 \\ -12 & 6 & 2\end{array}\right]$

The transpose of Aij

$\left[\begin{array}{rrr}-14 & 10 & -12 \\ 19 & -9 & 6 \\ 1 & -3 & 2\end{array}\right]$