Solve ([5,1,-1],[12,-1,2],[9,1,1]) | 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

$$\left[\begin{matrix} 5 & 1 & -1 \\12 & -1 & 2 \\9 & 1 & 1 \end{matrix}\right]$$

Calculate Determinant

$-30$

Steps Using Sarrus' Rule

Steps Using Expansion by Minors

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&1&-1\\12&-1&2\\9&1&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&1&-1&5&1\\12&-1&2&12&-1\\9&1&1&9&1\end{matrix}\right)$$

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

$$5\left(-1\right)+2\times 9-12=1$$

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

$$9\left(-1\right)\left(-1\right)+2\times 5+12=31$$

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

$$1-31$$

Subtract $31$ from $1$.

$$-30$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&1&-1\\12&-1&2\\9&1&1\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}-1&2\\1&1\end{matrix}\right))-det(\left(\begin{matrix}12&2\\9&1\end{matrix}\right))-det(\left(\begin{matrix}12&-1\\9&1\end{matrix}\right))$$

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

$$5\left(-1-2\right)-\left(12-9\times 2\right)-\left(12-9\left(-1\right)\right)$$

Simplify.

$$5\left(-3\right)-\left(-6\right)-21$$

Add the terms to obtain the final result.

$$-30$$

Evaluate

$\left(\begin{matrix}5&1&-1\\12&-1&2\\9&1&1\end{matrix}\right)$

Invert Matrix

$\left(\begin{matrix}\frac{1}{10}&\frac{1}{15}&-\frac{1}{30}\\-\frac{1}{5}&-\frac{7}{15}&\frac{11}{15}\\-\frac{7}{10}&-\frac{2}{15}&\frac{17}{30}\end{matrix}\right)$

Calculate Trace

$5$

Transpose Matrix

$\left(\begin{matrix}5&12&9\\1&-1&1\\-1&2&1\end{matrix}\right)$

Matrix Size

$3,3$

Reduce Matrix

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

Adjoint Matrix

$\left[\begin{array}{rrr}-3 & -2 & 1 \\ 6 & 14 & -22 \\ 21 & 4 & -17\end{array}\right]$

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 5

$A_{11}=\left|\begin{array}{rr}-1 & 2 \\ 1 & 1\end{array}\right|=-3$

Cofactor of 1

$A_{12}=-\left|\begin{array}{rr}12 & 2 \\ 9 & 1\end{array}\right|=6$

Cofactor of -1

$A_{13}=\left|\begin{array}{rr}12 & -1 \\ 9 & 1\end{array}\right|=21$

Cofactor of 12

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

Cofactor of -1

$A_{22}=\left|\begin{array}{rr}5 & -1 \\ 9 & 1\end{array}\right|=14$

Cofactor of 2

$A_{23}=-\left|\begin{array}{rr}5 & 1 \\ 9 & 1\end{array}\right|=4$

Cofactor of 9

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

Cofactor of 1

$A_{32}=-\left|\begin{array}{rr}5 & -1 \\ 12 & 2\end{array}\right|=-22$

Cofactor of 1

$A_{33}=\left|\begin{array}{rr}5 & 1 \\ 12 & -1\end{array}\right|=-17$

The cofactor matrix of A is [Aij]

$\left[\begin{array}{rrr}-3 & 6 & 21 \\ -2 & 14 & 4 \\ 1 & -22 & -17\end{array}\right]$

The transpose of Aij

$\left[\begin{array}{rrr}-3 & -2 & 1 \\ 6 & 14 & -22 \\ 21 & 4 & -17\end{array}\right]$