Solve ([1,5,-1],[3,12,2],[1,9,1]) | 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

$$\left[\begin{matrix} 1 & 5 & -1 \\3 & 12 & 2 \\1 & 9 & 1 \end{matrix}\right]$$

Calculate Determinant

$-26$

Steps Using Sarrus' Rule

Steps Using Expansion by Minors

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&5&-1\\3&12&2\\1&9&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&5&-1&1&5\\3&12&2&3&12\\1&9&1&1&9\end{matrix}\right)$$

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

$$12+5\times 2-3\times 9=-5$$

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

$$12\left(-1\right)+9\times 2+3\times 5=21$$

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

$$-5-21$$

Subtract $21$ from $-5$.

$$-26$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&5&-1\\3&12&2\\1&9&1\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}12&2\\9&1\end{matrix}\right))-5det(\left(\begin{matrix}3&2\\1&1\end{matrix}\right))-det(\left(\begin{matrix}3&12\\1&9\end{matrix}\right))$$

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

$$12-9\times 2-5\left(3-2\right)-\left(3\times 9-12\right)$$

Simplify.

$$-6-5-15$$

Add the terms to obtain the final result.

$$-26$$

Evaluate

$\left(\begin{matrix}1&5&-1\\3&12&2\\1&9&1\end{matrix}\right)$

Invert Matrix

$\left(\begin{matrix}\frac{3}{13}&\frac{7}{13}&-\frac{11}{13}\\\frac{1}{26}&-\frac{1}{13}&\frac{5}{26}\\-\frac{15}{26}&\frac{2}{13}&\frac{3}{26}\end{matrix}\right)$

Calculate Trace

$14$

Transpose Matrix

$\left(\begin{matrix}1&3&1\\5&12&9\\-1&2&1\end{matrix}\right)$

Matrix Size

$3,3$

Reduce Matrix

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

Adjoint Matrix

$\left[\begin{array}{rrr}-6 & -14 & 22 \\ -1 & 2 & -5 \\ 15 & -4 & -3\end{array}\right]$

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 1

$A_{11}=\left|\begin{array}{rr}12 & 2 \\ 9 & 1\end{array}\right|=-6$

Cofactor of 5

$A_{12}=-\left|\begin{array}{rr}3 & 2 \\ 1 & 1\end{array}\right|=-1$

Cofactor of -1

$A_{13}=\left|\begin{array}{rr}3 & 12 \\ 1 & 9\end{array}\right|=15$

Cofactor of 3

$A_{21}=-\left|\begin{array}{rr}5 & -1 \\ 9 & 1\end{array}\right|=-14$

Cofactor of 12

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

Cofactor of 2

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

Cofactor of 1

$A_{31}=\left|\begin{array}{rr}5 & -1 \\ 12 & 2\end{array}\right|=22$

Cofactor of 9

$A_{32}=-\left|\begin{array}{rr}1 & -1 \\ 3 & 2\end{array}\right|=-5$

Cofactor of 1

$A_{33}=\left|\begin{array}{rr}1 & 5 \\ 3 & 12\end{array}\right|=-3$

The cofactor matrix of A is [Aij]

$\left[\begin{array}{rrr}-6 & -1 & 15 \\ -14 & 2 & -4 \\ 22 & -5 & -3\end{array}\right]$

The transpose of Aij

$\left[\begin{array}{rrr}-6 & -14 & 22 \\ -1 & 2 & -5 \\ 15 & -4 & -3\end{array}\right]$