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

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

Calculate Determinant

$-13$

Steps Using Sarrus' Rule

Steps Using Expansion by Minors

Steps Using Sarrus' Rule

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

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

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

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

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

$$-9+12+5\times 3\times 2=33$$

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

$$-5+2\times 12+9\times 3=46$$

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

$$33-46$$

Subtract $46$ from $33$.

$$-13$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-33-15+5\times 7$$

Add the terms to obtain the final result.

$$-13$$

Evaluate

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

Invert Matrix

$\left(\begin{matrix}\frac{33}{13}&-\frac{1}{13}&-\frac{17}{13}\\\frac{15}{13}&-\frac{4}{13}&-\frac{3}{13}\\-\frac{7}{13}&\frac{1}{13}&\frac{4}{13}\end{matrix}\right)$

Calculate Trace

$9$

Transpose Matrix

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

Matrix Size

$3,3$

Reduce Matrix

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

Adjoint Matrix

$\left[\begin{array}{rrr}-33 & 1 & 17 \\ -15 & 4 & 3 \\ 7 & -1 & -4\end{array}\right]$

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

$A_{11}=\left|\begin{array}{rr}-1 & 12 \\ 2 & 9\end{array}\right|=-33$

Cofactor of 1

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

Cofactor of 5

$A_{13}=\left|\begin{array}{rr}3 & -1 \\ 1 & 2\end{array}\right|=7$

Cofactor of 3

$A_{21}=-\left|\begin{array}{rr}1 & 5 \\ 2 & 9\end{array}\right|=1$

Cofactor of -1

$A_{22}=\left|\begin{array}{rr}1 & 5 \\ 1 & 9\end{array}\right|=4$

Cofactor of 12

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

Cofactor of 1

$A_{31}=\left|\begin{array}{rr}1 & 5 \\ -1 & 12\end{array}\right|=17$

Cofactor of 2

$A_{32}=-\left|\begin{array}{rr}1 & 5 \\ 3 & 12\end{array}\right|=3$

Cofactor of 9

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

The cofactor matrix of A is [Aij]

$\left[\begin{array}{rrr}-33 & -15 & 7 \\ 1 & 4 & -1 \\ 17 & 3 & -4\end{array}\right]$

The transpose of Aij

$\left[\begin{array}{rrr}-33 & 1 & 17 \\ -15 & 4 & 3 \\ 7 & -1 & -4\end{array}\right]$