Solve ([-1,3,4],[1,9,6],[2,4,8]) | 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

$$\left[\begin{matrix} -1 & 3 & 4 \\1 & 9 & 6 \\2 & 4 & 8 \end{matrix}\right]$$

Calculate Determinant

$-92$

Steps Using Sarrus' Rule

Steps Using Expansion by Minors

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-1&3&4\\1&9&6\\2&4&8\end{matrix}\right))$$

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

$$\left(\begin{matrix}-1&3&4&-1&3\\1&9&6&1&9\\2&4&8&2&4\end{matrix}\right)$$

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

$$-9\times 8+3\times 6\times 2+4\times 4=-20$$

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

$$2\times 9\times 4+4\times 6\left(-1\right)+8\times 3=72$$

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

$$-20-72$$

Subtract $72$ from $-20$.

$$-92$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-1&3&4\\1&9&6\\2&4&8\end{matrix}\right))$$

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.

$$-det(\left(\begin{matrix}9&6\\4&8\end{matrix}\right))-3det(\left(\begin{matrix}1&6\\2&8\end{matrix}\right))+4det(\left(\begin{matrix}1&9\\2&4\end{matrix}\right))$$

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

$$-\left(9\times 8-4\times 6\right)-3\left(8-2\times 6\right)+4\left(4-2\times 9\right)$$

Simplify.

$$-48-3\left(-4\right)+4\left(-14\right)$$

Add the terms to obtain the final result.

$$-92$$

Evaluate

$\left(\begin{matrix}-1&3&4\\1&9&6\\2&4&8\end{matrix}\right)$

Invert Matrix

$\left(\begin{matrix}-\frac{12}{23}&\frac{2}{23}&\frac{9}{46}\\-\frac{1}{23}&\frac{4}{23}&-\frac{5}{46}\\\frac{7}{46}&-\frac{5}{46}&\frac{3}{23}\end{matrix}\right)$

Calculate Trace

$16$

Transpose Matrix

$\left(\begin{matrix}-1&1&2\\3&9&4\\4&6&8\end{matrix}\right)$

Matrix Size

$3,3$

Reduce Matrix

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

Adjoint Matrix

$\left[\begin{array}{rrr}48 & -8 & -18 \\ 4 & -16 & 10 \\ -14 & 10 & -12\end{array}\right]$

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

$A_{11}=\left|\begin{array}{rr}9 & 6 \\ 4 & 8\end{array}\right|=48$

Cofactor of 3

$A_{12}=-\left|\begin{array}{rr}1 & 6 \\ 2 & 8\end{array}\right|=4$

Cofactor of 4

$A_{13}=\left|\begin{array}{rr}1 & 9 \\ 2 & 4\end{array}\right|=-14$

Cofactor of 1

$A_{21}=-\left|\begin{array}{rr}3 & 4 \\ 4 & 8\end{array}\right|=-8$

Cofactor of 9

$A_{22}=\left|\begin{array}{rr}-1 & 4 \\ 2 & 8\end{array}\right|=-16$

Cofactor of 6

$A_{23}=-\left|\begin{array}{rr}-1 & 3 \\ 2 & 4\end{array}\right|=10$

Cofactor of 2

$A_{31}=\left|\begin{array}{rr}3 & 4 \\ 9 & 6\end{array}\right|=-18$

Cofactor of 4

$A_{32}=-\left|\begin{array}{rr}-1 & 4 \\ 1 & 6\end{array}\right|=10$

Cofactor of 8

$A_{33}=\left|\begin{array}{rr}-1 & 3 \\ 1 & 9\end{array}\right|=-12$

The cofactor matrix of A is [Aij]

$\left[\begin{array}{rrr}48 & 4 & -14 \\ -8 & -16 & 10 \\ -18 & 10 & -12\end{array}\right]$

The transpose of Aij

$\left[\begin{array}{rrr}48 & -8 & -18 \\ 4 & -16 & 10 \\ -14 & 10 & -12\end{array}\right]$