Solve ([-4,9,-4],[9,-9,2],[-4,5,-3]) | 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

$$\left[\begin{matrix} -4 & 9 & -4 \\9 & -9 & 2 \\-4 & 5 & -3 \end{matrix}\right]$$

Calculate Determinant

$67$

Steps Using Sarrus' Rule

Steps Using Expansion by Minors

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-4&9&-4\\9&-9&2\\-4&5&-3\end{matrix}\right))$$

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

$$\left(\begin{matrix}-4&9&-4&-4&9\\9&-9&2&9&-9\\-4&5&-3&-4&5\end{matrix}\right)$$

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

$$-4\left(-9\right)\left(-3\right)+9\times 2\left(-4\right)-4\times 9\times 5=-360$$

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

$$-4\left(-9\right)\left(-4\right)+5\times 2\left(-4\right)-3\times 9\times 9=-427$$

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

$$-360-\left(-427\right)$$

Subtract $-427$ from $-360$.

$$67$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-4&9&-4\\9&-9&2\\-4&5&-3\end{matrix}\right))$$

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.

$$-4det(\left(\begin{matrix}-9&2\\5&-3\end{matrix}\right))-9det(\left(\begin{matrix}9&2\\-4&-3\end{matrix}\right))-4det(\left(\begin{matrix}9&-9\\-4&5\end{matrix}\right))$$

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

$$-4\left(-9\left(-3\right)-5\times 2\right)-9\left(9\left(-3\right)-\left(-4\times 2\right)\right)-4\left(9\times 5-\left(-4\left(-9\right)\right)\right)$$

Simplify.

$$-4\times 17-9\left(-19\right)-4\times 9$$

Add the terms to obtain the final result.

$$67$$

Evaluate

$\left(\begin{matrix}-4&9&-4\\9&-9&2\\-4&5&-3\end{matrix}\right)$

Invert Matrix

$\left(\begin{matrix}\frac{17}{67}&\frac{7}{67}&-\frac{18}{67}\\\frac{19}{67}&-\frac{4}{67}&-\frac{28}{67}\\\frac{9}{67}&-\frac{16}{67}&-\frac{45}{67}\end{matrix}\right)$

Calculate Trace

$-16$

Transpose Matrix

$\left(\begin{matrix}-4&9&-4\\9&-9&5\\-4&2&-3\end{matrix}\right)$

Matrix Size

$3,3$

Reduce Matrix

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

Adjoint Matrix

$\left[\begin{array}{rrr}17 & 7 & -18 \\ 19 & -4 & -28 \\ 9 & -16 & -45\end{array}\right]$

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 -4

$A_{11}=\left|\begin{array}{rr}-9 & 2 \\ 5 & -3\end{array}\right|=17$

Cofactor of 9

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

Cofactor of -4

$A_{13}=\left|\begin{array}{rr}9 & -9 \\ -4 & 5\end{array}\right|=9$

Cofactor of 9

$A_{21}=-\left|\begin{array}{rr}9 & -4 \\ 5 & -3\end{array}\right|=7$

Cofactor of -9

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

Cofactor of 2

$A_{23}=-\left|\begin{array}{rr}-4 & 9 \\ -4 & 5\end{array}\right|=-16$

Cofactor of -4

$A_{31}=\left|\begin{array}{rr}9 & -4 \\ -9 & 2\end{array}\right|=-18$

Cofactor of 5

$A_{32}=-\left|\begin{array}{rr}-4 & -4 \\ 9 & 2\end{array}\right|=-28$

Cofactor of -3

$A_{33}=\left|\begin{array}{rr}-4 & 9 \\ 9 & -9\end{array}\right|=-45$

The cofactor matrix of A is [Aij]

$\left[\begin{array}{rrr}17 & 19 & 9 \\ 7 & -4 & -16 \\ -18 & -28 & -45\end{array}\right]$

The transpose of Aij

$\left[\begin{array}{rrr}17 & 7 & -18 \\ 19 & -4 & -28 \\ 9 & -16 & -45\end{array}\right]$