All categories

2 years ago

What is the determinant of the coefficient matrix of the system –11 –2 0 55

Mathematics

2 answers:

Alina [70]2 years ago

7 0

The first two rows of coefficients are identical, so by inspection, the determinant is 0.

Mademuasel [1]2 years ago

7 0

Answer:

The determinant of coefficient matrix of the given system is 0.

Step-by-step explanation:

Given system of equation

$-x-y-z=-3$

$-x-y-z=8$

$3x+2y+z=0$

The coefficient matrix of the system

$\left[\begin{array}{ccc}-1&-1&-1\\-1&-1&-1\\3&2&1\end{array}\right]$

Let A= $\left[\begin{array}{ccc}-1&-1&-1\\-1&-1&-1\\3&2&1\end{array}\right]$

X= $\left[\begin{array}{}x&y&z\end{array}\right]$

B= $\left[\begin{array}-3 &8&0\end{array}\right]$

Therefore , we can AX=B

$\left[\begin{array}{ccc}-1&-1&-1\\1&-1&-1\\3&2&1\end{array}\right]$ $\left[\begin{array}{}x&y&z \end{array}\right]$ = $\left[\begin{array}{}-3&8&0\end{array}\right]$

The determinant of coefficient matrix of the given system is given by

$\begin{vmatrix}-1&-1&-1\\-1&-1&-1\\3&2&1\end{vmatrix}$

By using determinant property : when two rowsor two columns are identical then the value of determinant is equal to zero .

$\therefore \begin{vmatrix}A\end{vmatrix}=0$

You might be interested in

Round each time to the nearest whole second. Then estimate how many seconds bailey won the race by.

Orlov [11]

The whole second is the first number to the left of the decimal (_.), so : 35.78=36 because 7>5. 39.51=40 because 5=5. 40-36=4. Bailey won the race by about 4 seconds. Hoped I helped!

8 0

2 years ago

Read 2 more answers

You have been observing a population of swallowtail butterflies for the last 10 years. the size of the population has varied as

EastWind [94]

The answer to this questions is = 17

4 0

2 years ago

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$