Solve the following systems of linear equations by computing...

Solve the following systems of linear equations by computing the inverses of their coefficient matrices. (a) $$ \begin{aligned} & x+2 y=1, \\ & x-2 y=-2 . \end{aligned} $$ (b) $$ \begin{aligned} 3 u-2 v & =2, \\ u+5 v & =12 . \end{aligned} $$ (c) $$ x-y+3 z=3, $$ $$ y+5 z=3, $$ (d) $$ x-2 y+z=2 . $$ $$ \begin{gathered} =-2, \quad(d) x-y+3 z=-1, \\ =2 . \quad-2 x+3 y=5 . \\ x-2 y+z+2 u=-2, \\ x-y+z-u=3, \\ 2 x-y+z+u=3, \end{gathered} $$ $$ x+y=4, $$ (e) $$ \begin{aligned} 2 x+7 y-2 z & =5, \\ -x-5 y+2 z & =-7 . \end{aligned} $$ (f) $$ \begin{aligned} 2 x+3 y-w & =11, \\ -y-z+w & =-7, \\ z-w & =6 . \end{aligned} $$ (g) $$ -x+3 y-2 z-u=2 . $$

Solve the following systems of linear equations by computing the inverses of their coefficient matrices.
(a)
$$
\begin{aligned}
& x+2 y=1, \\
& x-2 y=-2 .
\end{aligned}
$$
(b)
$$
\begin{aligned}
3 u-2 v & =2, \\
u+5 v & =12 .
\end{aligned}
$$
(c)
$$
x-y+3 z=3,
$$
$$
y+5 z=3,
$$
(d)
$$
x-2 y+z=2 .
$$
$$
\begin{gathered}
=-2, \quad(d) x-y+3 z=-1, \\
=2 . \quad-2 x+3 y=5 . \\
x-2 y+z+2 u=-2, \\
x-y+z-u=3, \\
2 x-y+z+u=3,
\end{gathered}
$$
$$
x+y=4,
$$
(e)
$$
\begin{aligned}
2 x+7 y-2 z & =5, \\
-x-5 y+2 z & =-7 .
\end{aligned}
$$
(f)
$$
\begin{aligned}
2 x+3 y-w & =11, \\
-y-z+w & =-7, \\
z-w & =6 .
\end{aligned}
$$
(g)
$$
-x+3 y-2 z-u=2 .
$$

Key Concepts

Systems of Linear Equations

A system of linear equations consists of multiple linear equations involving the same set of variables. The goal is to determine the values of the variables that satisfy all equations simultaneously. These systems can be represented in various forms, including standard, augmented, and matrix formats.

Coefficient Matrix

The coefficient matrix is a compact representation of the constants multiplying the variables in a system of linear equations. It captures the structure of the system and is formed by arranging the coefficients of each variable into rows and columns, which is essential for applying matrix operations in solving the system.

Matrix Equation Representation

Systems of linear equations can be expressed in the form Ax = b, where A is the coefficient matrix, x is the column vector of the variables, and b is the column vector of constants from the right-hand side of each equation. This formulation enables the use of linear algebra techniques to solve the system.

Matrix Inverse

The matrix inverse of a square matrix A, denoted A?¹, is a matrix that, when multiplied by A, yields the identity matrix. In the context of solving systems of linear equations, if A is invertible, then the system Ax = b can be solved uniquely by computing x = A?¹b.

Determinant and Invertibility

A square matrix is invertible if and only if its determinant is nonzero. The determinant is a scalar value that provides important information about the matrix, including whether the matrix has an inverse and the nature of the solution set of the corresponding system of linear equations. A zero determinant indicates that the matrix is singular, meaning the system may either have no solution or infinitely many solutions.

Transcript

00:01 For this problem, we're asked to find a solution to this system of equations using the inverse of coefficient matrix.

00:10 So in order to find the inverse of the coefficient matrix, we need to rewrite this system of equations in the matrix form.

00:19 So therefore, in our matrix a is going to contain all the coefficients for x, y, z, and w.

00:26 So from the first equation, my coefficients are 1, 1, and you notice that there's no z, so make sure you put 0 for z and then 2.

00:37 And the second equation is 2 -1 -1 -1 -1, 3 -3 -2 -2, 2 -9 -2.

00:46 And last equation is 1 -2 -1.

00:49 Again, there's no term for the w, so make sure you put 0.

00:53 And then x is going to, x matrix is going to contain the variables.

00:59 So we have four variables here, x, y, z, and w.

01:05 And then matrix b is going to be the solution, the answer after the equal sign numbers.

01:14 So we have 3, 3, 3, 5, 3.

01:18 So we already know that if we multiply matrix a times matrix x, that will give me the matrix b.

01:28 But our goal is to find out what the matrix x is.

01:32 So we know that to solve for x, x equals to the inverse of matrix a times b.

01:41 And for this problem, you already found out what inverse of matrix a is from problem 29.

01:49 So therefore, we know that the inverse of matrix a times b is equal to, again, inverse of matrix a, you found it in problem 29, and we found that it's this.

02:38 Okay, so again, we got this inverse of matrix a from problem 29.

02:43 And now you're going to multiply this by matrix b, which was 3353...

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