Answer to Question #111653 in Linear Algebra for Prosper Mawuli

Question

3. If the matrices A and B are given by A= and B =
a. Calculate 3A -2B+7I
b. 3A + 4X = B is an equation for an unknown matrix X. Find the matrix X
c. Calculate the matrix product AB
4. Solve the following set of equations by Gaussian elimination:
3x + 2y + z =2
4x + 2y + 2z =8
x – y + z = 8

Answer 1

let$A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $B=\begin{bmatrix} i & j\\ k & l \end{bmatrix}$

then

a. Calculate 3A -2B+7I

$3A=\begin{bmatrix} 3a & 3b \\ 3c & 3d \end{bmatrix}\\2B=\begin{bmatrix} 2i & 2j\\ 2k & 2l \end{bmatrix}\\7I=\begin{bmatrix} 7& 0 \\ 0& 7 \end{bmatrix}\\3A-2B+7I=\begin{bmatrix} 3a & 3b \\ 3c & 3d \end{bmatrix}-\begin{bmatrix} 2i & 2j\\ 2k & 2l \end{bmatrix}+\begin{bmatrix} 7 &0\\ 0& 7 \end{bmatrix}$

$3A-2B+7I=\begin{bmatrix} 3a-2i+7 & 3b-2j \\ 3c-2k & 3d-2l+7 \end{bmatrix}$

b.) 3A + 4X = B is an equation for an unknown matrix X. Find the matrix X

let $X=\begin{bmatrix} u&v\\ w& z \end{bmatrix}$

then

$\begin{bmatrix} 3a & 3b \\ 3c & 3d \end{bmatrix}+\begin{bmatrix} 4u& 4v \\ 4w & 4z \end{bmatrix}=\begin{bmatrix} i & j \\ k & l \end{bmatrix}$

forming system of linear equations

$3a+4u=i\\3b+4v=j\\3c+4w=k\\3d+4z=l$

solving for u, v, w and z

$u=\frac{i-3a}{4}\\v=\frac{j-3b}{4}\\w=\frac{k-3c}{4}\\z=\frac{l-3d}{4}$

hence

$X=\begin{bmatrix} \frac{i-3a}{4}& \frac{j-3b}{4} \\ \frac{k-3c}{4}& \frac{l-3d}{4} \end{bmatrix}$

c.) Calculate the matrix product AB

$A=\begin{bmatrix} a & b \\ c & d \end{bmatrix};B=\begin{bmatrix} i& j\\ k & l \end{bmatrix}$

then

$\\\\\\AB=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} i& j\\ k & l \end{bmatrix}\\\\ AB=\begin{bmatrix} ai+bk& aj+bl\\ ci+dk &cj+dl \end{bmatrix}$

4.) Solve the following set of equations by Gaussian elimination:

3x + 2y + z =2

4x + 2y + 2z =8

x – y + z = 8

extract augmented matrix

$\begin{bmatrix} 3 & 2&1&2 \\ 4& 2&2&8\\ 1&-1&1&8 \end{bmatrix}\\R_{1} to R _{3}\begin{bmatrix} 3 & 2&1&2 \\ 4& 2&2&8\\ 1&-1&1&8 \end{bmatrix} \\\begin{bmatrix} 1 & -1&1&8 \\ 4& 2&2&8\\ 3&2&1&2 \end{bmatrix}$

divide R2 by 2

$\begin{bmatrix} 1 & -1&1&8 \\ 2& 1&1&4\\ 3&2&1&2 \end{bmatrix}$

using row1 to reduce row2 and row3

$\begin{bmatrix} 1 & -1&1&8 \\ 0& 3&-1&-12\\ 0&5&-2&-22 \end{bmatrix}$

using row2 to reduce row3 and keep row1

$\begin{bmatrix} 1 & 0&2&12 \\ 0& 3&-1&-12\\ 0&0&-1&-6 \end{bmatrix}$

using row3 to reduce row1and keep row2

$\begin{bmatrix} 1 & 0&0&0 \\ 0& 3&0&-6\\ 0&0&-1&-6 \end{bmatrix}$

dividing R2 by 3 and R3 by -1

$\begin{bmatrix} 1 &0&0&0 \\ 0& 1&0&-2\\ 0&0&1&6 \end{bmatrix}$

hence

$x=0\\y=-2\\z=6$