Answer to Question #262380 in Linear Algebra for zid

Question

Let A =

(8 -1 2

2 0 -5)

B =

(-1 7

3 -2

1 5)

and C =

(2 1

3 5)

(a) Calculate AB and A + B if they exist.

(b) Verify that (AB)C = A(BC).

(c) Calculate C-1 A.

Answer 1

a)

(A+B) does not exist, since matrices are of different sizes (2x3 and 3x2)

$AB=\begin{pmatrix} 8 & -1 &2\\ 2 & 0&-5 \end{pmatrix}\begin{pmatrix} -1 & 7 \\ 3 & -2\\ 1&5 \end{pmatrix}=\begin{pmatrix} -8-3+2 & 56+2+10 \\ -2-5 & 14-25 \end{pmatrix}=\begin{pmatrix} -9 & 68 \\ -7 & -11 \end{pmatrix}$

b)

$(AB)C=\begin{pmatrix} -9 & 68 \\ -7 & -11 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 3 & 5 \end{pmatrix}=\begin{pmatrix} -18+204 & -9+340 \\ -14-33 & -7-55 \end{pmatrix}=\begin{pmatrix} 186 & 331 \\ -47 & -62 \end{pmatrix}$

$BC=\begin{pmatrix} -1 & 7 \\ 3 & -2\\ 1&5 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 3 & 5 \end{pmatrix}=\begin{pmatrix} -2+21 & -1+35 \\ 6-6 & 3-10\\ 2+15&1+25 \end{pmatrix}=\begin{pmatrix} 19 & 34 \\ 0 & -7\\ 17&26 \end{pmatrix}$

$A(BC)=\begin{pmatrix} 8 & -1 &2\\ 2 & 0&-5 \end{pmatrix}\begin{pmatrix} 19 & 34 \\ 0 & -7\\ 17&26 \end{pmatrix}=\begin{pmatrix} 152+34 & 272+7+52 \\ 38-85 & 68-130 \end{pmatrix}=$

$=\begin{pmatrix} 186 & 331 \\ -47 & -62 \end{pmatrix}$

c)

for C^-1:

$C_*=\begin{pmatrix} 5 & -3 \\ -1 & 2 \end{pmatrix}$ , $C_*^T=\begin{pmatrix} 5 & -1 \\ -3 & 2 \end{pmatrix}$

$|C|=10-3=7$

$C^{-1}=\frac{C_*^T}{|C|}=\frac{}{}=\frac{1}{7}\begin{pmatrix} 5 & -1 \\ -3 & 2 \end{pmatrix}$

$C^{-1}A=\frac{1}{7}\begin{pmatrix} 5 & -1 \\ -3 & 2 \end{pmatrix}\begin{pmatrix} 8 & -1 &2\\ 2 & 0&-5 \end{pmatrix}=\frac{1}{7}\begin{pmatrix} 40-2 & -5 &10+5\\ -24+4 & 3&-6-10 \end{pmatrix}=$

$=\frac{1}{7}\begin{pmatrix} 38 & -5 &15\\ -20 & 3&-16 \end{pmatrix}$