Question

5. (a) Two matrices, are given as $\begin{pmatrix} 4 & -1 & 6 \ 0 & -5 & 2 \ 1 & 0 & 2 \end{pmatrix}$ $A = \begin{pmatrix} 10 & -2 & -28 \ -2 & -2 & 8 \ -5 & 1 & 20 \end{pmatrix}$ $B =$ i) By finding AB, deduce that $A^{-1} = \frac{1}{12}B$. ii) Hence, or otherwise, solve the system of equations given as $\begin{pmatrix} 4 & -1 & 6 \ 0 & -5 & 2 \ 1 & 0 & 2 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} -35 \ -5 \ 12 \end{pmatrix}$

          5. (a) Two matrices, are given as
$\begin{pmatrix} 4 & -1 & 6 \ 0 & -5 & 2 \ 1 & 0 & 2 \end{pmatrix}$ $A = \begin{pmatrix} 10 & -2 & -28 \ -2 & -2 & 8 \ -5 & 1 & 20 \end{pmatrix}$
$B =$
i) By finding AB, deduce that $A^{-1} = \frac{1}{12}B$.
ii) Hence, or otherwise, solve the system of equations given as
$\begin{pmatrix} 4 & -1 & 6 \ 0 & -5 & 2 \ 1 & 0 & 2 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} -35 \ -5 \ 12 \end{pmatrix}$

5. (a) Two matrices, are given as
< p m a t r i x > A =
< p m a t r i x >
B =
i) By finding AB, deduce that A^-1 = (1)/(12)B.
ii) Hence, or otherwise, solve the system of equations given as
< p m a t r i x >

< p m a t r i x >
=
< p m a t r i x >

Added by Abigail V.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Daniel Carr

Step 1

First, let's find the product of matrices A and B: A = [10 -2] [-28 2] B = [-28 51] [20 02] Show more…

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Two matrices are given as: 10 -2 -28 2 -28 51 20 02 By finding AB, deduce that A = B. Hence, or otherwise, solve the system of equations given as: [s] [4]