1. matrix

Say i have a matrix ,

$\begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}+5$

is it correct if i do it this way ,

$\begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}+5\begin{bmatrix}{1}&{0}\\{0}& { 1}\end{bmatrix}$

$=\begin{bmatrix}{9}&{3}\\{-1}&{12}\end{bmatrix}$
is 5 a scalar = 5I where I is an identity matrix ?

2. Originally Posted by thereddevils
Say i have a matrix ,

$\begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}+5$ Mr F says: This makes no sense. Is this what the question said? Are you sure it didn't say ${\color{red}\begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}+5{\color{blue}I} }$ ....?

is it correct if i do it this way ,

$\begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}+5\begin{bmatrix}{1}&{0}\\{0}& { 1}\end{bmatrix}$

$=\begin{bmatrix}{9}&{3}\\{-1}&{12}\end{bmatrix}$
is 5 a scalar = 5I where I is an identity matrix ?
..

3. Originally Posted by mr fantastic
..
GIven a matrix $A=\begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}$

And also given $A^2-5A+6I=0$ and i am supposed to deduce $A^{-1}$ from this .

so $A(A-5)=-6I$

$A(\frac{A-5}{-6})=I$

so $A^{-1}=\frac{A-5}{-6}$

and my problem is here .. the (A-5) part . However , i did what i showed in my previous post and managed to get the answers .

4. Why don't you just use the fact that

If $A = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]$

then $A^{-1} = \frac{1}{ad - bc}\left[\begin{matrix}d & -b \\ -c & a\end{matrix}\right]$?

5. Originally Posted by Prove It
Why don't you just use the fact that

If $A = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]$

then $A^{-1} = \frac{1}{ad - bc}\left[\begin{matrix}d & -b \\ -c & a\end{matrix}\right]$?
the question restricts me from doing that , it says hence deduce the inverse from that equation .

6. Originally Posted by thereddevils
GIven a matrix $A=\begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}$

And also given $A^2-5A+6I=0$ and i am supposed to deduce $A^{-1}$ from this .

so $A(A-5)=-6I$

$A(\frac{A-5}{-6})=I$

so $A^{-1}=\frac{A-5}{-6}$

and my problem is here .. the (A-5) part . However , i did what i showed in my previous post and managed to get the answers .
$A^2-5A+6I=0$

Post multiply both sides by $A^{-1}$:

$A - 5I + 6A^{-1} = 0$

$\Rightarrow A^{-1} = \frac{1}{6} (5I - A)$.