Show that B is the inverse of A.

$A=\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right],B=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]$

Reeves
2020-11-02
Answered

Show that B is the inverse of A.

$A=\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right],B=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]$

You can still ask an expert for help

Nathanael Webber

Answered 2020-11-03
Author has **117** answers

Step 1

Given the matrices

$A=\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right],B=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]$

Show that B is the inverse of A.

Step 2

To show that two matrices are inverse it must satisfy the following condition

AB=BA=I

where I is the identity matrix

$AB=\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]$

$AB=\left[\begin{array}{cc}(1\cdot 2)+(-1\cdot 1)& (1\cdot 1)+(-1\cdot 1)\\ (-1\cdot 2)+(2\cdot 1)& (-1\cdot 1)+(2\cdot 1)\end{array}\right]$

$AB=\left[\begin{array}{cc}2-1& 1-1\\ 1-1& 2-1\end{array}\right]$

$AB=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$

$BA=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]$

$BA=\left[\begin{array}{cc}(2\cdot 1)+(1\cdot -1)& (2\cdot -1)+(1\cdot 2)\\ (1\cdot 1)+(1\cdot -1)& (1\cdot -1)+(1\cdot 2)\end{array}\right]$

$BA=\left[\begin{array}{cc}2-1& -2+2\\ 1-1& -1+2\end{array}\right]$

$BA=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$

So AB=BA=I

hence B is the inverse of A.

Given the matrices

Show that B is the inverse of A.

Step 2

To show that two matrices are inverse it must satisfy the following condition

AB=BA=I

where I is the identity matrix

So AB=BA=I

hence B is the inverse of A.

Jeffrey Jordon

Answered 2022-01-24
Author has **2313** answers

Answer is given below (on video)

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$\text{Basis}=\{\left[\begin{array}{cc}& \\ & \end{array}\right],\left[\begin{array}{cc}& \\ & \end{array}\right]\}$

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Let B be a 4x4 matrix to which we apply the following operations:

1. double column 1,

2. halve row 3,

3. add row 3 to row 1,

4. interchange columns 1 and 4,

5. subtract row 2 from each of the other rows,

6. replace column 4 by column 3,

7. delete column 1 (column dimension is reduced by 1).

(a) Write the result as a product of eight matrices.

(b) Write it again as a product of ABC (same B) of three matrices.

1. double column 1,

2. halve row 3,

3. add row 3 to row 1,

4. interchange columns 1 and 4,

5. subtract row 2 from each of the other rows,

6. replace column 4 by column 3,

7. delete column 1 (column dimension is reduced by 1).

(a) Write the result as a product of eight matrices.

(b) Write it again as a product of ABC (same B) of three matrices.

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Write the uncoded row matrices for the message.

Message: SELL CONSOLIDATED

Row Matrix Size:$1\times 3$

Encoding Matrix:$A=\left[\begin{array}{ccc}1& -1& 0\\ 1& 0& -1\\ -6& 3& 2\end{array}\right]$

SEL=$\left[\begin{array}{ccc}19& 5& 12\end{array}\right]$

L-C=$\left[\begin{array}{ccc}12& 0& 3\end{array}\right]$

ONS=$\left[\begin{array}{ccc}15& 14& 19\end{array}\right]$

OLI=$\left[\begin{array}{ccc}15& 12& 9\end{array}\right]$

DAT=$\left[\begin{array}{ccc}4& 1& 20\end{array}\right]$

ED=$\left[\begin{array}{ccc}5& 4& 0\end{array}\right]$

Message: SELL CONSOLIDATED

Row Matrix Size:

Encoding Matrix:

SEL=

L-C=

ONS=

OLI=

DAT=

ED=

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Given n students participating in a contest of m questions. At each stage, a student may choose to do the question in English or German or skip it. For every two questions, there exists a student who chooses to do both the questions and do them in different languages. What is the largest value that m can take.

Hint: The answer to this question is $m\le {2}^{n}$, you might expect where the binary system is used

Hint: The answer to this question is $m\le {2}^{n}$, you might expect where the binary system is used

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Let $g:{\mathbb{R}}^{\mathbb{2}}\to \mathbb{R}$ be totally differentiable in the point $a\in {\mathbb{R}}^{\mathbb{2}}$

Let ${v}_{1}:=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)$ and ${v}_{2}:=\frac{1}{5}\left(\begin{array}{c}4\\ -3\end{array}\right)$ be normalized direction vectors in ${\mathbb{R}}^{\mathbb{2}}$

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With the info given above, how can one find out $\mathrm{\nabla}g(a)$?

Let ${v}_{1}:=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)$ and ${v}_{2}:=\frac{1}{5}\left(\begin{array}{c}4\\ -3\end{array}\right)$ be normalized direction vectors in ${\mathbb{R}}^{\mathbb{2}}$

It is $\frac{\mathrm{\partial}g}{\mathrm{\partial}{v}_{1}}(a)=5\sqrt{2}$ and $\frac{\mathrm{\partial}g}{\mathrm{\partial}{v}_{2}}(a)=1$

With the info given above, how can one find out $\mathrm{\nabla}g(a)$?