Question

if A= [1/0=0/1] find A^2-A​

Answer 1

Answer:

(1/0=0/1)

Step-by-step explanation:

(1/0=0/1) ^2 - (1/0=0/1)

(1/0=0/1)

Answer 2

Appropriate Question :-

If $\begin{gathered} \rm A = \begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix} \end{gathered}$ find A² - A

Answer :-

$\rm{A {}^{2} - A = 0}$

Step by step explanation :-

We know that,

➦ A² = A × A

Now, Firstly we find A²

$\small\rm:\longmapsto{\begin{gathered} \rm A² = \begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix} \end{gathered} \times \begin{gathered}\begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix} \end{gathered} }$

Let's learn,

• How to multiply matrix 2×2

$\small\rm{A × B = \begin{gathered} \rm \begin{bmatrix} \rm a& \rm b \\ \rm c & \rm d \end{bmatrix} \times \begin{gathered} \rm \begin{bmatrix} \rm e& \rm f \\ \rm g & \rm h \end{bmatrix} \end{gathered} \end{gathered} = \begin{gathered} \rm \begin{bmatrix} \rm ae + bg & \rm af + bh\\ \rm ce + dg&\rm cf + dh \end{bmatrix} \end{gathered}}$

Now,

$\small\rm:\longmapsto{A² =\begin{gathered} \begin{bmatrix} (1 \times 1) + (0 \times 0)& (1 \times 0) + (0 \times 1) \\ (0 \times 1) + (1 \times 0) & (0 \times 0) + (1 \times 1) \end{bmatrix} \end{gathered} } \\ \\ \small\rm:\longmapsto{A² =\begin{gathered} \begin{bmatrix} 1 + 0& 0 + 0 \\ 0 + 0 & 0 + 1 \end{bmatrix} \end{gathered} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \small\rm:\longmapsto \red{A² =\begin{gathered} \begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix} \end{gathered} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now, we find A² - A

And, we have

$\small\rm:\longmapsto \orange{A² =\begin{gathered} \begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix} \end{gathered} }$

Let's learn,

• How to subtract matrix

$\small\rm{A - B = \begin{gathered} \rm \begin{bmatrix} \rm a& \rm b \\ \rm c & \rm d \end{bmatrix} - \begin{gathered} \rm \begin{bmatrix} \rm e& \rm f \\ \rm g & \rm h \end{bmatrix} \end{gathered} \end{gathered} = \begin{gathered} \rm \begin{bmatrix} \rm a - e & \rm b - f\\ \rm c - g&\rm d - h \end{bmatrix} \end{gathered}}$

We have to find

$\rm:\longmapsto{A² - A = \begin{gathered} \rm \begin{bmatrix} 1 - 1& 0 -0 \\ 0 - 0& 1 - 1\end{bmatrix} \end{gathered} } \\ \\ \rm:\longmapsto \red{A² - A = \begin{gathered} \rm \begin{bmatrix} 0& 0 \\ 0 & 0 \end{bmatrix} \end{gathered} } \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Hence,

$\rm:\longmapsto{A² - A = 0}$

Where,

• 0 is the null matrix.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬