Question

If A= [1 3 -2 4] show that A^2-5A+10I=0​

Answer 1

Step-by-step explanation:

[1 3 -2 4]×[1 3 -2 4]=A^2

[1×1 3×3 -2×-2 4×4]=A^2

[1 9 4 16]=A^2

-5A=[-5 -15 10 -20]

10I=[1 0 0 0]×10

10I=[10 0 0 0]

A^2-5A+10I=[1 9 4 16] +[-5 -15 10 -20]+[10 0 0 0]

A^2-5A+10I=0

Answer 2

Proved the expression $A^2-5A+10I=0$

Step-by-step explanation:

Given: Matrix $A = \left[\begin{array}{cc}1&3\\-2&4\end{array}\right]$

To Prove: $A^2-5A+10I=0$

Solution:

• Poof of A² - 5A + 10I = 0

$\Rightarrow A^2 = AA = \left[\begin{array}{cc}1&3\\-2&4\end{array}\right]\left[\begin{array}{cc}1&3\\-2&4\end{array}\right]$

$\Rightarrow A^2 = \left[\begin{array}{cc}1-6&3+12\\-2-8&-6+16\end{array}\right]$

$\Rightarrow A^2 = \left[\begin{array}{cc}-5&15\\-10&10\end{array}\right] \ \cdots \cdots (1)$

and, $\Rightarrow 5A = \left[\begin{array}{cc}5&15\\-10&20\end{array}\right] \ \cdots \cdots (2)$

To prove $A^2-5A+10I=0$, we have,

$\Rightarrow \left[\begin{array}{cc}-5&15\\-10&10\end{array}\right] - \left[\begin{array}{cc}5&15\\-10&20\end{array}\right] +10 \left[\begin{array}{cc}1&0\\0&1\end{array}\right] =0$

$\Rightarrow \left[\begin{array}{cc}-10&0\\0&-10\end{array}\right] + \left[\begin{array}{cc}10&0\\0&10\end{array}\right] =0$

$\Rightarrow \left[\begin{array}{cc}0&0\\0&0\end{array}\right] =0$

Hence, proved the expression $A^2-5A+10I=0$