Question

Please help me. How can I solve this? ​

Answer 1

Answer:

see explanation

Step-by-step explanation:

Given matrix A = $\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right]$, then

detA = ad - bc

Given

A = $\left[\begin{array}{ccc}5&1\\a+1&a\\\end{array}\right]$

detA = 5a - (a + 1) = 5a - a - 1 = 4a - 1

and

4a - 1 = 7 ( add 1 to both sides )

4a = 8 ( divide both sides by 4 )

a = 2

Thus

A = $\left[\begin{array}{ccc}5&1\\3&2\\\end{array}\right]$

A² - 7A + 7I

A² = $\left[\begin{array}{ccc}5&1\\3&2\\\end{array}\right]$ $\left[\begin{array}{ccc}5&1\\3&2\\\end{array}\right]$

    = $\left[\begin{array}{ccc}5(5)+1(3)&5(1)+1(2)\\3(5)+2(3)&3(1)+2(2)\\\end{array}\right]$ = $\left[\begin{array}{ccc}28&7\\21&7\\\end{array}\right]$

7A= 7 $\left[\begin{array}{ccc}5&1\\3&2\\\end{array}\right]$ = $\left[\begin{array}{ccc}7(5)&7(1)\\7(3)&7(2)\\\end{array}\right]$ = $\left[\begin{array}{ccc}35&7\\21&14\\\end{array}\right]$

7I = 7 $\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]$ = $\left[\begin{array}{ccc}7&0\\0&7\\\end{array}\right]$

Thus

$\left[\begin{array}{ccc}28&7\\21&7\\\end{array}\right]$ - $\left[\begin{array}{ccc}35&7\\21&14\\\end{array}\right]$ + $\left[\begin{array}{ccc}7&0\\0&7\\\end{array}\right]$

= $\left[\begin{array}{ccc}(28-35+7)&(7-7+0)\\(21-21+0)&(7-14+7)\\\end{array}\right]$ = $\left[\begin{array}{ccc}0&0\\0&0\\\end{array}\right]$

Hence

7A² - 7A + 7I = 0