Question

If A = $\left[\begin{array}{ccc}1&0&0\\2&3&4\\5&-6&x\end{array}\right]$ and det A = 45 then find x.

Answer 1

Determinant of a Matrix

Here we are given a question relating Determinants to Matrices. We have an unknown value inside a matrix A, which we are supposed to find using the given value of determinant.

$\displaystyle\sf A=\left[\begin{array}{ccc}1&0&0\\2&3&4\\5&-6&x\end{array}\right] \\\\\\ \implies det\ A = \left|\begin{array}{ccc}1&0&0\\2&3&4\\5&-6&x\end{array}\right| = 45 \\\\\\ \implies 1.\left|\begin{array}{cc}3&4\\-6&x\end{array}\right| + 0.\left|\begin{array}{cc}2&4\\5&x\end{array}\right|+0.\left|\begin{array}{cc}2&3\\5&-6\end{array}\right|=45\\\\\\ \implies 1(3.x-(4.(-6)))+0=45\\\\\\\implies 3x+24=45\\\\\\\implies 3x=45-24\\\\\\\implies 3x=21\\\\\\ \implies \huge \boxed{\sf x=7}$

Hence, the value of x is 7.

Answer 2

Answer:

Given data:

The first 100 natural numbers are taken into consideration. (1, 2 3,..., 100)

To find:

The Units digit of the sum of the third powers (cubes) of the first 100 natural numbers.

Solution:

We know that :

Starting from 1, the sum of cubes of first n natural numbers is: (\frac{n(n+1)}{2})^2(

2

n(n+1)

)

2

Here, n=100.

The sum of the cubes of first 100 natural numbers i.e.,

1³+2³+3³+...+100³ = (\frac{100(100+1)}{2})^2(

2

100(100+1)

)

2

= (\frac{100*101}{2})^2(

2

100∗101

)

2

= (50 x 101)²

= 5050²

As we need to calculate the units digit only, there is no need to calculate the complete value of 5050².

As the last digit of 5050 is 0, Squaring the number 5050 results the ending or units digit as 0.

Therefore, The Units digit of the sum of the third powers (cubes) of the first 100 natural numbers i.e.,

The units digit of 1³+2³+3³+...+100³ is 0.

Check:

5050² = 25502500

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