Yes, we can obtain a diagonal matrix by multiplying two non diagonal matrix.
Consider the matrix multiplication below
![\left[\begin{array}{cc}a&b\\c&d\end{array}\right] \left[\begin{array}{cc}e&f\\g&h\end{array}\right] = \left[\begin{array}{cc}a e+b g&a f+b h\\c e+d g&c f+d h\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%20e%2Bb%20g%26a%20f%2Bb%20h%5C%5Cc%20e%2Bd%20g%26c%20f%2Bd%20h%5Cend%7Barray%7D%5Cright%5D%20)
For the product to be a diagonal matrix,
a f + b h = 0 ⇒ a f = -b h
and c e + d g = 0 ⇒ c e = -d g
Consider the following sets of values
The the matrix product becomes:
![\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}\frac{1}{3}&-1\\-\frac{1}{4}&\frac{1}{2}\end{array}\right] = \left[\begin{array}{cc}\frac{1}{3}-\frac{1}{2}&-1+1\\1-1&-3+2\end{array}\right]= \left[\begin{array}{cc}-\frac{1}{6}&0\\0&-1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7B1%7D%7B3%7D%26-1%5C%5C-%5Cfrac%7B1%7D%7B4%7D%26%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7B1%7D%7B3%7D-%5Cfrac%7B1%7D%7B2%7D%26-1%2B1%5C%5C1-1%26-3%2B2%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-%5Cfrac%7B1%7D%7B6%7D%260%5C%5C0%26-1%5Cend%7Barray%7D%5Cright%5D)
Thus, as can be seen we can obtain a diagonal matrix that is a product of non diagonal matrices.
8(8x)
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Step-by-step explanation:
We dimension of the smallest cube to be made from cuboids of sides 3 m , 4 m and 5 m will be the least common multiple of 3 m, 4 m and 5 m i.e. 60 m
12 cuboidal should be stacked along 5 m edge to 60 m, 15 cuboids should be stacked along 4 m edge and 20. cuboids should be stacked along edge to make a cube of 60 m edge, Hence number of cuboids are 12× 15 ×20=3600
hope it helps
Answer:
1 is congruent 2 is neither and 3 is similar
Step-by-step explanation:
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Answer:
- Total surface area of cylinderal can is 150.8 cm²
<u>Step-by-step </u><u>explanation</u><u>:</u><u> </u>
Given that a cylinderal can has base with a diameter of 6 cm and its height measures 5 cm.
So,
Radius = Diameter/2 = 6/2 = 3 cm
We know that,
- TSA of cylinder = 2πr(h + r)
Substituting required values:
➝ TSA = 2 × 22/7 × 3(5 + 3)
➝ TSA = 44/ 7 × 3 × 8
➝ TSA = 1056/7
➝ TSA = 150.8 cm²
Hence,
- <u>Total </u><u>surface </u><u>area </u><u>of </u><u>cylinderal</u><u> </u><u>can </u><u>is </u><u>1</u><u>5</u><u>0</u><u>.</u><u>8</u><u> </u><u>cm²</u>
