Matrix

[dagenny]

Consider a square matrix having order 2. If we assume A²=81I we can always conclude that

a. det A = 0
b. all the other statements are false
c. A = 9I
d. |det A| = 9
e. det A = 9

The right answer is b), but why is that?

 

[stapel]

Consider a square matrix having order 2. If we assume A²=81I ...

What is the meaning of "81/"? Does the exercise really assign "81/" as the "value" (??) of the square of the matrix A? Or does it actually refer to the determinant of A?

...we can always conclude that

a. det A = 0
b. all the other statements are false
c. A = 9I
d. |det A| = 9
e. det A = 9

What is the meaning of "9/"? In option (d), are they taking the absolute value of the determinant?

 

[dagenny]

I stands for Identity Matrix
And yes, in option d they're taking the absolute value of the determinant.

What is the meaning of "81/"? Does the exercise really assign "81/" as the "value" (??) of the square of the matrix A? Or does it actually refer to the determinant of A?


What is the meaning of "9/"? In option (d), are they taking the absolute value of the determinant?

 

[dagenny]

Yes, the exercise assigns 81I as the value, not the determinant.

What is the meaning of "81/"? Does the exercise really assign "81/" as the "value" (??) of the square of the matrix A? Or does it actually refer to the determinant of A?


What is the meaning of "9/"? In option (d), are they taking the absolute value of the determinant?

 

[ksdhart2]

Consider a square matrix [\(A\)] having order 2

[...]

I stands for Identity Matrix

If that's the case, then it should be easy to see that the given condition is really:

\(\displaystyle A^2 =
\begin{bmatrix}
81 & 0 \\
0 & 81
\end{bmatrix}
\)

Even though we're dealing with matrices now, this is still a form of algebra and so the single most important rule in all of algebra still applies - if you don't know the value of a variable, give it a name so you can talk about it and work with it. Let's assign the matrix \(A\) a generic value:

\(\displaystyle A =
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{bmatrix}
\)

Synthesizing that with the given information:

\(\displaystyle
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{bmatrix} \times
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{bmatrix} =
\begin{bmatrix}
81 & 0 \\
0 & 81
\end{bmatrix}
\)

What now? Where does this lead you?

 

[Dr.Peterson]

Consider a square matrix having order 2. If we assume A²=81I we can always conclude that

a. det A = 0
b. all the other statements are false
c. A = 9I
d. |det A| = 9
e. det A = 9

The right answer is b), but why is that?

Just dealing with the notation question: The sans-serif font makes this italic capital i (I) look a lot like a slash (/), though not in this case like an italic lowercase L (l). I couldn't make any sense out of it. Probably it would be clearer (apart from using LaTeX) to use bold for matrices:

Consider a square matrix having order 2. If we assume A²=81I we can always conclude that​

​

a. det A = 0​

b. all the other statements are false​

c. A = 9I​

d. |det A| = 9​

e. det A = 9​

As for the problem itself, one could consider how multiplication affects determinants. What is det(I)? What is det(A²)? What might det(A) be?