How to compute this determinant as quickly as possible (without using any software or calculator)?

Question

I sat an Algebra test yesterday, which consisted of 30 questions and a total time of 45 minutes (an average of 1 min 30 secs per question). One question of the test was this:

Given the matrix: $$A=\begin{bmatrix} -2 & 4 & 2 & 1 \\ 4 & 2 & 1 & -2 \\ 2 & 1 & -2 & 4 \\ 1 & -2 & 4 & 2 \end{bmatrix}$$ Which of the following options is correct?

(A) $A^{-1}=\dfrac{A}{25}$

(B) $A^{-1}=\dfrac{A}{5}$

(C) $A^{-1}=\dfrac{A}{15}$

(D) It has no inverse.

I do know how to compute the inverse of a matrix. For example, using the Gauss-Jordan elimination method. Or for example, using this formula: $$A^{-1}=\dfrac{\text{Adj}(A^T)}{\text{det}(A)}$$

I calculated the determinant and it is $625$. However, this won't help me pick the correct option (of course I can eliminate option D, which is false).

How in the world am I supposed to solve this problem in around 90-100 seconds, without using a calculator? Is there any shortcut or trick or something I missed? 90 seconds was the average time per question in the test. Given how little time I was given to solve the problem, this leads me to think that the structure of A could simplify the answer.

Answer 1

If $A^{-1}=\dfrac{A}{25}$ then $A^2=25I$. Similar for other options. Now you need to calculate only one diagonal element of $A^2$ to find the correct option.

Answer 2

NB you don't need to compute $A^{-1}$ in order to solve this problem, you just need to determine which option is correct---presumably the question also tests your recognition of this fact.

If $A$ is invertible, then multiplying both sides of each of the first three options, $A^{-1} = \frac{A}{\lambda}$, gives an equivalent equation $$A^2 = \lambda I .$$ If $A$ is not invertible, then there is no (nonzero) $\lambda$ for which the equation holds. So, to determine the answer, it's enough to compute $A^2$, which is computationally much cheaper than applying G.-J. elimination.

Answer 3

You're there with the computation of the determinant. The $\det A^2 = 5^8$, so the answer must be (A) in order for $\det \left(A\cdot A^{-1}\right) = 1$

But PJK's answer provides a more efficient approach.

Answer 4

The matrix has the form $A=\begin{bmatrix} C & D \\ D & C \end {bmatrix}$

For this kind of block matrices,

$det (A) = det( C.C -D.C^{-1}.D.C) = $

$ = det (C.C + D.D ) = det \left ( \begin{bmatrix} 20 & 0 \\ 0 & 20 \end {bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end {bmatrix} \right ) = 25^2 $

since $ C.D = -D.C $

$A^2=\begin{bmatrix} C & D \\ D & C \end {bmatrix} \begin{bmatrix} C & D \\ D & C \end {bmatrix} = \begin{bmatrix} C^2 + D^2 & 0 \\ 0 & C^2 + D^2 \end {bmatrix} = \begin{bmatrix} 25 & 0 & 0 & 0 \\ 0 & 25 & 0 & 0 \\ 0 & 0 & 25 & 0 \\ 0 & 0 & 0 & 25 \end{bmatrix} $

Answer 5

You write

I calculated the determinant and it is 625

which is enough to conclusively answer.

For option $\mathbf A$ we have $\frac{1}{det(A)} = det(A^{-1}) {\space\overset{\mathbf A}{=}\space} det(\frac{A}{25}) = \frac{det(A)}{25^4}$. Similarly for the other answers.

Answer 6

We can also solve this problem quickly without using the multiple choice options as hints to cut short any computations. So, suppose that we're given the matrix $A$ and asked to compute its inverse, without being told that the inverse is proportional to $A$ itself. Then observe that the columns of the matrix are orthogonal, so the matrix is an orthogonal matrix up to a normalization factor. The norm of the columns is 5. The inverse of $A/5$ is therefore equal to its transpose, and the transpose is easily seen to be equal to the matrix itself. So, the inverse of $A$ is $A/25$.

Answer 7

EDIT: I realized after writing that the essence of my answer is already in the answer provided by Count Iblis. Perhaps one can consider this answer as containing supplementary material to his.

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Deriving the answer

We will make use of the observation that in the given options, multiplying $A$ on both sides gives the result $\frac{1}{c}I$ for some constant $c$.

First, notice that $A$ is a circulant real symmetric matrix. Because it is a circulant matrix, all its rows have the same norm of 5.

We can therefore write $A = 5 B$ where B is also real symmetric and each row (and column) of B has unit norm. This gives us the following

$A^2 = 25B^2 = 25 B^TB$.

At this point, this should provide enough hint to inspect the matrix $A$ more carefully and you will notice that it is actually orthogonal. Therefore $B$ is orthonormal so $B^TB = I$ and we have $A^2 = 25I$ as required.

If the orthogonality of $A$ still eludes you, we can use the spectral theorem and write

$A^2 = 25B^TB = 25 U\Lambda^2U^T = 25(\Lambda^{\frac{1}{2}}U^T)^T(\Lambda^{\frac{1}{2}}U^T) \implies B = \Lambda^{\frac{1}{2}}U^T$.

Since $U^T$ is orthonormal and each row and column of $B$ has unit norm, this means $\Lambda^{\frac{1}{2}} = I$ and $B^2 = B^TB = I$ as required.

Also, notice that $\Lambda = I$ so all the eigenvalues of $A$ are non-zero (they are all 1) so $A$ is invertible and this excludes the last option.

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Determining the determinant

Now that we know $B$ is orthonormal, we can obtain the determinant of $A$ easily using the property $\text{det}(cM) = c^n \text{det}(M)$ for $M \in \mathbb{R}^{n \times n}$ and det($Q$) = 1 if $Q$ is a square orthonormal matrix

$A = 5B \implies \text{det}(A) = 5^4 \text{det}(B) = 625$

Answer 8

Here is a very simple although not as elegant answer:

Whatever the inverse would be, multiplication must be $1$ for diagonal elements, in particular for the first element. We can call it $c_{11}$.

This shields $\frac{25}{x}$ where $x$ is the one between options that makes this element 1. So it's a or d. In any case, you need to calculate all elements for choosing between this two options.