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

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.

• Strategy: take the RHS's and multiply by $A$ to see if you get $I$. Jun 19, 2018 at 14:57
• The short answer is $(-2)^2 + 4^2 + 2^2 + 1^2 = 25$. (For reasons given below.) Jun 20, 2018 at 0:40
• You're assuming that because you have, on average, 90 seconds to solve a problem that each problem must be solved in 90 seconds. Some will undoubtedly take less than 90 seconds. Some will undoubtedly take more. Do your best. Jun 20, 2018 at 16:15

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.

• If $A$ has no inverse, there will be no (nonzero) solution $\lambda$ to $A^2 = \lambda I$, so computing $A^2$ gives this information, too. Jun 19, 2018 at 15:01
• Indeed it will but it would require to compute more than just one diagonal element of $A^2$ (the answer states that "you need to calculate only one diagonal element of $A^2$ to find the correct option."). Sorry for being so petty :). Jun 19, 2018 at 15:11
• @alephzero: To be specific, if you calculate one diagonal element and it's 25, that doesn't tell you whether $A^2=25I$ or $A^2=\begin{bmatrix}25 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}$ ... Jun 19, 2018 at 21:23
• For checking if $A^{-1}$ exists, this matrix is very amenable to the trick of computing the determinant modulo 2
– jld
Jun 19, 2018 at 22:47
• @JoseLopezGarcia for an integer-valued matrix, the determinant is non-zero if the determinant with every entry taken modulo 2 is also non-zero. So in your case $$A \mod 2=\begin{bmatrix}0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}$$and this matrix is clearly full rank and so has non-zero determinant (it will certainly be $\pm 1$ but the actual value isn't important so you can stop as soon as you recognize it's not singular).
– jld
Jun 20, 2018 at 13:54

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 $A/25$.

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 $C.D = -D.C$

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

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.

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.

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.

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.

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

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

• nice answer: the matrix in fact is symmetric and orthogonal (+1) Jun 20, 2018 at 15:03

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.

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.