ISS 2016 Statistics Paper-2 Solution: Question 7 (Generalized Inverse of a Matrix)
Continuing our question-by-question walk through the ISS Statistics Paper-2 previous year papers, today we pick up right where the last post left off and tackle Question 7 from the 2016 paper. If you have been following along, you already know the drill: read the question carefully, work every step out by hand, and check the final answer against the official key before moving on.
Quick Summary
Topic: Generalized inverse of a matrix (existence and uniqueness properties)
Question reference: ISS 2016, Statistics Paper-2, Question 7
Correct answer: Option (b) — statements 1, 2 and 4 are correct
The Question
Which of the following statements are correct about generalized inverse of a matrix A?
A generalized inverse of a matrix is defined as AA-A = A.
A generalized inverse is not unique.
A matrix to have generalized inverse must be non-singular.
A matrix to have generalized inverse may be either square matrix or rectangular matrix.
Select the correct answer using the code given below:
(a) 1, 2 and 3
(b) 1, 2 and 4
(c) 2, 3 and 4
(d) 1, 3 and 4
Step 1: What Exactly Is a Generalized Inverse?
Before checking each of the four statements, we need a clear working definition. For an ordinary square, non-singular matrix A, the inverse A-1 satisfies AA-1 = A-1A = I. But most matrices you meet in statistics — for example, the X'X matrix in a regression model with collinear columns, or the coefficient matrix of a two-way ANOVA with an over-parameterised model — are singular, or not even square. For such matrices, an ordinary inverse simply does not exist.
To handle this, statisticians use a generalized inverse: given any matrix A of order m x n, a matrix G of order n x m is called a generalized inverse of A if it satisfies
AGA = A.
This single equation is the entire definition. Nothing is said in it about A being square, or about A being non-singular. That one observation is really the key to answering this whole question, so keep it in mind as we go through each statement.
Step 2: Checking Statement 1 — Is AGA = A the Correct Definition?
Statement 1 says: "A generalized inverse of a matrix is defined as AA-A = A," where A- denotes the generalized inverse (often written as G in textbooks to avoid confusing it with an ordinary inverse). This is exactly the standard definition used in every treatment of generalized inverses in linear model theory (Rao, Mitra, and others). So statement 1 is correct.
Step 3: Checking Statement 2 — Is the Generalized Inverse Unique?
This is where a worked numerical example is far more convincing than just quoting a rule, so let's build one from scratch.
Take the simple 2 x 2 matrix
A = [[1, 0], [0, 0]]
Notice this A is singular — its determinant is 1 x 0 - 0 x 0 = 0, so an ordinary inverse does not exist for it. Let's still look for a matrix G = [[g11, g12], [g21, g22]] satisfying AGA = A.
First compute AG:
AG = [[1,0],[0,0]] x [[g11,g12],[g21,g22]] = [[g11, g12],[0, 0]]
Now multiply this by A again to get AGA:
AGA = [[g11, g12],[0, 0]] x [[1,0],[0,0]] = [[g11, 0],[0, 0]]
We need this to equal A = [[1,0],[0,0]]. Comparing entries, the only condition this forces is g11 = 1. The entries g12, g21 and g22 can be absolutely anything — 0, 5, -100, it does not matter, the equation AGA = A is still satisfied.
So both G1 = [[1,0],[0,1]] (the identity matrix) and G2 = [[1,7],[3,-9]] are valid generalized inverses of the same A. Since we found two different matrices that both work, the generalized inverse cannot be unique in general. Statement 2 is correct.
Pro Tip: Whenever a question asks you to judge a claim about generalized inverses, don't just recall the rule from memory — build a 2 x 2 singular example like the one above and test it. Working through one concrete case by hand, the way it's typically drilled into students until the algebra becomes automatic, makes the abstract theorem impossible to forget, and it also happens to double up as your proof for statement 3 below.
Step 4: Checking Statement 3 — Must the Matrix Be Non-Singular?
We already have everything we need for this one. In Step 3, the matrix A = [[1,0],[0,0]] is singular (determinant zero), yet we explicitly found generalized inverses for it. So a matrix does not need to be non-singular to possess a generalized inverse — in fact, generalized inverses are most useful precisely in the singular case, since that is exactly where the ordinary inverse fails to exist. If every matrix we cared about were already non-singular, nobody would have needed to invent the concept of a generalized inverse at all. Statement 3 is false.
Step 5: Checking Statement 4 — Square or Rectangular, Does It Matter?
Now let's push the idea further and check a matrix that is not even square. Take the 2 x 3 matrix
A = [[1, 0, 0], [0, 1, 0]]
We look for a 3 x 2 matrix G such that AGA = A. Try
G = [[1, 0], [0, 1], [0, 0]]
Compute AG first (a 2 x 3 matrix times a 3 x 2 matrix gives a 2 x 2 matrix):
Row 1 of AG: (1)(1) + (0)(0) + (0)(0) = 1, and (1)(0) + (0)(1) + (0)(0) = 0
Row 2 of AG: (0)(1) + (1)(0) + (0)(0) = 0, and (0)(0) + (1)(1) + (0)(0) = 1
So AG = [[1,0],[0,1]], which is simply the 2 x 2 identity matrix. Multiplying by A once more, AGA = I x A = A. So G does satisfy the defining equation, and A here is rectangular — it is not even a square matrix, so terms like "singular" or "non-singular" do not even apply to it, since only square matrices have a determinant in the first place. This confirms that a generalized inverse can exist for rectangular matrices just as well as for square ones. Statement 4 is correct.
Step 6: Combining the Results
Putting the four checks together:
Statement 1 (AGA = A is the definition): correct
Statement 2 (not unique): correct
Statement 3 (must be non-singular): false
Statement 4 (square or rectangular): correct
So the correct statements are 1, 2 and 4, which is exactly option (b). Checking the official answer key for this paper confirms the key also lists B for Question 7, so our derivation matches the official answer exactly.
Why This Question Matters
Generalized inverses are not just an abstract algebra curiosity in the ISS syllabus — they sit underneath a large part of linear model theory. Whenever a design matrix is not of full column rank (which happens routinely in ANOVA models written with all the redundant treatment effects included, or in any regression with exactly collinear regressors), the normal equations X'Xb = X'y cannot be solved using an ordinary inverse of X'X, because X'X is singular. Instead, estimators like BLUEs of estimable functions are constructed using a generalized inverse of X'X. Students who get comfortable constructing small examples like the ones above, rather than only memorizing the properties as a list, tend to find questions on estimability, BLUEs and the Gauss-Markov model in the rest of the paper noticeably easier, since they all lean on exactly this same idea.
Frequently Asked Questions
What is the difference between a generalized inverse and the ordinary matrix inverse?
The ordinary inverse A-1 exists only for square, non-singular matrices and satisfies both AA-1 = I and A-1A = I. A generalized inverse G only needs to satisfy the single, weaker condition AGA = A, and it exists for every matrix — square or rectangular, singular or not.
Why isn't the generalized inverse unique?
Because the defining equation AGA = A is a weaker condition than requiring AG = GA = I. As the worked example in Step 3 shows, once you fix the entries of G that are forced by the equation, any remaining entries are free to take any value, so infinitely many matrices can satisfy the same equation.
Does every matrix have a generalized inverse?
Yes. Every matrix of every shape and every rank has at least one generalized inverse. This is precisely what makes the concept so useful in statistics, where singular and rectangular matrices show up constantly.
Is the Moore-Penrose pseudo-inverse the same as a generalized inverse?
The Moore-Penrose pseudo-inverse is one specific generalized inverse, singled out because it additionally satisfies three more conditions beyond AGA = A (it also satisfies GAG = G, and makes AG and GA symmetric). Because of those extra conditions, it is the one generalized inverse that turns out to be unique for any given A, but it is only one member of the much larger family of matrices that satisfy AGA = A.
Where does this topic come up again later in the ISS syllabus?
It reappears heavily in the theory of linear estimation — estimability of linear parametric functions, the Gauss-Markov theorem, and the analysis of variance for less-than-full-rank models. Any time you see a claim about "a" or "any" generalized inverse of X'X, you are working with exactly the idea tested in this question.
How should I revise this topic before the exam?
Rather than memorizing the list of properties, work through two or three small numerical examples by hand — one singular square matrix and one rectangular matrix, exactly like the ones solved above. Once you have constructed a generalized inverse yourself even once, statements about existence, uniqueness and shape stop being facts to recall and become things you can re-derive on the spot.
If anything in this derivation felt unclear, or if you would solve one of the steps differently, drop a comment below — doubts like these are exactly what this series is meant to help with. And if you know a fellow ISS aspirant working through the same papers, feel free to share this post with them.
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