Q.If 12323a111 is a non-singular matrix and a∈A, then the set A is:
(a) R
(b) {0}
(c) {4}
(d) R−{4}
CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★
✓ Free question
Concept understanding — Determinant Range Analysis
Determinant Range Analysis
The determinant of a square matrix is a single number that tells you a lot: whether the matrix is invertible, how it scales area or volume, and so on. Now suppose the entries are not fixed but are each allowed to move between two limits. Then the determinant is not free to be anything either — it lives inside a definite range.
Determinant range analysis answers exactly this: given bounds on the entries, what are the smallest and largest possible values of the determinant?
Note
Why it matters: if the determinant can never reach 0, the matrix stays invertible no matter how the entries wobble. If the range includes0, the matrix could become singular — a warning sign for the system it describes.
The intuition with a 2×2 example
Take A=(acbd), so det(A)=ad−bc, and suppose every entry lies in [0,1].
Largest value: make ad big and bc small — set a=d=1, b=c=0, giving det=1.
Smallest value: make ad small and bc big — set a=d=0, b=c=1, giving det=−1.
So the range is [−1,1]. Since 0 is inside it, some matrices in this family are singular.
The key fact: extremes sit at the corners
The determinant is linear in each entry when the others are held fixed (this is the row/column-linearity of determinants). A linear function on an interval always attains its largest and smallest values at the endpoints. So to find the range you push each variable entry to one of its two bounds and compare — you never need an interior value.
Watch out
Do not just plug each entry's extreme into the formula independently and assume that gives the range. The terms of a determinant share entries, so those choices interact — always track which combination is actually achievable.
A worked, exam-style problem
Problem. Let A=(x11y) with 0≤x≤2 and 0≤y≤2. Find the range of det(A).
Solution.det(A)=xy−1. Here x and y are independent and non-negative, so xy ranges from 0 to 4. Hence xy−1 ranges from −1 to 3.
det(A)∈[−1,3].
Tip
When only a couple of entries vary, treat the determinant as a plain function of those variables and find its minimum and maximum over the given intervals — usually faster than checking every corner.
That is the whole idea: the determinant is a window into invertibility and scaling, and its range tells you whether the matrix is guaranteed well-behaved or might slip to singular.
Finding the range of a determinant whose entries vary within given bounds is a favourite JEE Main and JEE Advanced problem type, building on the NCERT Class 12 Determinants syllabus rather than appearing as a named chapter topic itself. Students searching 'determinant range problems JEE' or 'maximum value of determinant class 12' will recognize this corner-point, linear-in-each-entry reasoning as the standard approach.
Concept: Non-singular matrix means determinant =0.
Step 1: Compute the determinant.
Δ=12323a111
Using R2→R2−2R1 and R3→R3−3R1:
Δ=1002−1a−61−1−2
Step 2: Expand along first column.
Δ=1⋅−1a−6−1−2=(−1)(−2)−(−1)(a−6)=2+(a−6)=a−4
Step 3: For non-singular, Δ=0⟹a−4=0⟹a=4.
Thus A=R−{4}.
✓Final answer
The set A is R−{4}, which corresponds to option (d).
A matrix is non-singular when its determinant is non-zero. For this 3×3 matrix, the determinant simplifies to a−4, so the matrix is non-singular for all a=4. Thus A=R−{4}.
The key idea here is simple: a square matrix is called non-singular (or invertible) exactly when its determinant is not zero. If the determinant is zero, the matrix is singular — it has no inverse, and its rows (or columns) are linearly dependent.
So the problem reduces to: find all real numbers a for which the determinant of the given matrix is non-zero. Then the set A is precisely that collection of a values.
Let’s compute the determinant.
Write the matrix:
M=12323a111
Compute det(M) using expansion along the first row (or any row/column). I’ll expand along the first row:
det(M)=1⋅3a11−2⋅2311+1⋅233a
Evaluate each 2×2 determinant:
3a11=(3)(1)−(1)(a)=3−a
2311=(2)(1)−(1)(3)=2−3=−1
233a=(2)(a)−(3)(3)=2a−9
Substitute back:
det(M)=1⋅(3−a)−2⋅(−1)+1⋅(2a−9)
Simplify:
det(M)=3−a+2+2a−9
Combine like terms:
det(M)=(3+2−9)+(−a+2a)=(−4)+(a)=a−4
Tip
Notice how the first and third columns are identical except for the middle entry? That’s a quick check: if a=4, the second and third rows become proportional? Actually no — but the determinant simplifies so cleanly that you could also spot that the matrix has a pattern: subtracting row 1 from row 2 and row 3 gives a simpler form. Try it: R2→R2−2R1, R3→R3−3R1 yields 1002−1a−61−1−2, and the determinant is 1⋅−1a−6−1−2=(−1)(−2)−(−1)(a−6)=2+a−6=a−4. Same result, faster.
The matrix is non-singular when det(M)=0, i.e.:
a−4=0⇒a=4
Since a is a real number (the problem implies a∈R from the options), the set of all such a is all real numbers except 4:
A=R−{4}
Watch out
A common mistake is to think "non-singular" means the determinant is positive, or that it must be something other than zero in some special sense. No — it simply means not zero. Also, don’t confuse this with the matrix being "non-invertible" — that’s the same as singular. So a=4 is the only value that makes the matrix singular.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2024Set 65/1/11 mark
Q.Assertion (A): For the matrix A=1−cosθ−1cosθ1−cosθ1cosθ1, where θ∈[0,2π], ∣A∣∈[2,4]. Reason (R): cosθ∈[−1,1],∀θ∈[0,2π]. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
›Reveal solutionSolution
We calculate the determinant ∣A∣ as 2+2cos2θ. Using the fundamental range of cosθ (given in Reason (R)), we find that ∣A∣ lies in [2,4]. Both Assertion (A) and Reason (R) are true, and (R) correctly explains (A).
The problem asks us to evaluate an Assertion-Reason pair. This requires us to perform three distinct checks:
Determine if Assertion (A) is true.
Determine if Reason (R) is true.
If both are true, determine if Reason (R) provides a correct explanation for Assertion (A).
The core concept here involves calculating the determinant of a 3×3 matrix and then finding the range of a trigonometric expression. The intuition is that once we express the determinant in terms of cosθ, the known bounds of cosθ (which is what Reason (R) states) will directly help us find the bounds of the determinant.
Evaluate Reason (R):
Reason (R) states: cosθ∈[−1,1],∀θ∈[0,2π].
This is a fundamental property of the cosine function. For any real value of θ, the cosine function's output is always between −1 and 1, inclusive. The interval θ∈[0,2π] covers all possible values of cosθ.
Therefore, Reason (R) is true.
Calculate the Determinant of Matrix A:
The given matrix is A=1−cosθ−1cosθ1−cosθ1cosθ1.
We calculate the determinant ∣A∣ using cofactor expansion along the first row:
Determine the Range of ∣A∣ using Reason (R):
We have found that ∣A∣=2+2cos2θ.
From Reason (R), we know that cosθ∈[−1,1].
To find the range of cos2θ:
Since cosθ can be any value between −1 and 1, its square, cos2θ, will be non-negative.
The minimum value of cos2θ occurs when cosθ=0, giving cos2θ=0.
The maximum value of cos2θ occurs when cosθ=1 or cosθ=−1, giving cos2θ=1.
Thus, the range of cos2θ is [0,1].
0≤cos2θ≤1
Now, we build up the expression for $|A|$:
Multiply by 2:
2⋅0≤2cos2θ≤2⋅1
0≤2cos2θ≤2
Add 2 to all parts of the inequality:
2+0≤2+2cos2θ≤2+2
2≤∣A∣≤4
This means that $|A| \in [2, 4]$.
4. Evaluate Assertion (A):
Assertion (A) states: For the given matrix A, ∣A∣∈[2,4].
From our calculation in step 3, we found that ∣A∣ indeed lies in the interval [2,4].
Therefore, Assertion (A) is true.
Determine if Reason (R) is the Correct Explanation for Assertion (A):
We used the fact that cosθ∈[−1,1] (which is Reason (R)) to determine the range of cos2θ, which was then directly used to establish the range of ∣A∣. Without the knowledge of the range of cosθ, we could not have determined the range of ∣A∣. Thus, Reason (R) is directly instrumental in proving Assertion (A).
Therefore, Reason (R) is the correct explanation of Assertion (A).
✓Final answer
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A), so the correct option is (A).
CBSE 2023Set 65/2/11 markMCQ
Q.If 12323a111 is a non-singular matrix and a∈A, then the set A is:
(a) R
(b) {0}
(c) {4}
(d) R−{4}
›Reveal solutionSolution
A matrix is non-singular when its determinant is non-zero. For this 3×3 matrix, the determinant simplifies to a−4, so the matrix is non-singular for all a=4. Thus A=R−{4}.
The key idea here is simple: a square matrix is called non-singular (or invertible) exactly when its determinant is not zero. If the determinant is zero, the matrix is singular — it has no inverse, and its rows (or columns) are linearly dependent.
So the problem reduces to: find all real numbers a for which the determinant of the given matrix is non-zero. Then the set A is precisely that collection of a values.
Let’s compute the determinant.
Write the matrix:
M=12323a111
Compute det(M) using expansion along the first row (or any row/column). I’ll expand along the first row:
det(M)=1⋅3a11−2⋅2311+1⋅233a
Evaluate each 2×2 determinant:
3a11=(3)(1)−(1)(a)=3−a
2311=(2)(1)−(1)(3)=2−3=−1
233a=(2)(a)−(3)(3)=2a−9
Substitute back:
det(M)=1⋅(3−a)−2⋅(−1)+1⋅(2a−9)
Simplify:
det(M)=3−a+2+2a−9
Combine like terms:
det(M)=(3+2−9)+(−a+2a)=(−4)+(a)=a−4
Tip
Notice how the first and third columns are identical except for the middle entry? That’s a quick check: if a=4, the second and third rows become proportional? Actually no — but the determinant simplifies so cleanly that you could also spot that the matrix has a pattern: subtracting row 1 from row 2 and row 3 gives a simpler form. Try it: R2→R2−2R1, R3→R3−3R1 yields 1002−1a−61−1−2, and the determinant is 1⋅−1a−6−1−2=(−1)(−2)−(−1)(a−6)=2+a−6=a−4. Same result, faster.
The matrix is non-singular when det(M)=0, i.e.:
a−4=0⇒a=4
Since a is a real number (the problem implies a∈R from the options), the set of all such a is all real numbers except 4:
A=R−{4}
Watch out
A common mistake is to think "non-singular" means the determinant is positive, or that it must be something other than zero in some special sense. No — it simply means not zero. Also, don’t confuse this with the matrix being "non-invertible" — that’s the same as singular. So a=4 is the only value that makes the matrix singular.