Q.If A is symmetric matrix, then B′AB is _________.
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Symmetric and Skew-Symmetric Matrices
These are two special kinds of square matrices, defined by how a matrix compares with its own transpose A′ (the matrix with rows and columns swapped). They are among the most-tested ideas in the Matrices chapter.
Symmetric matrix
A square matrix A is symmetric if it equals its transpose:
A′=A,that isaij=aji for all i,j.
Entries are mirror images across the main diagonal. For example,
A=147425753,a12=a21=4, a13=a31=7.
Skew-symmetric matrix
A square matrix A is skew-symmetric if its transpose is its negative:
A′=−A,that isaij=−aji for all i,j.
Putting i=j gives aii=−aii, so 2aii=0 — every diagonal entry of a skew-symmetric matrix is 0. For example,
B=0−3230−5−250,bij=−bji.
Both definitions demand a square matrix — the condition aij=±aji only makes sense when both entries exist.
Key facts
- For any square matrix A, the matrix A+A′ is always symmetric and A−A′ is always skew-symmetric. (Check: (A+A′)′=A′+A=A+A′.)
- If A is skew-symmetric of odd order, then detA=0. …
Test whether B′AB equals its own transpose. Using the reversal law (XYZ)′=Z′Y′X′ and (B′)′=B:
(B′AB)′=B′A′(B′)′=B′A′B. …
Transposing B′AB gives back B′AB (because A′=A), so B′AB is symmetric.
What to check
A matrix M is symmetric when M′=M. So set M=B′AB and compute its transpose.
Apply the reversal law
The transpose of a product reverses the factor order: (XYZ)′=Z′Y′X′. With X=B′, Y=A, Z=B,
(B′AB)′=B′A′(B′)′.
Now simplify each piece:
- (B′)′=B (transposing twice returns the original), so the last factor becomes B;
- A′=A, because A is symmetric (given).
Hence
(B′AB)′=B′AB=B′AB.
Conclusion …
Method: Test the symmetry of a triple product B′AB
Use this for congruence-type expressions where a matrix is sandwiched as B′AB.
Steps
Step 1: Set M=B′AB and transpose it.
Symmetry is decided by whether M′=M.
Step 2: Apply the reversal law to all three factors.
(B′AB)′=B′A′(B′)′=B′A′B.
Note (B′)′=B. …
Common Mistakes
Mistake 1: Believing B must be symmetric (or square) for B′AB to be symmetric.
Why it's wrong: the result holds for any B making the product defined; only A's symmetry is used. Correct approach: transpose and rely on A′=A alone.
Mistake 2: Mishandling the triple transpose order. …
- COMEDK 2021Set 2021-B1 markMCQQ.If A=01−2−10−3230, then A+2A′= (A) −A′ (B) −2A′ (C) A′ (D) 0
›Reveal solutionSolution
A+2A′=A′.
The matrix A=01−2−10−3230 is skew-symmetric: its transpose A′=0−12103−2−30=−A. …
- COMEDK 2025Set 2025-A1 markMCQQ.If A=01−2−10−3230, then A+2AT= (A) −AT (B) AT (C) 2A2 (D) A
›Reveal solutionSolution
The matrix A is skew-symmetric (AT=−A), so A+2AT=A−2A=−A=AT. The correct option is (B).
We start by noticing the structure of A. It has zeros on the diagonal, and the off-diagonal entries appear in opposite pairs with opposite signs:
- A12=−1 and A21=1
- A13=2 and A31=−2
- A23=3 and A32=−3
This is the classic pattern of a skew-symmetric matrix, where AT=−A. Let’s verify quickly:
AT=0−12103−2−30=−A.
Yes, every entry flips sign.
Now the expression is A+2AT. Using AT=−A, we substitute:
A+2(−A)=A−2A=−A.
But −A is exactly AT (since AT=−A). So the result is AT.
Let’s check the options:
- (A) −AT would be −(−A)=A, not our result.
- (B) AT matches.
- (C) 2A2 is something else entirely (and A2 is symmetric, not equal to −A).
- (D) A is not equal to −A unless A=0.
Thus the answer is clear. …
- COMEDK 2026Set 2026-A1 markMCQQ.If the matrix M=x+5−2ca06−4by+1 is a skew symmetric matrix, the value of the expression ab+c2−xy is: (A) −33 (B) −9 (C) 0 (D) −1
›Reveal solutionSolution
The skew-symmetric condition gives a=2, b=−6, c=4, x=−5, y=−1, so ab+c2−xy=−12+16−5=−1 — option (D).
Concept
A matrix M is skew-symmetric when MT=−M. Entrywise this forces
mii=0(diagonal),mji=−mij(off-diagonal).
Applying these to the given entries determines all unknowns.
Solution
- Diagonal entries vanish:
x+5=0⇒x=−5,y+1=0⇒y=−1.
- Off-diagonal relations (with M=x+5−2ca06−4by+1):
m21=−m12: −2=−a⇒a=2,
m31=−m13: c=−(−4)=4,
m32=−m23: 6=−b⇒b=−6.
- Evaluate the expression:
ab=(2)(−6)=−12,c2=16,xy=(−5)(−1)=5,
ab+c2−xy=−12+16−5=−1. …
- KCET 2026Set UNKNOWN1 markMCQQ.Match List-I with List-II List-Ia) A matrix which is not a square matrixb) A square matrix A′=Ac) The diagonal elements of a diagonal matrix are samed) A matrix which is both symmetric and skew symmetric List-IIi) Symmetric matrixii) Null matrixiii) Rectangular matrixiv) Scalar matrix Codes: (A) a - iii, b - i, c - iv, d - ii (B) a - iii, b - ii, c - iv, d - i (C) a - i, b - ii, c - iv, d - iii (D) a - iii, b - iv, c - i, d - ii
›Reveal solutionSolution
Match each List-I description to the standard matrix-type definition in List-II: rectangular, symmetric, scalar, and null matrices.
Step 1 — Match item (a)
"A matrix which is not a square matrix" is, by definition, a rectangular matrix (number of rows = number of columns). So a→iii.
Step 2 — Match item (b)
"A square matrix A′=A" (the transpose equals the matrix itself) is exactly the definition of a symmetric matrix. So b→i.
Step 3 — Match item (c)
"The diagonal elements of a diagonal matrix are same" — a diagonal matrix whose diagonal entries are all equal is called a scalar matrix. So c→iv.
Step 4 — Match item (d) …
- KCET 2026Set UNKNOWN1 markMCQQ.Consider the following statements: Statement I : If A is a non-singular matrix, then A−1 exists. Statement II : If A and B are symmetric matrices of same order, then (AB−BA) is a skew symmetric matrix. Choose the correct option. (A) Statement I is true and Statement II is false (B) Statement I is false and Statement II is false (C) Statement I is true and Statement II is true (D) Statement I is false and Statement II is true
›Reveal solutionSolution
Statement I is a direct restatement of the invertibility theorem; Statement II follows from checking that (AB−BA)T=−(AB−BA) for symmetric A,B.
Step 1 — Statement I: existence of A−1
By the fundamental theorem on matrix inverses, a square matrix A is invertible (i.e. A−1 exists) if and only if A is non-singular, i.e. ∣A∣=0. Since A is given as non-singular, A−1 exists. Statement I is true.
Step 2 — Statement II: is AB−BA skew symmetric?
Since A and B are symmetric, AT=A and BT=B. Consider …
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