Q.Find A2−5A+6I, if A=22101−1130.
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Matrix Polynomial Evaluation
You know how to evaluate a polynomial like p(x)=2x2−3x+5 at a number: plug in x, get a number out. Now plug in a square matrix A instead. The variable becomes A, and — crucially — the constant term becomes a multiple of the identity matrix I, because you cannot add a bare number to a matrix.
The Definition
For p(x)=anxn+⋯+a1x+a0 and a square matrix A,
p(A)=anAn+an−1An−1+⋯+a1A+a0I.
Here Ak is k-fold matrix multiplication, akAk is scalar multiplication, and a0I replaces the constant. The result is a square matrix of the same size as A.
There is no ambiguity from non-commutativity here: a polynomial only ever multiplies A by itself, and A always commutes with A.
A Worked Example
Let p(x)=x2−4x+3 and A=(2013).
A2=(4059),−4A=(−80−4−12),3I=(3003).
Adding term by term,
p(A)=(−1010).
A Shortcut for Diagonal Matrices
If A=(λ100λ2), then Ak=(λ1k00λ2k), so
p(A)=(p(λ1)00p(λ2)).
You simply evaluate p at each diagonal entry. …
Concept: Matrix Polynomial Evaluation — substitute the matrix into the polynomial, compute powers and scalar multiples, then combine.
First, compute A2:
A2=A⋅A=22101−113022101−1130=4+0+14+2+32−2+00+0−10+1−30−1+02+0+02+3+01−3+0=590−1−2−125−2
Now compute 5A and 6I:
5A=1010505−55150,6I=600060006
Finally, combine: …
Compute the matrix polynomial by finding A2, then combining with −5A and 6I entry-by-entry. The result is A2−5A+6I=1−1−5−1−14−3−104.
Here A acts like a variable, but with matrix multiplication and addition. The constant term 6I is 6 times the 3×3 identity, which keeps every term a 3×3 matrix.
Step 1: Compute A2=A⋅A
A=22101−1130
Each entry (i,j) is row i of A dotted with column j of A:
- (1,1)=2⋅2+0⋅2+1⋅1=5
- (1,2)=2⋅0+0⋅1+1⋅(−1)=−1
- (1,3)=2⋅1+0⋅3+1⋅0=2
- (2,1)=2⋅2+1⋅2+3⋅1=9
- (2,2)=2⋅0+1⋅1+3⋅(−1)=−2
- (2,3)=2⋅1+1⋅3+3⋅0=5
- (3,1)=1⋅2+(−1)⋅2+0⋅1=0
- (3,2)=1⋅0+(−1)⋅1+0⋅(−1)=−1
- (3,3)=1⋅1+(−1)⋅3+0⋅0=−2
A2=590−1−2−125−2
Step 2: Compute −5A and 6I
−5A=−10−10−50−55−5−150,6I=600060006 …
Method: Evaluating a matrix polynomial p(A)
To compute an expression like A2−5A+6I, treat A as the variable but replace the constant term with that constant times the identity matrix, then add the terms entry-wise.
Steps
Step 1: Compute the required powers of A
Find A2=A⋅A by matrix multiplication (not by squaring individual entries), and any higher powers needed.
Step 2: Form each term to the same order …
Common Mistakes
Mistake 1: Squaring entries instead of multiplying the matrix by itself
Why it's wrong: A2 means A⋅A via the row-by-column rule, not squaring each entry; the two give completely different matrices. Correct approach: perform the full matrix multiplication.
Mistake 2: Adding the bare constant 6 instead of 6I …
- AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQQ.Let A=[23−51] what is f(A)=? where f(x)=x3−2x2−5. (A) [−50427036] (B) [−504270−36] (C) [−50−4270−36] (D) [−50−427036]
›Reveal solutionSolution
Cayley–Hamilton reduces f(A) to −14A−22I, giving [−50−4270−36].
Concept and Intuition
Rather than computing A3 by brute-force matrix multiplication three times, use the Cayley–Hamilton theorem: every square matrix satisfies its own characteristic equation, letting us reduce any polynomial in A to a linear expression αA+βI.
Step-by-Step Solution
- A=[23−51]: trace =3, det=2(1)−(−5)(3)=2+15=17.
- Characteristic equation: A2−3A+17I=0⇒A2=3A−17I.
- A3=A⋅A2=A(3A−17I)=3A2−17A=3(3A−17I)−17A=9A−51I−17A=−8A−51I.
- f(A)=A3−2A2−5I=(−8A−51I)−2(3A−17I)−5I=−8A−51I−6A+34I−5I=−14A−22I. …
- AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQQ.If P, Q and R are 3×3 matrices such that 3x2+x+35x2+3x+23x2+2x+52x2−x+44x2−2x−14x2−x−27x2+8x+57x2+5x+83x2+8x+7=Px2+Qx+R, then detR= (A) 0 (B) 136 (C) 48 (D) −72
›Reveal solutionSolution
R is just the constant-term matrix extracted from each quadratic-in-x entry; expanding its determinant gives 136.
Concept and Intuition
Since Px2+Qx+R must equal the given matrix of quadratics for every x (matching coefficients entry-by-entry), R is simply the matrix formed by the x0 (constant) coefficient of every entry — no matrix algebra with P,Q is even needed to answer this.
Step-by-Step Solution
- Read off constant terms entry-by-entry: Row 1: 3,4,5; Row 2: 2,−1,8; Row 3: 5,−2,7. So R=3254−1−2587.
- Expand along row 1: detR=3[(−1)(7)−(8)(−2)]−4[(2)(7)−(8)(5)]+5[(2)(−2)−(−1)(5)]. …
- AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQQ.If A=123012234, then A2−5A+6I= (A) 8344800412 (B) 8344600414 (C) 8326800414 (D) 8344800414
›Reveal solutionSolution
Straightforward matrix arithmetic: compute A2, then combine A2−5A+6I entrywise.
Concept and Intuition
This is a direct test of matrix multiplication and linear combination of matrices — no shortcut (like Cayley–Hamilton) is needed since we're not told A's characteristic polynomial matches this exact expression; computing directly is safest and fastest here.
Step-by-Step Solution
- A=123012234.
- Compute A2 row by row:
- Row 1: (1⋅1+0⋅2+2⋅3, 1⋅0+0⋅1+2⋅2, 1⋅2+0⋅3+2⋅4)=(7,4,10)
- Row 2: (2⋅1+1⋅2+3⋅3, 2⋅0+1⋅1+3⋅2, 2⋅2+1⋅3+3⋅4)=(13,7,19)
- Row 3: (3⋅1+2⋅2+4⋅3, 3⋅0+2⋅1+4⋅2, 3⋅2+2⋅3+4⋅4)=(19,10,28) So A2=713194710101928.
- 5A=510150510101520, and 6I=600060006.
- Combine entrywise: A2−5A+6I: …
- AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQQ.If A=(i00−i), B=(01−10) and C=(0ii0), then (A) A2+B2+C2=3A2B2C2 (B) A2+B2+C2=3ABC (C) A2+B2+C2=3I (D) A2+B2+C2=2ABC
›Reveal solutionSolution
Each of A,B,C squares to −I (they behave like quaternion-style imaginary units), and their product ABC=I. Matching A2+B2+C2=−3I against the options shows it equals 3A2B2C2.
Concept and Intuition
These three matrices are 2×2 representations reminiscent of Pauli-matrix-like objects, each satisfying M2=−I. Rather than manipulate symbols, direct matrix multiplication settles the identity quickly and unambiguously.
Step-by-Step Solution
- A2=(i00−i)2=(i200(−i)2)=(−100−1)=−I.
- B2=(01−10)2=(−100−1)=−I (a 90∘ rotation squared is a 180∘ rotation, i.e. −I).
- C2=(0ii0)2=(i⋅i00i⋅i)=(−100−1)=−I.
- So A2+B2+C2=−I−I−I=−3I. …
- AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQQ.If A=112173397 then Tr(A2−A)= (A) 0 (B) -12 (C) 152 (D) 125
›Reveal solutionSolution
Using the linearity of trace, Tr(A2−A)=Tr(A2)−Tr(A), computed directly from A's diagonal entries and the diagonal entries of A2 (each needing only one row-column dot product).
Concept and Intuition
Trace is linear: Tr(M+N)=Tr(M)+Tr(N) for matrices of the same size, so Tr(A2−A)=Tr(A2)−Tr(A). This avoids computing the full matrix A2 — we only need its three diagonal entries, each of which is just the dot product of the corresponding row of A with the corresponding column of A.
Step-by-Step Solution
- A=112173397. Tr(A)=1+7+7=15.
- Columns of A: column 1 =(1,1,2), column 2 =(1,7,3), column 3 =(3,9,7).
- (A2)11= row 1 ⋅ column 1 =1(1)+1(1)+3(2)=1+1+6=8.
- (A2)22= row 2 ⋅ column 2 =1(1)+7(7)+9(3)=1+49+27=77.
- (A2)33= row 3 ⋅ column 3 =2(3)+3(9)+7(7)=6+27+49=82.
- Tr(A2)=8+77+82=167. …
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