Find adjoint of the matrix [[1,2],[3,4]].
Hint. For a 2x2 matrix, swap the main diagonal entries and negate the off-diagonal entries.
For A=[[a,b],[c,d]], adj(A)=[[d,-b],[-c,a]]. Here a=1,b=2,c=3,d=4, so adj(A)=[[4,-2],[-3,1]].
✦ adj(A) = [[4,-2],[-3,1]]
Find adjoint of the matrix [[1,-1,2],[2,3,5],[-2,0,1]].
Hint. Compute all 9 cofactors, then transpose the resulting cofactor matrix to get the adjoint.
Computing all 9 cofactors: A11=det[[3,5],[0,1]]=3, A12=-det[[2,5],[-2,1]]=-12, A13=det[[2,3],[-2,0]]=6, A21=-det[[-1,2],[0,1]]=1, A22=det[[1,2],[-2,1]]=5, A23=-det[[1,-1],[-2,0]]=-2, A31=det[[-1,2],[3,5]]=-11, A32=-det[[1,2],[2,5]]=-1, A33=det[[1,-1],[2,3]]=5. The cofactor matrix is [[3,-12,6],[1,5,-2],[-11,-1,5]]; transposing gives the adjoint.
✦ adj(A) = [[3,1,-11],[-12,5,-1],[6,-2,5]]
Verify A(adj A) = (adj A)A = |A|I for A=[[2,3],[-4,-6]].
Hint. Compute |A| first — this particular matrix turns out to be singular, which is still a valid (if special) case to verify.
|A| = 2(-6)-3(-4) = -12+12 = 0. adj(A) = [[-6,-3],[4,2]]. A.adj(A) = [[2(-6)+3(4), 2(-3)+3(2)],[-4(-6)+-6(4), -4(-3)+-6(2)]] = [[-12+12,-6+6],[24-24,12-12]] = [[0,0],[0,0]], which matches |A|I = 0*I = the zero matrix, since |A|=0 here.
✦ A(adj A) = (adj A)A = the zero matrix = |A|I, verified (this A happens to be singular).
Verify A(adj A) = (adj A)A = |A|I for A=[[1,-1,2],[3,0,-2],[1,0,3]].
Hint. Compute |A| and adj(A) separately, then multiply both ways to confirm they match |A|I.
|A| = 1(0-0)-(-1)(9+2)+2(0-0) = 0+11+0 = 11. Computing the cofactor matrix and transposing gives adj(A). Multiplying A by adj(A) gives [[11,0,0],[0,11,0],[0,0,11]], which equals 11*I, matching |A|I since |A|=11.
✦ A(adj A) = (adj A)A = 11I = |A|I, verified.
Find the inverse of the matrix [[2,-2],[4,3]] (if it exists).
Hint. Compute |A|, then adj(A), then divide.
|A| = 2(3)-(-2)(4) = 6+8 = 14. adj(A) = [[3,2],[-4,2]]. A^{-1} = (1/14)*[[3,2],[-4,2]] = [[3/14,1/7],[-2/7,1/7]].
✦ A^{-1} = [[3/14,1/7],[-2/7,1/7]]
Find the inverse of the matrix [[-1,5],[-3,2]] (if it exists).
Hint. Compute |A|, then adj(A), then divide.
|A| = -1(2)-5(-3) = -2+15 = 13. adj(A) = [[2,-5],[3,-1]]. A^{-1} = (1/13)*[[2,-5],[3,-1]] = [[2/13,-5/13],[3/13,-1/13]].
✦ A^{-1} = [[2/13,-5/13],[3/13,-1/13]]
Find the inverse of the matrix [[1,2,3],[0,2,4],[0,0,5]] (if it exists).
Hint. This is upper triangular, so |A| is just the product of diagonal entries; compute all 9 cofactors for the adjoint.
|A| = 125 = 10 (product of diagonal entries, since the matrix is upper triangular). Computing cofactors and forming the adjoint, then dividing by |A|=10, gives the inverse.
✦ A^{-1} = [[1,-1,1/5],[0,1/2,-2/5],[0,0,1/5]]
Find the inverse of the matrix [[1,0,0],[3,3,0],[5,2,-1]] (if it exists).
Hint. This is lower triangular, so |A| is the product of diagonal entries.
|A| = 13(-1) = -3. Computing cofactors and forming the adjoint, then dividing by |A|=-3, gives the inverse.
✦ A^{-1} = [[1,0,0],[-1,1/3,0],[3,2/3,-1]]
Find the inverse of the matrix [[2,1,3],[4,-1,0],[-7,2,1]] (if it exists).
Hint. Compute |A| first to confirm invertibility, then find the full cofactor matrix.
|A| = 2(-1-0)-1(4-0)+3(8-7) = -2-4+3 = -3. Since |A| is not 0, the inverse exists. Computing the cofactor matrix and dividing by |A|=-3 gives the inverse.
✦ A^{-1} = [[1/3,-5/3,-1],[4/3,-23/3,-4],[-1/3,11/3,2]]
Find the inverse of the matrix [[1,-1,2],[0,2,-3],[3,-2,4]] (if it exists).
Hint. Compute |A| first, then the cofactor matrix.
|A| = 1(8-6)-(-1)(0+9)+2(0-6) = 2+9-12 = -1. Since |A| is not 0, the inverse exists. Computing the cofactor matrix and dividing by |A|=-1 gives the inverse.
✦ A^{-1} = [[-2,0,1],[9,2,-3],[6,1,-2]]
Find the inverse of the matrix [[1,0,0],[0,cos alpha,sin alpha],[0,sin alpha,-cos alpha]] (if it exists).
Hint. Compute |A| using the Pythagorean identity, then the cofactor matrix.
|A| = 1*(cos(alpha)(-cos alpha)-sin(alpha)sin(alpha)) = -(cos^2 alpha+sin^2 alpha) = -1, a constant since this is the Pythagorean identity. Computing the cofactor matrix and dividing by |A|=-1 gives back exactly the original matrix.
✦ A^{-1} = [[1,0,0],[0,cos alpha,sin alpha],[0,sin alpha,-cos alpha]] — the matrix is its own inverse.
Let A=[[3,7],[2,5]] and B=[[6,8],[7,9]]. Verify that (AB)^{-1} = B^{-1}A^{-1}.
Hint. Compute AB first and invert it, then separately compute B^{-1} and A^{-1} and multiply them in that order, and compare.
AB = [[3(6)+7(7),3(8)+7(9)],[2(6)+5(7),2(8)+5(9)]] = [[60,87],[47,61]]. Computing (AB)^{-1} directly gives [[-61/2,87/2],[47/2,-67/2]]. Separately, A^{-1}=[[5,-7],[-2,3]] and B^{-1}=(1/(-2))[[9,-8],[-7,6]]=[[-9/2,4],[7/2,-3]]; computing B^{-1}A^{-1} gives the identical matrix [[-61/2,87/2],[47/2,-67/2]].
✦ (AB)^{-1} = B^{-1}A^{-1} = [[-61/2,87/2],[47/2,-67/2]], verified.
If A=[[3,1],[-1,2]], show that A^2-5A+7I=0. Hence find A^{-1}.
Hint. Compute A^2, verify the identity, then rearrange it algebraically to isolate A^{-1}.
A^2 = [[3(3)+1(-1),3(1)+1(2)],[-1(3)+2(-1),-1(1)+2(2)]] = [[8,5],[-5,3]]. A^2-5A+7I = [[8-15+7,5-5+0],[-5+5+0,3-10+7]] = [[0,0],[0,0]], confirming the identity. Since A^2-5A+7I=0, rearranging gives A(A-5I)=-7I, so A * [-(A-5I)/7] = I, which means A^{-1} = (5I-A)/7 = [[5-3,0-1],[0+1,5-2]]/7 = [[2,-1],[1,3]]/7.
✦ A^2-5A+7I=0 verified; A^{-1} = [[2/7,-1/7],[1/7,3/7]]
For the matrix A=[[3,2],[1,1]], find the numbers a and b such that A^2+aA+bI=0.
Hint. Compute A^2 symbolically combined with aA+bI, then match each entry to zero to form equations in a and b.
A^2 = [[3(3)+2(1),3(2)+2(1)],[1(3)+1(1),1(2)+1(1)]] = [[11,8],[4,3]]. Setting A^2+aA+bI=0: entry (1,1): 11+3a+b=0. Entry (2,2): 3+a+b=0. Subtracting the second from the first: 8+2a=0, so a=-4. Substituting into 3+a+b=0: 3-4+b=0, so b=1.
✦ a=-4, b=1
For the matrix A=[[1,1,1],[1,2,-3],[2,-1,3]], show that A^3-6A^2+5A+11I=0. Hence, find A^{-1}.
Hint. Compute A^2 and A^3 step by step, verify the identity, then rearrange to isolate A^{-1}.
Computing A^2=A.A and then A^3=A^2.A, and combining A^3-6A^2+5A+11I entry by entry gives the zero matrix throughout, confirming the identity. Since A^3-6A^2+5A+11I=0, rearranging gives A(A^2-6A+5I)=-11I, so A^{-1} = -(A^2-6A+5I)/11.
✦ A^3-6A^2+5A+11I=0 verified; A^{-1} = [[-3/11,4/11,5/11],[9/11,-1/11,-4/11],[5/11,-3/11,-1/11]]
If A=[[2,-1,1],[-1,2,-1],[1,-1,2]], verify that A^3-6A^2+9A-4I=0 and hence find A^{-1}.
Hint. Compute A^2 and A^3, verify the identity, then rearrange to isolate A^{-1}.
Computing A^2 and A^3 step by step, and combining A^3-6A^2+9A-4I entry by entry gives the zero matrix, confirming the identity. Since A^3-6A^2+9A-4I=0, rearranging gives A(A^2-6A+9I)=4I, so A^{-1} = (A^2-6A+9I)/4.
✦ A^3-6A^2+9A-4I=0 verified; A^{-1} = [[3/4,1/4,-1/4],[1/4,3/4,1/4],[-1/4,1/4,3/4]]
Let A be a nonsingular square matrix of order 3x3. Then |adj A| is equal to (A) |A| (B) |A|^2 (C) |A|^3 (D) 3|A|.
Hint. Recall the identity |adj(A)| = |A|^(n-1) for an n x n matrix.
For an n x n matrix, |adj(A)| = |A|^(n-1). Here n=3, so |adj(A)| = |A|^(3-1) = |A|^2.
✦ (B) |A|^2
If A is an invertible matrix of order 2, then det(A^{-1}) is equal to (A) det(A) (B) 1/det(A) (C) 1 (D) 0.
Hint. Use the identity det(A)det(A^{-1})=det(I)=1.
Since A.A^{-1}=I, taking determinants of both sides: det(A).det(A^{-1}) = det(I) = 1. So det(A^{-1}) = 1/det(A).
✦ (B) 1/det(A)
Solutions written by the tuition.in editorial team and checked against the NCERT Class 12 Mathematics textbook, Reprint 2026-27 (lemh104.pdf) — Exercise 4.1 (8 questions), Exercise 4.2 (5 questions), Exercise 4.3 (5 questions), Exercise 4.4 (18 questions), Exercise 4.5 (14 questions), plus the chapter's Miscellaneous Exercise (9 questions), 59 questions total, the largest exercise set in Class 12 Mathematics so far. Every exercise page was rendered as a 300dpi image to read the determinant bar notation and matrix brackets accurately. The old stub taught Cramer's Rule as a core technique, but the current book never mentions it anywhere — the only method taught for solving a linear system is X=A^{-1}B (the matrix method). The old stub also collapsed all six real exercises into one invented 35-question group with no solutions file behind it. A genuine book-level inconsistency was also found and preserved as a note rather than silently corrected: the Miscellaneous Exercise's own instruction lines reference 'Exercises 11 to 15' and 'Exercise 17 to 19', which no longer exist after rationalisation trimmed the exercise down to 9 questions — confirmed by reading straight through the exercise and finding no such numbered questions. Every determinant, adjoint, inverse, and system-solution computation in this file was independently verified with a Python script using sympy, catching zero discrepancies against the hand-derived working, including the two symbolic matrix-cubed identities (Q15, Q16 of Exercise 4.4) and the parametrised consistency question (Q4 of Exercise 4.5).. Questions are referenced from the NCERT textbook for identification.
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