Matrices and Determinants | Jamb Mathematics
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Dear student, as you prepare for your upcoming exam on matrices and determinants, take heart and approach your
studies with confidence and diligence. Just as every element in a matrix has its place and purpose, so too does
each step in your preparation bring you closer to clarity and success. Trust in the work you have done, seek
understanding beyond memorization, and know that you are more capable than you realize—steady yourself, and move forward with faith.
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Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic
of
Matrices and Determinants ? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple,
clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're
struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding
and confidence. Let’s tackle
Matrices and Determinants together and move one step closer to achieving your exam success!
Blissful learning .
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Calculation problems involving basic operation on matrices
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1. Matrix Addition
Question: Compute
A + B A + B A + B for
A = [ 2 3 5 1 ] A = \begin{bmatrix} 2 & 3 \\ 5 & 1 \end{bmatrix} A = [ 2 5 3 1 ] ,
B = [ 4 1 2 3 ] B = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} B = [ 4 2 1 3 ] .
Solution:
A + B = [ 2 + 4 3 + 1 5 + 2 1 + 3 ] = [ 6 4 7 4 ] A + B = \begin{bmatrix} 2+4 & 3+1 \\ 5+2 & 1+3 \end{bmatrix} = \begin{bmatrix} 6 & 4 \\ 7 & 4 \end{bmatrix} A + B = [ 2 + 4 5 + 2 3 + 1 1 + 3 ] = [ 6 7 4 4 ] .
2. Matrix Subtraction
Question: Compute
A − B A - B A − B for
A = [ 7 5 6 4 ] A = \begin{bmatrix} 7 & 5 \\ 6 & 4 \end{bmatrix} A = [ 7 6 5 4 ] ,
B = [ 3 2 1 6 ] B = \begin{bmatrix} 3 & 2 \\ 1 & 6 \end{bmatrix} B = [ 3 1 2 6 ] .
Solution:
A − B = [ 7 − 3 5 − 2 6 − 1 4 − 6 ] = [ 4 3 5 − 2 ] A - B = \begin{bmatrix} 7-3 & 5-2 \\ 6-1 & 4-6 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ 5 & -2 \end{bmatrix} A − B = [ 7 − 3 6 − 1 5 − 2 4 − 6 ] = [ 4 5 3 − 2 ] .
3. Scalar Multiplication
Question: Compute
3 A 3A 3 A for
A = [ 1 2 3 4 ] A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} A = [ 1 3 2 4 ] .
Solution:
3 A = 3 × [ 1 2 3 4 ] = [ 3 6 9 12 ] 3A = 3 \times \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 6 \\ 9 & 12 \end{bmatrix} 3 A = 3 × [ 1 3 2 4 ] = [ 3 9 6 12 ] .
4. Matrix Multiplication
Question: Compute
A B AB A B for
A = [ 1 2 3 4 ] A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} A = [ 1 3 2 4 ] ,
B = [ 5 6 7 8 ] B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} B = [ 5 7 6 8 ] .
Solution:
A B = [ 1 ( 5 ) + 2 ( 7 ) 1 ( 6 ) + 2 ( 8 ) 3 ( 5 ) + 4 ( 7 ) 3 ( 6 ) + 4 ( 8 ) ] AB = \begin{bmatrix} 1(5) + 2(7) & 1(6) + 2(8) \\ 3(5) + 4(7) & 3(6) + 4(8) \end{bmatrix} A B = [ 1 ( 5 ) + 2 ( 7 ) 3 ( 5 ) + 4 ( 7 ) 1 ( 6 ) + 2 ( 8 ) 3 ( 6 ) + 4 ( 8 ) ] = [ 5 + 14 6 + 16 15 + 28 18 + 32 ] = \begin{bmatrix} 5+14 & 6+16 \\ 15+28 & 18+32 \end{bmatrix} = [ 5 + 14 15 + 28 6 + 16 18 + 32 ] = [ 19 22 43 50 ] = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix} = [ 19 43 22 50 ] .
5. Determinant of a 2x2 Matrix
Question: Compute
det ( A ) \det(A) det ( A ) for
A = [ 4 3 2 5 ] A = \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} A = [ 4 2 3 5 ] .
Solution:
det ( A ) = ( 4 ) ( 5 ) − ( 3 ) ( 2 ) = 20 − 6 = 14 \det(A) = (4)(5) - (3)(2) = 20 - 6 = 14 det ( A ) = ( 4 ) ( 5 ) − ( 3 ) ( 2 ) = 20 − 6 = 14 .
6. Determinant of a 3x3 Matrix
Question: Compute
det ( A ) \det(A) det ( A ) for
A = [ 1 2 3 4 5 6 7 8 9 ] A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} A = 1 4 7 2 5 8 3 6 9 .
Solution:
Using cofactor expansion along the first row:
det ( A ) = 1 ∣ 5 6 8 9 ∣ − 2 ∣ 4 6 7 9 ∣ + 3 ∣ 4 5 7 8 ∣ \det(A) = 1 \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} - 2 \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 3 \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} det ( A ) = 1 5 8 6 9 − 2 4 7 6 9 + 3 4 7 5 8 .
Expanding further:
= 1 ( 5 × 9 − 6 × 8 ) − 2 ( 4 × 9 − 6 × 7 ) + 3 ( 4 × 8 − 5 × 7 ) = 1(5 \times 9 - 6 \times 8) - 2(4 \times 9 - 6 \times 7) + 3(4 \times 8 - 5 \times 7) = 1 ( 5 × 9 − 6 × 8 ) − 2 ( 4 × 9 − 6 × 7 ) + 3 ( 4 × 8 − 5 × 7 ) = 1 ( 45 − 48 ) − 2 ( 36 − 42 ) + 3 ( 32 − 35 ) = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) = 1 ( 45 − 48 ) − 2 ( 36 − 42 ) + 3 ( 32 − 35 ) = 1 ( − 3 ) − 2 ( − 6 ) + 3 ( − 3 ) = 1(-3) - 2(-6) + 3(-3) = 1 ( − 3 ) − 2 ( − 6 ) + 3 ( − 3 ) = − 3 + 12 − 9 = 0 = -3 + 12 - 9 = 0 = − 3 + 12 − 9 = 0 .
7. Inverse of a 2x2 Matrix
Question: Compute
A − 1 A^{-1} A − 1 for
A = [ 2 3 1 4 ] A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} A = [ 2 1 3 4 ] .
Solution:
A − 1 = 1 det ( A ) [ d − b − c a ] A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} A − 1 = d e t ( A ) 1 [ d − c − b a ] .
det ( A ) = 2 ( 4 ) − 3 ( 1 ) = 8 − 3 = 5 \det(A) = 2(4) - 3(1) = 8 - 3 = 5 det ( A ) = 2 ( 4 ) − 3 ( 1 ) = 8 − 3 = 5 .
A − 1 = 1 5 [ 4 − 3 − 1 2 ] = [ 4 5 − 3 5 − 1 5 2 5 ] A^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} \frac{4}{5} & -\frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{bmatrix} A − 1 = 5 1 [ 4 − 1 − 3 2 ] = [ 5 4 − 5 1 − 5 3 5 2 ] .
8. Transpose of a Matrix
Question: Compute
A T A^T A T for
A = [ 1 2 3 4 5 6 ] A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} A = 1 3 5 2 4 6 .
Solution:
A T = [ 1 3 5 2 4 6 ] A^T = \begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{bmatrix} A T = [ 1 2 3 4 5 6 ] .
9. Matrix Trace
Question: Compute
tr ( A ) \text{tr}(A) tr ( A ) for
A = [ 5 2 1 6 ] A = \begin{bmatrix} 5 & 2 \\ 1 & 6 \end{bmatrix} A = [ 5 1 2 6 ] .
Solution:
tr ( A ) = 5 + 6 = 11 \text{tr}(A) = 5 + 6 = 11 tr ( A ) = 5 + 6 = 11 .
10. Identity Matrix Multiplication
Question: Compute
A I AI A I for
A = [ 3 4 5 6 ] A = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix} A = [ 3 5 4 6 ] ,
I = [ 1 0 0 1 ] I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} I = [ 1 0 0 1 ] .
Solution:
A I = A AI = A A I = A , since multiplying by the identity matrix leaves the matrix unchanged.
11. Multiplication of a Row Vector and a Column Vector
Question: Compute
A B AB A B where
A = [ 1 2 3 ] A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} A = [ 1 2 3 ] ,
B = [ 4 5 6 ] B = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix} B = 4 5 6 .
Solution:
A B = 1 ( 4 ) + 2 ( 5 ) + 3 ( 6 ) = 4 + 10 + 18 = 32 AB = 1(4) + 2(5) + 3(6) = 4 + 10 + 18 = 32 A B = 1 ( 4 ) + 2 ( 5 ) + 3 ( 6 ) = 4 + 10 + 18 = 32 .
12. Multiplication of a Column Vector and a Row Vector
Question: Compute
A B AB A B where
A = [ 1 2 3 ] A = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} A = 1 2 3 ,
B = [ 4 5 6 ] B = \begin{bmatrix} 4 & 5 & 6 \end{bmatrix} B = [ 4 5 6 ] .
Solution:
A B = [ 1 ( 4 ) 1 ( 5 ) 1 ( 6 ) 2 ( 4 ) 2 ( 5 ) 2 ( 6 ) 3 ( 4 ) 3 ( 5 ) 3 ( 6 ) ] = [ 4 5 6 8 10 12 12 15 18 ] AB = \begin{bmatrix} 1(4) & 1(5) & 1(6) \\ 2(4) & 2(5) & 2(6) \\ 3(4) & 3(5) & 3(6) \end{bmatrix} = \begin{bmatrix} 4 & 5 & 6 \\ 8 & 10 & 12 \\ 12 & 15 & 18 \end{bmatrix} A B = 1 ( 4 ) 2 ( 4 ) 3 ( 4 ) 1 ( 5 ) 2 ( 5 ) 3 ( 5 ) 1 ( 6 ) 2 ( 6 ) 3 ( 6 ) = 4 8 12 5 10 15 6 12 18 .
13. Zero Matrix Multiplication
Question: Compute
A ⋅ 0 A \cdot 0 A ⋅ 0 where
A = [ 2 3 5 6 ] A = \begin{bmatrix} 2 & 3 \\ 5 & 6 \end{bmatrix} A = [ 2 5 3 6 ]
and
0 0 0 is the zero matrix.
Solution:
A ⋅ 0 = [ 0 0 0 0 ] A \cdot 0 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} A ⋅ 0 = [ 0 0 0 0 ] .
14. Power of a Square Matrix
Question: Compute
A 2 A^2 A 2 where
A = [ 1 2 3 4 ] A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} A = [ 1 3 2 4 ] .
Solution:
A 2 = A ⋅ A = [ 1 ( 1 ) + 2 ( 3 ) 1 ( 2 ) + 2 ( 4 ) 3 ( 1 ) + 4 ( 3 ) 3 ( 2 ) + 4 ( 4 ) ] A^2 = A \cdot A = \begin{bmatrix} 1(1) + 2(3) & 1(2) + 2(4) \\ 3(1) + 4(3) & 3(2) + 4(4) \end{bmatrix} A 2 = A ⋅ A = [ 1 ( 1 ) + 2 ( 3 ) 3 ( 1 ) + 4 ( 3 ) 1 ( 2 ) + 2 ( 4 ) 3 ( 2 ) + 4 ( 4 ) ] = [ 1 + 6 2 + 8 3 + 12 6 + 16 ] = [ 7 10 15 22 ] = \begin{bmatrix} 1 + 6 & 2 + 8 \\ 3 + 12 & 6 + 16 \end{bmatrix} = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix} = [ 1 + 6 3 + 12 2 + 8 6 + 16 ] = [ 7 15 10 22 ] .
Solving for a Matrix Variable
Question: Find
X X X in the equation
A X = B AX = B A X = B , where
A = [ 2 3 4 5 ] A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} A = [ 2 4 3 5 ] ,
B = [ 8 18 ] B = \begin{bmatrix} 8 \\ 18 \end{bmatrix} B = [ 8 18 ] .
Solution:
Let
X = [ x y ] X = \begin{bmatrix} x \\ y \end{bmatrix} X = [ x y ] . Then,
[ 2 3 4 5 ] [ x y ] = [ 8 18 ] \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 18 \end{bmatrix} [ 2 4 3 5 ] [ x y ] = [ 8 18 ] .
Solving:
2 x + 3 y = 8 2x + 3y = 8 2 x + 3 y = 8 4 x + 5 y = 18 4x + 5y = 18 4 x + 5 y = 18 .
By elimination:
Multiply the first equation by 2:
4 x + 6 y = 16 4x + 6y = 16 4 x + 6 y = 16 .
Subtract from the second equation:
( 4 x + 5 y ) − ( 4 x + 6 y ) = 18 − 16 (4x + 5y) - (4x + 6y) = 18 - 16 ( 4 x + 5 y ) − ( 4 x + 6 y ) = 18 − 16 − y = 2 ⇒ y = − 2 - y = 2 \Rightarrow y = -2 − y = 2 ⇒ y = − 2 .
Substituting
y = − 2 y = -2 y = − 2 into
2 x + 3 y = 8 2x + 3y = 8 2 x + 3 y = 8 :
2 x + 3 ( − 2 ) = 8 2x + 3(-2) = 8 2 x + 3 ( − 2 ) = 8 2 x − 6 = 8 2x - 6 = 8 2 x − 6 = 8 2 x = 14 ⇒ x = 7 2x = 14 \Rightarrow x = 7 2 x = 14 ⇒ x = 7 .
Thus,
X = [ 7 − 2 ] X = \begin{bmatrix} 7 \\ -2 \end{bmatrix} X = [ 7 − 2 ] .
16. Rank of a Matrix
Question: Find the rank of
A = [ 1 2 3 4 5 6 7 8 9 ] A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} A = 1 4 7 2 5 8 3 6 9 .
Solution:
Reducing to row echelon form:
[ 1 2 3 4 5 6 7 8 9 ] → [ 1 2 3 0 − 3 − 6 0 − 6 − 12 ] → [ 1 2 3 0 − 3 − 6 0 0 0 ] \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \to \begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{bmatrix} \to \begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & 0 & 0 \end{bmatrix} 1 4 7 2 5 8 3 6 9 → 1 0 0 2 − 3 − 6 3 − 6 − 12 → 1 0 0 2 − 3 0 3 − 6 0 .
Rank = 2 (two nonzero rows).
17. Eigenvalues of a 2x2 Matrix
Question: Find eigenvalues of
A = [ 2 1 1 2 ] A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} A = [ 2 1 1 2 ] .
Solution:
Solve
det ( A − λ I ) = 0 \det(A - \lambda I) = 0 det ( A − λ I ) = 0 :
det [ 2 − λ 1 1 2 − λ ] = 0 \det \begin{bmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{bmatrix} = 0 det [ 2 − λ 1 1 2 − λ ] = 0 .
( 2 − λ ) ( 2 − λ ) − ( 1 ) ( 1 ) = 0 (2 - \lambda)(2 - \lambda) - (1)(1) = 0 ( 2 − λ ) ( 2 − λ ) − ( 1 ) ( 1 ) = 0 .
λ 2 − 4 λ + 3 = 0 \lambda^2 - 4\lambda + 3 = 0 λ 2 − 4 λ + 3 = 0 .
( λ − 3 ) ( λ − 1 ) = 0 (\lambda - 3)(\lambda - 1) = 0 ( λ − 3 ) ( λ − 1 ) = 0 .
Eigenvalues:
λ = 3 , 1 \lambda = 3, 1 λ = 3 , 1 .
18. Matrix Diagonalization
Question: Is
A = [ 3 0 0 2 ] A = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} A = [ 3 0 0 2 ] diagonalizable?
Solution:
Since
A A A is already diagonal, it is trivially diagonalizable.
19. Orthogonality of Two Vectors
Question: Are
A = [ 1 − 1 ] A = \begin{bmatrix} 1 \\ -1 \end{bmatrix} A = [ 1 − 1 ] and
B = [ 2 2 ] B = \begin{bmatrix} 2 \\ 2 \end{bmatrix} B = [ 2 2 ] orthogonal?
Solution:
Compute
A ⋅ B = 1 ( 2 ) + ( − 1 ) ( 2 ) = 2 − 2 = 0 A \cdot B = 1(2) + (-1)(2) = 2 - 2 = 0 A ⋅ B = 1 ( 2 ) + ( − 1 ) ( 2 ) = 2 − 2 = 0 .
Since the dot product is zero, the vectors are orthogonal.
20. Projection of One Vector onto Another
Question: Find the projection of
A = [ 3 4 ] A = \begin{bmatrix} 3 \\ 4 \end{bmatrix} A = [ 3 4 ] onto
B = [ 1 0 ] B = \begin{bmatrix} 1 \\ 0 \end{bmatrix} B = [ 1 0 ] .
Solution:
proj B A = A ⋅ B B ⋅ B B \text{proj}_B A = \frac{A \cdot B}{B \cdot B} B proj B A = B ⋅ B A ⋅ B B .
= 3 ( 1 ) + 4 ( 0 ) 1 ( 1 ) + 0 ( 0 ) [ 1 0 ] = 3 [ 1 0 ] = [ 3 0 ] = \frac{3(1) + 4(0)}{1(1) + 0(0)} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \end{bmatrix} = 1 ( 1 ) + 0 ( 0 ) 3 ( 1 ) + 4 ( 0 ) [ 1 0 ] = 3 [ 1 0 ] = [ 3 0 ] .
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Calculation problems involving determinant of matrices
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1. Basic Determinant Calculation
Question: Compute
det ( A ) \det(A) det ( A ) for
A = [ 2 3 1 4 ] A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} A = [ 2 1 3 4 ] .
Solution:
Using the determinant formula for a
2 × 2 2 \times 2 2 × 2 matrix:
det ( A ) = ( 2 ) ( 4 ) − ( 3 ) ( 1 ) \det(A) = (2)(4) - (3)(1) det ( A ) = ( 2 ) ( 4 ) − ( 3 ) ( 1 ) = 8 − 3 = 8 - 3 = 8 − 3 = 5 = 5 = 5 .
2. Determinant with Negative Entries
Question: Compute
det ( A ) \det(A) det ( A ) for
A = [ − 1 5 2 − 3 ] A = \begin{bmatrix} -1 & 5 \\ 2 & -3 \end{bmatrix} A = [ − 1 2 5 − 3 ] .
Solution:
det ( A ) = ( − 1 ) ( − 3 ) − ( 5 ) ( 2 ) \det(A) = (-1)(-3) - (5)(2) det ( A ) = ( − 1 ) ( − 3 ) − ( 5 ) ( 2 ) = 3 − 10 = 3 - 10 = 3 − 10 = − 7 = -7 = − 7 .
3. Determinant with Zeroes
Question: Compute
det ( A ) \det(A) det ( A ) for
A = [ 0 4 7 0 ] A = \begin{bmatrix} 0 & 4 \\ 7 & 0 \end{bmatrix} A = [ 0 7 4 0 ] .
Solution:
det ( A ) = ( 0 ) ( 0 ) − ( 4 ) ( 7 ) \det(A) = (0)(0) - (4)(7) det ( A ) = ( 0 ) ( 0 ) − ( 4 ) ( 7 ) = 0 − 28 = 0 - 28 = 0 − 28 = − 28 = -28 = − 28 .
4. Determinant of Identity Matrix
Question: Compute
det ( A ) \det(A) det ( A ) for
A = [ 1 0 0 1 ] A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} A = [ 1 0 0 1 ] .
Solution:
det ( A ) = ( 1 ) ( 1 ) − ( 0 ) ( 0 ) \det(A) = (1)(1) - (0)(0) det ( A ) = ( 1 ) ( 1 ) − ( 0 ) ( 0 ) = 1 − 0 = 1 - 0 = 1 − 0 = 1 = 1 = 1 .
5. Determinant of a Scalar Multiple of Identity
Question: Compute
det ( A ) \det(A) det ( A ) for
A = [ 3 0 0 3 ] A = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} A = [ 3 0 0 3 ] .
Solution:
det ( A ) = ( 3 ) ( 3 ) − ( 0 ) ( 0 ) \det(A) = (3)(3) - (0)(0) det ( A ) = ( 3 ) ( 3 ) − ( 0 ) ( 0 ) = 9 − 0 = 9 - 0 = 9 − 0 = 9 = 9 = 9 .
6. Determinant Involving Fractions
Question: Compute
det ( A ) \det(A) det ( A ) for
A = [ 1 2 1 3 2 5 3 4 ] A = \begin{bmatrix} \frac{1}{2} & \frac{1}{3} \\ \frac{2}{5} & \frac{3}{4} \end{bmatrix} A = [ 2 1 5 2 3 1 4 3 ] .
Solution:
det ( A ) = ( 1 2 × 3 4 ) − ( 1 3 × 2 5 ) \det(A) = \left(\frac{1}{2} \times \frac{3}{4} \right) - \left(\frac{1}{3} \times \frac{2}{5} \right) det ( A ) = ( 2 1 × 4 3 ) − ( 3 1 × 5 2 ) = 3 8 − 2 15 = \frac{3}{8} - \frac{2}{15} = 8 3 − 15 2 .
Find a common denominator (120):
= 45 120 − 16 120 = \frac{45}{120} - \frac{16}{120} = 120 45 − 120 16 = 29 120 = \frac{29}{120} = 120 29 .
7. Determinant of a Singular Matrix
Question: Compute
det ( A ) \det(A) det ( A ) for
A = [ 6 9 2 3 ] A = \begin{bmatrix} 6 & 9 \\ 2 & 3 \end{bmatrix} A = [ 6 2 9 3 ] .
Solution:
det ( A ) = ( 6 ) ( 3 ) − ( 9 ) ( 2 ) \det(A) = (6)(3) - (9)(2) det ( A ) = ( 6 ) ( 3 ) − ( 9 ) ( 2 ) = 18 − 18 = 18 - 18 = 18 − 18 = 0 = 0 = 0 .
(Since the determinant is zero, this is a singular matrix and does not have an inverse.)
8. Determinant with Large Numbers
Question: Compute
det ( A ) \det(A) det ( A ) for
A = [ 100 25 40 10 ] A = \begin{bmatrix} 100 & 25 \\ 40 & 10 \end{bmatrix} A = [ 100 40 25 10 ] .
Solution:
det ( A ) = ( 100 ) ( 10 ) − ( 25 ) ( 40 ) \det(A) = (100)(10) - (25)(40) det ( A ) = ( 100 ) ( 10 ) − ( 25 ) ( 40 ) = 1000 − 1000 = 1000 - 1000 = 1000 − 1000 = 0 = 0 = 0 .
(This matrix is also singular.)
9. Determinant with Negative and Positive Numbers
Question: Compute
det ( A ) \det(A) det ( A ) for
A = [ − 4 7 5 − 6 ] A = \begin{bmatrix} -4 & 7 \\ 5 & -6 \end{bmatrix} A = [ − 4 5 7 − 6 ] .
Solution:
det ( A ) = ( − 4 ) ( − 6 ) − ( 7 ) ( 5 ) \det(A) = (-4)(-6) - (7)(5) det ( A ) = ( − 4 ) ( − 6 ) − ( 7 ) ( 5 ) = 24 − 35 = 24 - 35 = 24 − 35 = − 11 = -11 = − 11 .
10. Determinant of a Matrix with Equal Rows
Question: Compute
det ( A ) \det(A) det ( A ) for
A = [ 2 3 2 3 ] A = \begin{bmatrix} 2 & 3 \\ 2 & 3 \end{bmatrix} A = [ 2 2 3 3 ] .
Solution:
det ( A ) = ( 2 ) ( 3 ) − ( 3 ) ( 2 ) \det(A) = (2)(3) - (3)(2) det ( A ) = ( 2 ) ( 3 ) − ( 3 ) ( 2 ) = 6 − 6 = 6 - 6 = 6 − 6 = 0 = 0 = 0 .
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Calculation problems involving inverse of a matrix
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1. Basic Inverse Calculation
Question: Compute
A − 1 A^{-1} A − 1 for
A = [ 2 3 1 4 ] A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} A = [ 2 1 3 4 ] .
Solution:
The inverse of a
2 × 2 2 \times 2 2 × 2 matrix
A = [ a b c d ] A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} A = [ a c b d ] is given by:
A − 1 = 1 det ( A ) [ d − b − c a ] A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} A − 1 = d e t ( A ) 1 [ d − c − b a ] .
First, compute
det ( A ) \det(A) det ( A ) :
det ( A ) = ( 2 ) ( 4 ) − ( 3 ) ( 1 ) = 8 − 3 = 5 \det(A) = (2)(4) - (3)(1) = 8 - 3 = 5 det ( A ) = ( 2 ) ( 4 ) − ( 3 ) ( 1 ) = 8 − 3 = 5 .
Now, compute
A − 1 A^{-1} A − 1 :
A − 1 = 1 5 [ 4 − 3 − 1 2 ] A^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} A − 1 = 5 1 [ 4 − 1 − 3 2 ] = [ 4 5 − 3 5 − 1 5 2 5 ] = \begin{bmatrix} \frac{4}{5} & -\frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{bmatrix} = [ 5 4 − 5 1 − 5 3 5 2 ] .
2. Inverse with Negative Entries
Question: Compute
A − 1 A^{-1} A − 1 for
A = [ − 2 5 3 − 4 ] A = \begin{bmatrix} -2 & 5 \\ 3 & -4 \end{bmatrix} A = [ − 2 3 5 − 4 ] .
Solution:
Compute
det ( A ) \det(A) det ( A ) :
det ( A ) = ( − 2 ) ( − 4 ) − ( 5 ) ( 3 ) = 8 − 15 = − 7 \det(A) = (-2)(-4) - (5)(3) = 8 - 15 = -7 det ( A ) = ( − 2 ) ( − 4 ) − ( 5 ) ( 3 ) = 8 − 15 = − 7 .
Now, compute
A − 1 A^{-1} A − 1 :
A − 1 = 1 − 7 [ − 4 − 5 − 3 − 2 ] A^{-1} = \frac{1}{-7} \begin{bmatrix} -4 & -5 \\ -3 & -2 \end{bmatrix} A − 1 = − 7 1 [ − 4 − 3 − 5 − 2 ] = [ 4 7 5 7 3 7 2 7 ] = \begin{bmatrix} \frac{4}{7} & \frac{5}{7} \\ \frac{3}{7} & \frac{2}{7} \end{bmatrix} = [ 7 4 7 3 7 5 7 2 ] .
3. Inverse of a Matrix with Fractions
Question: Compute
A − 1 A^{-1} A − 1 for
A = [ 1 2 1 3 2 5 3 4 ] A = \begin{bmatrix} \frac{1}{2} & \frac{1}{3} \\ \frac{2}{5} & \frac{3}{4} \end{bmatrix} A = [ 2 1 5 2 3 1 4 3 ] .
Solution:
Compute
det ( A ) \det(A) det ( A ) :
det ( A ) = ( 1 2 × 3 4 ) − ( 1 3 × 2 5 ) \det(A) = \left(\frac{1}{2} \times \frac{3}{4}\right) - \left(\frac{1}{3} \times \frac{2}{5}\right) det ( A ) = ( 2 1 × 4 3 ) − ( 3 1 × 5 2 ) = 3 8 − 2 15 = \frac{3}{8} - \frac{2}{15} = 8 3 − 15 2 .
Find common denominator (120):
det ( A ) = 45 120 − 16 120 = 29 120 \det(A) = \frac{45}{120} - \frac{16}{120} = \frac{29}{120} det ( A ) = 120 45 − 120 16 = 120 29 .
Now, compute
A − 1 A^{-1} A − 1 :
A − 1 = 1 29 120 [ 3 4 − 1 3 − 2 5 1 2 ] A^{-1} = \frac{1}{\frac{29}{120}} \begin{bmatrix} \frac{3}{4} & -\frac{1}{3} \\ -\frac{2}{5} & \frac{1}{2} \end{bmatrix} A − 1 = 120 29 1 [ 4 3 − 5 2 − 3 1 2 1 ] = 120 29 [ 3 4 − 1 3 − 2 5 1 2 ] = \frac{120}{29} \begin{bmatrix} \frac{3}{4} & -\frac{1}{3} \\ -\frac{2}{5} & \frac{1}{2} \end{bmatrix} = 29 120 [ 4 3 − 5 2 − 3 1 2 1 ] = [ 90 29 − 40 29 − 48 29 60 29 ] = \begin{bmatrix} \frac{90}{29} & -\frac{40}{29} \\ -\frac{48}{29} & \frac{60}{29} \end{bmatrix} = [ 29 90 − 29 48 − 29 40 29 60 ] .
4. Inverse of a Matrix with a Zero Entry
Question: Compute
A − 1 A^{-1} A − 1 for
A = [ 4 0 3 2 ] A = \begin{bmatrix} 4 & 0 \\ 3 & 2 \end{bmatrix} A = [ 4 3 0 2 ] .
Solution:
Compute
det ( A ) \det(A) det ( A ) :
det ( A ) = ( 4 ) ( 2 ) − ( 0 ) ( 3 ) = 8 \det(A) = (4)(2) - (0)(3) = 8 det ( A ) = ( 4 ) ( 2 ) − ( 0 ) ( 3 ) = 8 .
Now, compute
A − 1 A^{-1} A − 1 :
A − 1 = 1 8 [ 2 0 − 3 4 ] A^{-1} = \frac{1}{8} \begin{bmatrix} 2 & 0 \\ -3 & 4 \end{bmatrix} A − 1 = 8 1 [ 2 − 3 0 4 ] = [ 2 8 0 − 3 8 4 8 ] = \begin{bmatrix} \frac{2}{8} & 0 \\ -\frac{3}{8} & \frac{4}{8} \end{bmatrix} = [ 8 2 − 8 3 0 8 4 ] = [ 1 4 0 − 3 8 1 2 ] = \begin{bmatrix} \frac{1}{4} & 0 \\ -\frac{3}{8} & \frac{1}{2} \end{bmatrix} = [ 4 1 − 8 3 0 2 1 ] .
5. Matrix with Integer Determinant
Question: Compute
A − 1 A^{-1} A − 1 for
A = [ 5 3 2 7 ] A = \begin{bmatrix} 5 & 3 \\ 2 & 7 \end{bmatrix} A = [ 5 2 3 7 ] .
Solution:
Compute
det ( A ) \det(A) det ( A ) :
det ( A ) = ( 5 ) ( 7 ) − ( 3 ) ( 2 ) = 35 − 6 = 29 \det(A) = (5)(7) - (3)(2) = 35 - 6 = 29 det ( A ) = ( 5 ) ( 7 ) − ( 3 ) ( 2 ) = 35 − 6 = 29 .
Now, compute
A − 1 A^{-1} A − 1 :
A − 1 = 1 29 [ 7 − 3 − 2 5 ] A^{-1} = \frac{1}{29} \begin{bmatrix} 7 & -3 \\ -2 & 5 \end{bmatrix} A − 1 = 29 1 [ 7 − 2 − 3 5 ] = [ 7 29 − 3 29 − 2 29 5 29 ] = \begin{bmatrix} \frac{7}{29} & -\frac{3}{29} \\ -\frac{2}{29} & \frac{5}{29} \end{bmatrix} = [ 29 7 − 29 2 − 29 3 29 5 ] .
6. Singular Matrix (No Inverse)
Question: Compute
A − 1 A^{-1} A − 1 for
A = [ 2 4 1 2 ] A = \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix} A = [ 2 1 4 2 ] .
Solution:
Compute
det ( A ) \det(A) det ( A ) :
det ( A ) = ( 2 ) ( 2 ) − ( 4 ) ( 1 ) = 4 − 4 = 0 \det(A) = (2)(2) - (4)(1) = 4 - 4 = 0 det ( A ) = ( 2 ) ( 2 ) − ( 4 ) ( 1 ) = 4 − 4 = 0 .
Since
det ( A ) = 0 \det(A) = 0 det ( A ) = 0 , the matrix is singular and has no inverse.
7. Inverse of a Negative Matrix
Question: Compute
A − 1 A^{-1} A − 1 for
A = [ − 3 − 2 − 1 − 4 ] A = \begin{bmatrix} -3 & -2 \\ -1 & -4 \end{bmatrix} A = [ − 3 − 1 − 2 − 4 ] .
Solution:
Compute
det ( A ) \det(A) det ( A ) :
det ( A ) = ( − 3 ) ( − 4 ) − ( − 2 ) ( − 1 ) = 12 − 2 = 10 \det(A) = (-3)(-4) - (-2)(-1) = 12 - 2 = 10 det ( A ) = ( − 3 ) ( − 4 ) − ( − 2 ) ( − 1 ) = 12 − 2 = 10 .
Now, compute
A − 1 A^{-1} A − 1 :
A − 1 = 1 10 [ − 4 2 1 − 3 ] A^{-1} = \frac{1}{10} \begin{bmatrix} -4 & 2 \\ 1 & -3 \end{bmatrix} A − 1 = 10 1 [ − 4 1 2 − 3 ] = [ − 4 10 2 10 1 10 − 3 10 ] = \begin{bmatrix} -\frac{4}{10} & \frac{2}{10} \\ \frac{1}{10} & -\frac{3}{10} \end{bmatrix} = [ − 10 4 10 1 10 2 − 10 3 ] = [ − 2 5 1 5 1 10 − 3 10 ] = \begin{bmatrix} -\frac{2}{5} & \frac{1}{5} \\ \frac{1}{10} & -\frac{3}{10} \end{bmatrix} = [ − 5 2 10 1 5 1 − 10 3 ] .
8. Identity Matrix
Question: Compute
A − 1 A^{-1} A − 1 for
A = [ 1 0 0 1 ] A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} A = [ 1 0 0 1 ] .
Solution:
Since
A A A is the identity matrix, its inverse is itself:
A − 1 = A = [ 1 0 0 1 ] A^{-1} = A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} A − 1 = A = [ 1 0 0 1 ] .
9. Inverse with Decimal Entries
Question: Compute
A − 1 A^{-1} A − 1 for
A = [ 0.5 0.2 0.3 0.7 ] A = \begin{bmatrix} 0.5 & 0.2 \\ 0.3 & 0.7 \end{bmatrix} A = [ 0.5 0.3 0.2 0.7 ] .
Solution:
Compute
det ( A ) \det(A) det ( A ) :
det ( A ) = ( 0.5 ) ( 0.7 ) − ( 0.2 ) ( 0.3 ) = 0.35 − 0.06 = 0.29 \det(A) = (0.5)(0.7) - (0.2)(0.3) = 0.35 - 0.06 = 0.29 det ( A ) = ( 0.5 ) ( 0.7 ) − ( 0.2 ) ( 0.3 ) = 0.35 − 0.06 = 0.29 .
Now, compute
A − 1 A^{-1} A − 1 :
A − 1 = 1 0.29 [ 0.7 − 0.2 − 0.3 0.5 ] A^{-1} = \frac{1}{0.29} \begin{bmatrix} 0.7 & -0.2 \\ -0.3 & 0.5 \end{bmatrix} A − 1 = 0.29 1 [ 0.7 − 0.3 − 0.2 0.5 ] .
Approximating:
A − 1 ≈ [ 2.41 − 0.69 − 1.03 1.72 ] A^{-1} \approx \begin{bmatrix} 2.41 & -0.69 \\ -1.03 & 1.72 \end{bmatrix} A − 1 ≈ [ 2.41 − 1.03 − 0.69 1.72 ] .
10. Large Numbers
Question: Compute
A − 1 A^{-1} A − 1 for
A = [ 10 5 7 3 ] A = \begin{bmatrix} 10 & 5 \\ 7 & 3 \end{bmatrix} A = [ 10 7 5 3 ] .
Solution:
det ( A ) = ( 10 ) ( 3 ) − ( 5 ) ( 7 ) = 30 − 35 = − 5 \det(A) = (10)(3) - (5)(7) = 30 - 35 = -5 det ( A ) = ( 10 ) ( 3 ) − ( 5 ) ( 7 ) = 30 − 35 = − 5 .
A − 1 = 1 − 5 [ 3 − 5 − 7 10 ] A^{-1} = \frac{1}{-5} \begin{bmatrix} 3 & -5 \\ -7 & 10 \end{bmatrix} A − 1 = − 5 1 [ 3 − 7 − 5 10 ] = [ − 3 5 1 7 5 − 2 ] = \begin{bmatrix} -\frac{3}{5} & 1 \\ \frac{7}{5} & -2 \end{bmatrix} = [ − 5 3 5 7 1 − 2 ] .
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