Find the determinant of the 5×5 matrix using some formula of linear algebra:
| 2 0 0 0 4 |
| -1 1 2 -2 1 |
| 3 3 0 1 1 |
| 1 0 0 0 1 |
| 1 -1 1 2 3 |
The Correct Answer and Explanation is:
To find the determinant of the given 5×55 \times 5 matrix, we denote it as: A=[20004−112−2133011100011−1123]A = \begin{bmatrix} 2 & 0 & 0 & 0 & 4 \\ -1 & 1 & 2 & -2 & 1 \\ 3 & 3 & 0 & 1 & 1 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & -1 & 1 & 2 & 3 \end{bmatrix}
Step-by-Step Calculation
We use cofactor expansion (Laplace expansion), ideally along a row or column with many zeros to simplify. Let’s expand along Row 1: det(A)=2⋅det(M11)+0+0+0+4⋅det(M15)\det(A) = 2 \cdot \det(M_{11}) + 0 + 0 + 0 + 4 \cdot \det(M_{15})
Where:
- M11M_{11} is the 4×44 \times 4 matrix formed by removing the 1st row and 1st column.
- M15M_{15} is the 4×44 \times 4 matrix formed by removing the 1st row and 5th column.
Compute det(M11)\det(M_{11})
M11=[12−2130110001−1123]M_{11} = \begin{bmatrix} 1 & 2 & -2 & 1 \\ 3 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ -1 & 1 & 2 & 3 \end{bmatrix}
Use row operations or further cofactor expansion. But notice that Row 3 has three zeros, so expand along that row: det(M11)=0+0+0+1⋅(−1)3⋅det([12−2301−112])=−det(B)\det(M_{11}) = 0 + 0 + 0 + 1 \cdot (-1)^3 \cdot \det\left( \begin{bmatrix} 1 & 2 & -2 \\ 3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix} \right) = -\det(B)
Compute det(B)\det(B) by cofactor or row reduction: B=[12−2301−112]⇒det(B)=1(0⋅2−1⋅1)−2(3⋅2−1⋅−1)+(−2)(3⋅1−0⋅−1)=−1−2(6+1)−2(3)=−1−14−6=−21⇒det(M11)=−(−21)=21B = \begin{bmatrix} 1 & 2 & -2 \\ 3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix} \Rightarrow \det(B) = 1(0 \cdot 2 – 1 \cdot 1) – 2(3 \cdot 2 – 1 \cdot -1) + (-2)(3 \cdot 1 – 0 \cdot -1) \\ = -1 – 2(6 + 1) -2(3) = -1 -14 -6 = -21 \Rightarrow \det(M_{11}) = -(-21) = 21
Compute det(M15)\det(M_{15})
M15=[−112−2330110001−112]M_{15} = \begin{bmatrix} -1 & 1 & 2 & -2 \\ 3 & 3 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 1 & -1 & 1 & 2 \end{bmatrix}
Again, Row 3 has three zeros. Expand along Row 3: det(M15)=1⋅(−1)3⋅det([−1123301−12])=−det(C)\det(M_{15}) = 1 \cdot (-1)^3 \cdot \det\left( \begin{bmatrix} -1 & 1 & 2 \\ 3 & 3 & 0 \\ 1 & -1 & 2 \end{bmatrix} \right) = -\det(C)
Now compute: C=[−1123301−12]C = \begin{bmatrix} -1 & 1 & 2 \\ 3 & 3 & 0 \\ 1 & -1 & 2 \end{bmatrix}
Use cofactor expansion or Sarrus: det(C)=−1(3⋅2−0⋅−1)−1(3⋅2−0⋅1)+2(3⋅−1−3⋅1)=−1(6)−1(6)+2(−3−3)=−6−6+2(−6)=−12−12=−24⇒det(M15)=−(−24)=24\det(C) = -1(3 \cdot 2 – 0 \cdot -1) – 1(3 \cdot 2 – 0 \cdot 1) + 2(3 \cdot -1 – 3 \cdot 1) \\ = -1(6) – 1(6) + 2(-3 – 3) = -6 -6 + 2(-6) = -12 -12 = -24 \Rightarrow \det(M_{15}) = -(-24) = 24
Final Determinant
det(A)=2⋅21+4⋅24=42+96=138\det(A) = 2 \cdot 21 + 4 \cdot 24 = 42 + 96 = \boxed{138}
Explanation
To compute the determinant of a 5×55 \times 5 matrix, we use cofactor expansion, which involves breaking the matrix into smaller matrices recursively until reaching 2×22 \times 2 matrices. This method is practical when one row or column contains many zeros, which minimizes the number of terms.
In the given matrix, the first row has three zeros. This makes it ideal for expansion. According to the cofactor formula, the determinant is calculated as the sum of the products of each element in the row and the determinant of its corresponding minor (i.e., the submatrix formed by removing the element’s row and column), multiplied by (−1)i+j(-1)^{i+j}, which adjusts for position-based signs.
We expanded along the first row: only two non-zero elements contribute — 2 and 4. For each, we calculated the determinant of its corresponding 4×44 \times 4 submatrix. These were again expanded, this time using rows that contained many zeros (like the third row in both submatrices), making the process efficient.
In both cases, the 4×44 \times 4 matrices reduced to 3×33 \times 3 matrices, which we evaluated directly using the cofactor method. Each time, we carefully applied the formula and signed terms appropriately.
Finally, we summed the results from both cofactor paths to get the full determinant: 138\boxed{138}
This process illustrates how cofactor expansion and recognizing structure in the matrix (like zeros) greatly simplify determinant calculations.
