The inverse of matrix N = [[2, 3], [1, 4]] is

Mathematics
The inverse of matrix N = [[2, 3], [1, 4]] is

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Answer

\dfrac{1}{5}\begin{pmatrix}4&-3\-1&2\end{pmatrix}

Step 1: Compute the determinant of N=(2314)N = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}.

det⁡(N)=∣2314∣=2⋅4−3⋅1=8−3=5\det(N) = \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 2 \cdot 4 - 3 \cdot 1 = 8 - 3 = 5

Since det⁡(N)=5≠0\det(N) = 5 \neq 0, the inverse exists.

Step 2: Apply the formula for the inverse of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}:

(abcd)−1=1ad−bc(d−b−ca)\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

Substitute a=2a=2, b=3b=3, c=1c=1, d=4d=4, det⁡(N)=5\det(N)=5:

N−1=15(4−3−12)N^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}

Step 3: Verify by computing N⋅N−1N \cdot N^{-1}.

First row, first column: 2⋅45+3⋅−15=85−35=55=12 \cdot \frac{4}{5} + 3 \cdot \frac{-1}{5} = \frac{8}{5} - \frac{3}{5} = \frac{5}{5} = 1

First row, second column: 2⋅−35+3⋅25=−65+65=05=02 \cdot \frac{-3}{5} + 3 \cdot \frac{2}{5} = \frac{-6}{5} + \frac{6}{5} = \frac{0}{5} = 0

Second row, first column: 1⋅45+4⋅−15=45−45=05=01 \cdot \frac{4}{5} + 4 \cdot \frac{-1}{5} = \frac{4}{5} - \frac{4}{5} = \frac{0}{5} = 0

Second row, second column: 1⋅−35+4⋅25=−35+85=55=11 \cdot \frac{-3}{5} + 4 \cdot \frac{2}{5} = \frac{-3}{5} + \frac{8}{5} = \frac{5}{5} = 1

N⋅N−1=(1001)=IN \cdot N^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I

15(4−3−12)\boxed{\dfrac{1}{5}\begin{pmatrix}4&-3\\-1&2\end{pmatrix}}

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Compute the determinant of N = 2 & 3 \\ 1 & 4 . (N) = 2 & 3 \\ 1 & 4 = 2 · 4 - 3 · 1 = 8 - 3 = 5 Since (N) = 5 ≠ 0, the inverse exists.

The inverse of matrix N = [[2, 3], [1, 4]] is
Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

ScanSolve AI Answer
Step 1: Compute the determinant of N = 2 & 3 \\ 1 & 4 . (N) = 2 & 3 \\ 1 & 4 = 2 · 4 - 3 · 1 = 8 - 3 = 5 Since (N) = 5 ≠ 0, the inverse exists. Step 2: Apply the formula for the inverse of a 2 × 2 matrix a & b \\ c & d : a & b \\ c & d ^-1 = (1)/(ad - bc) d & -b \\ -c & a Substitute a=2, b=3, c=1, d=4, (N)=5: N^-1 = (1)/(5) 4 & -3 \\ -1 & 2 Step 3: Verify by computing N · N^-1. First row, first column: 2 · (4)/(5) + 3 · (-1)/(5) = (8)/(5) - (3)/(5) = (5)/(5) = 1 First row, second column: 2 · (-3)/(5) + 3 · (2)/(5) = (-6)/(5) + (6)/(5) = (0)/(5) = 0 Second row, first column: 1 · (4)/(5) + 4 · (-1)/(5) = (4)/(5) - (4)/(5) = (0)/(5) = 0 Second row, second column: 1 · (-3)/(5) + 4 · (2)/(5) = (-3)/(5) + (8)/(5) = (5)/(5) = 1 N · N^-1 = 1 & 0 \\ 0 & 1 = I (1)/(5)4&-3\\-1&2