Find the inverse of the matrix if it exists. [−748−5]

Step 1 Inverse of a matrix A denoted by A−1 is the matrix such that, ∀−1=A−1A=I where I is the identity matrix. A matrix A has an inverse if and only if its determinant is non-zero. The inverse of a matrix can be found out by elementary row or elementary column transformations. Step 2 The given matrix is, A=[−748−5] The determinant of the above matrix is, |A|=(−7×−5)−(4×8) =35−32 =3≠0 Hence A has an inverse. Step 3 To find the inverse of A: A=AI ⇒[[−7,4].[8,−5]]=A[1001] R1→R1+R2 [−7+84−58−5]=A[1+00+101] ⇒[[1,−1][8,−5]]=A[[1,1][0,1]] R2→R2−8R1 ⇒[1−18−1×8−5−(−1)×8]=A[110−1×81−1×8] ⇒[1−103]=A[11−8−7] R2→R23 ⇒[1−10333]=A[11−83−73] ⇒[1−101]=A[11−83−73] R1→R1+R2 [1+0−1+101]=A[1−831−73−83−73]

Step 1 Given: (−748−5) Find the determinant of the matrix. det((−748−5)) For the 2×2 matrix (abcd), the determinant is ad−bc −7(−5)−4×8 Multiply -7 times -5 35−4×8 Multiply 4 times 8 35−32 Subtract 32 from 35 Answer: 3

Step 1 Inverse of a matrix A denoted by A−1 is the matrix such that, ∀−1=A−1A=I where I is the identity matrix. A matrix A has an inverse if and only if its determinant is non-zero. The inverse of a matrix can be found out by elementary row or elementary column transformations. Step 2 The given matrix is, A=[−748−5] The determinant of the above matrix is, |A|=(−7×−5)−(4×8) =35−32 =3≠0 Hence A has an inverse. Step 3 To find the inverse of A: A=AI ⇒[[−7,4].[8,−5]]=A[1001] R1→R1+R2 [−7+84−58−5]=A[1+00+101] ⇒[[1,−1][8,−5]]=A[[1,1][0,1]] R2→R2−8R1 ⇒[1−18−1×8−5−(−1)×8]=A[110−1×81−1×8] ⇒[1−103]=A[11−8−7] R2→R23 ⇒[1−10333]=A[11−83−73] ⇒[1−101]=A[11−83−73] R1→R1+R2 [1+0−1+101]=A[1−831−73−83−73]

Step 1 Given: (−748−5) Find the determinant of the matrix. det((−748−5)) For the 2×2 matrix (abcd), the determinant is ad−bc −7(−5)−4×8 Multiply -7 times -5 35−4×8 Multiply 4 times 8 35−32 Subtract 32 from 35 Answer: 3

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2022-01-21

Find the inverse of the matrix if it exists.

[\begin{array}{cc} - 7 & 4 \\ 8 & - 5 \end{array}]

Answer & Explanation

seibesitoeu

Beginner2022-01-22Added 12 answers

Step 1
Inverse of a matrix A denoted by $A^{- 1}$ is the matrix such that, $\forall^{- 1} = A^{- 1} A = I$ where I is the identity matrix.
A matrix A has an inverse if and only if its determinant is non-zero.
The inverse of a matrix can be found out by elementary row or elementary column transformations.
Step 2
The given matrix is,
$A = [\begin{array}{cc} - 7 & 4 \\ 8 & - 5 \end{array}]$
The determinant of the above matrix is,
$| A | = (- 7 \times - 5) - (4 \times 8)$
$= 35 - 32$
$= 3 \neq 0$
Hence A has an inverse.
Step 3
To find the inverse of A:
$A = A I$
$\Rightarrow [[- 7, 4] . [8, - 5]] = A [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$
$R_{1} \to R_{1} + R_{2}$
$[\begin{array}{cc} - 7 + 8 & 4 - 5 \\ 8 & - 5 \end{array}] = A [\begin{array}{cc} 1 + 0 & 0 + 1 \\ 0 & 1 \end{array}]$
$\Rightarrow [[1, - 1] [8, - 5]] = A [[1, 1] [0, 1]]$
$R_{2} \to R_{2} - 8 R_{1}$
$\Rightarrow [\begin{array}{cc} 1 & - 1 \\ 8 - 1 \times 8 & - 5 - (- 1) \times 8 \end{array}] = A [\begin{array}{cc} 1 & 1 \\ 0 - 1 \times 8 & 1 - 1 \times 8 \end{array}]$
$\Rightarrow [\begin{array}{cc} 1 & - 1 \\ 0 & 3 \end{array}] = A [\begin{array}{cc} 1 & 1 \\ - 8 & - 7 \end{array}]$
$R_{2} \to \frac{R_{2}}{3}$
$\Rightarrow [\begin{array}{cc} 1 & - 1 \\ \frac{0}{3} & \frac{3}{3} \end{array}] = A [\begin{array}{cc} 1 & 1 \\ \frac{- 8}{3} & \frac{- 7}{3} \end{array}]$
$\Rightarrow [\begin{array}{cc} 1 & - 1 \\ 0 & 1 \end{array}] = A [\begin{array}{cc} 1 & 1 \\ \frac{- 8}{3} & \frac{- 7}{3} \end{array}]$
$R_{1} \to R_{1} + R_{2}$
$[\begin{array}{cc} 1 + 0 & - 1 + 1 \\ 0 & 1 \end{array}] = A [\begin{array}{cc} 1 - \frac{8}{3} & 1 - \frac{7}{3} \\ \frac{- 8}{3} & \frac{- 7}{3} \end{array}]$

search633504

Beginner2022-01-23Added 16 answers

Step 1
Given:

(\begin{array}{cc} - 7 & 4 \\ 8 & - 5 \end{array})

Find the determinant of the matrix.

det ((\begin{array}{cc} - 7 & 4 \\ 8 & - 5 \end{array}))

For the

2 \times 2

matrix

(\begin{array}{cc} a & b \\ c & d \end{array})

, the determinant is

a d - b c

- 7 (- 5) - 4 \times 8

Multiply -7 times -5

35 - 4 \times 8

Multiply 4 times 8

35 - 32

Subtract 32 from 35
Answer: 3

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