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  • Show that A equals 5 3 minus 1 minus 2 satisfies the equation A^2 minus 3A minus 7I = 0 and hence find A-1.

Show that \begin{bmatrix} 5 &3 \\-1 &-2 \end{bmatrix} satisfies the equation A^2 - 3A - 7I = 0 and hence find A^{-1}.

 

Answers (1)

We have the given matrix A, such that

\begin{bmatrix} 5 &3 \\-1 &-2 \end{bmatrix}

(i). We need to show that the matrix A satisfies the equation A\textsuperscript{2} -3A - 7I = 0.

(ii). Also, we need to find A\textsuperscript{-1}.

(i). Take L.H.S: A\textsuperscript{2} - 3A - 7I

First, compute A\textsuperscript{2}.

A\textsuperscript{2} = A.A

A^{2}=\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]

By convention, if we have to multiple matrix A and B then the number of columns in matrix A should be equal to the number of rows in matrix B.  Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

Multiply 1st row of matrix A by matching members of 1st column of matrix A, then sum them up.

\\(5, 3).(5, -1) = (5 $ \times $ 5) + (3 $ \times $ -1) \\$ \Rightarrow $ (5, 3).(5, -1) = 25 + (-3) \\$ \Rightarrow $ (5, 3).(5, -1) = 25 - 3 \\$ \Rightarrow $ (5, 3).(5, -1) = 22

\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right] = \begin{bmatrix} 22 & \\ & \end{bmatrix}

Multiply 1st row of matrix A by matching members of 2nd column of matrix A, then sum them up.

\\(5, 3).(3, -2) = (5 $ \times $ 3) + (3 $ \times $ -2) \\$ \Rightarrow $ (5, 3).(3, -2) = 15 + (-6) \\$ \Rightarrow $ (5, 3).(3, -2) = 15 - 6 \\$ \Rightarrow $ (5, 3).(3, -2) = 9

\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right] = \begin{bmatrix} 22 & 9\\ & \end{bmatrix}

Multiply 2nd row of matrix A by matching members of 1st column of matrix A, then sum them up.

\\ (-1, -2).(5, -1) = (-1 $ \times $ 5) + (-2 $ \times $ -1) \\$ \Rightarrow $ (-1, -2).(5, -1) = -5 + 2 \\$ \Rightarrow $ (-1, -2).(5, -1) = -3

\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right] = \begin{bmatrix} 22 & 9\\ -3& \end{bmatrix}

Multiply 2nd row of matrix A by matching members of 2nd column of matrix A, then sum them up.

\\ (-1, -2).(3, -2) = (-1 $ \times $ 3) + (-2 $ \times $ -2) \\$ \Rightarrow $ (-1, -2).(3, -2) = -3 + 4 \\$ \Rightarrow $ (-1, -2).(3, -2) = 1

\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right] = \begin{bmatrix} 22 & 9\\ -3& 1\end{bmatrix}

A^2 = \begin{bmatrix} 22 & 9\\ -3& 1\end{bmatrix}

Substitute values of A\textsuperscript{2}  and A in A\textsuperscript{2} - 3A - 7I.

A^{2}-3 A-7 I=\left[\begin{array}{ll} 22 & 9 \\ -3 & 1 \end{array}\right]-3\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]-7I

Also, since matrix A is of the order 2 × 2, then I will be the identity matrix of order 2 × 2 such that,

\begin{array}{l} \Rightarrow A^{2}-3 A-7 I=\left[\begin{array}{cc} 22 & 9 \\ -3 & 1 \end{array}\right]-3\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]-7\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \\\\ \Rightarrow A^{2}-3 A-7 I=\left[\begin{array}{cc} 22 & 9 \\ -3 & 1 \end{array}\right]-\left[\begin{array}{cc} 3 \times 5 & 3 \times 3 \\ 3 \times-1 & 3 \times-2 \end{array}\right]-\left[\begin{array}{cc} 7 \times 1 & 7 \times 0 \\ 7 \times 0 & 7 \times 1 \end{array}\right] \\\\ \Rightarrow A^{2}-3 A-7 I=\left[\begin{array}{cc} 22 & 9 \\ -3 & 1 \end{array}\right]-\left[\begin{array}{cc} 15 & 9 \\ -3 & -6 \end{array}\right]-\left[\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}\right] \\ \\\Rightarrow A^{2}-3 A-7 I=\left[\begin{array}{cc} 22-15-7 & 9-9-0 \\ -3-(-3)-0 & 1-(-6)-7 \end{array}\right] \\ \\\Rightarrow A^{2}-3 A-7 I=\left[\begin{array}{cc} 22-22 & 0 \\ -3+3 & 1+6-7 \end{array}\right] \\ \end{array}

\Rightarrow A^{2}-3 A-7 I=\left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right]

Hence proved,

L.H.S = R.H.S

Thus, we have shown that matrix A satisfy A\textsuperscript{2} - 3A - 7I = 0.

(ii). Now, let us find A\textsuperscript{-1}.

We know that, inverse of matrix A is A\textsuperscript{-1}. is true only when

A $ \times $ A\textsuperscript{-1} = A\textsuperscript{-1} $ \times $ A = I

Where, I = Identity matrix

We get,

A\textsuperscript{2} - 3A - 7I = 0

Multiply A\textsuperscript{-1}. on both sides, we get

\\A\textsuperscript{-1}(A\textsuperscript{2} - 3A - 7I) = A\textsuperscript{-1} $ \times $ 0 \\$ \Rightarrow $ A\textsuperscript{-1}.A\textsuperscript{2} - A\textsuperscript{-1}.3A - A\textsuperscript{-1}.7I = 0 \\$ \Rightarrow $ A\textsuperscript{-1}.A.A - 3A\textsuperscript{-1}.A - 7A\textsuperscript{-1}.I = 0 \\$ \Rightarrow $ (A\textsuperscript{-1}A)A - 3(A\textsuperscript{-1}A) - 7(A\textsuperscript{-1}I) = 0 \\ \text{And as } A\textsuperscript{-1}A = I \: \: and\: \: A\textsuperscript{-1}I = A\textsuperscript{-1} \\$ \Rightarrow $ IA - 3I - 7A\textsuperscript{-1} = 0 \\ \text{Since, IA = A} \\$ \Rightarrow $ A - 3I - 7A\textsuperscript{-1} = 0 \\$ \Rightarrow $ 7A\textsuperscript{-1} = A - 3I

\begin{aligned} &\Rightarrow \mathrm{A}^{-1}=\frac{1}{7}(\mathrm{~A}-3 \mathrm{I})\\ &\Rightarrow A^{-1}=\frac{1}{7}\left(\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]-3\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\right)_{[\because} A=\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right] \text { and } I=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\\ &\Rightarrow A^{-1}=\frac{1}{7}\left(\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]-\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]\right)\\ &\Rightarrow A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 5-3 & 3-0 \\ -1-0 & -2-3 \end{array}\right]\\ &\Rightarrow A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 2 & 3 \\ -1 & -5 \end{array}\right]\\ ,&A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 2 & 3 \\ -1 & -5 \end{array}\right]\\ \end{aligned}

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