# Find the values of a and b, if A = B, where:

\[A=\left( \begin{matrix}

a+b & 3b \\

8 & -6 \\

\end{matrix} \right),B=\left( \begin{matrix}

2a+2 & 3b \\

8 & {{b}^{2}}-10 \\

\end{matrix} \right)\]

Last updated date: 19th Mar 2023

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Answer

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304.5k+ views

Hint: In this question, we need to look at the definitions of matrix and then equality of matrices to solve the above question. Then we need to equate each and every corresponding term in both the matrices and then get the values of a and b.

Complete step-by-step Solution:

Let us look at the definitions first.

MATRIX: A matrix is a rectangular arrangement of numbers (real or complex) which may be represented as

\[A=\left( \begin{matrix}

{{a}_{11}} & \ldots & {{a}_{1n}} \\

\vdots & \ddots & \vdots \\

{{a}_{m1}} & \cdots & {{a}_{mn}} \\

\end{matrix} \right)\]

Matrix is enclosed by ( ) or [ ]

Compact from the above matrix is represented by \[{{\left[ {{a}_{ij}} \right]}_{m\times n}}\] or \[A=\left[ {{a}_{ij}} \right]\] .

ELEMENT OF A MATRIX: The numbers a11, a12, etc., in the above matrix are known as the elements of the matrix, generally represented as \[{{a}_{ij}}\], which denotes element in the

\[i\text{th row and }j\text{th column}\text{.}\]

ORDER OF A MATRIX: In above matrix has m rows and n columns, then A is of order

\[m\times n\].

TYPES OF MATRICES:

Row Matrix

Column Matrix

Rectangular Matrix

Horizontal Matrix

Vertical Matrix

Null / Zero Matrix

Square Matrix

Diagonal Matrix

Scalar Matrix

Unit / Identity Matrix

Upper Triangular Matrix

Lower Triangular Matrix

Sub Matrix

Equal Matrices

Principal Diagonal of a Matrix

Singular Matrix

EQUAL MATRICES: Two matrices A and B are said to be equal, if both are having the same order and corresponding elements of both the matrices are equal.

Now, by equating both the given matrices A and B.

\[\begin{align}

& \Rightarrow A=B \\

& \Rightarrow \left( \begin{matrix}

a+b & 3b \\

8 & -6 \\

\end{matrix} \right)=\left( \begin{matrix}

2a+2 & 3b \\

8 & {{b}^{2}}-10 \\

\end{matrix} \right) \\

\end{align}\]

On equating the corresponding elements in both the matrices we get,

\[\begin{align}

& \Rightarrow a+b=2a+2...............\left( 1 \right) \\

& \Rightarrow -6={{b}^{2}}-10...................\left( 2 \right) \\

\end{align}\]

Now, on simplifying the above equation (1) we get,

\[\Rightarrow b=a+2............\left( 3 \right)\]

On simplifying the above equation (2) we get,

\[\begin{align}

& \Rightarrow -6+10={{b}^{2}} \\

& \Rightarrow {{b}^{2}}=4 \\

& \therefore b=2 \\

\end{align}\]

Now, on substituting the value of b in equation (3) we get,

\[\begin{align}

& \Rightarrow b=a+2 \\

& \Rightarrow a=b-2 \\

& \Rightarrow a=2-2\text{ }\left[ \because b=2 \right] \\

& \therefore a=0 \\

\end{align}\]

Similarly, on considering the other possible case,

\[\begin{align}

& \Rightarrow -6+10={{b}^{2}} \\

& \Rightarrow {{b}^{2}}=4 \\

& \therefore b=-2 \\

\end{align}\]

Now, on substituting this value of b we get,

\[\begin{align}

& \Rightarrow b=a+2 \\

& \Rightarrow a=b-2 \\

& \Rightarrow a=-2-2\text{ }\left[ \because b=-2 \right] \\

& \therefore a=-4 \\

\end{align}\]

Note:

It is important to note that when we consider two matrices to be equal then in order to hold the equality every corresponding element in both the matrices should be equal.

In the above simplification if we consider other values of b instead of considering only the positive value then we get an alternate value of a. So, it is important to consider both the positive and negative values of b as it gives different values of a.

Even for the negative values and for the positive values the equality of both the matrices holds true in both the cases.

Complete step-by-step Solution:

Let us look at the definitions first.

MATRIX: A matrix is a rectangular arrangement of numbers (real or complex) which may be represented as

\[A=\left( \begin{matrix}

{{a}_{11}} & \ldots & {{a}_{1n}} \\

\vdots & \ddots & \vdots \\

{{a}_{m1}} & \cdots & {{a}_{mn}} \\

\end{matrix} \right)\]

Matrix is enclosed by ( ) or [ ]

Compact from the above matrix is represented by \[{{\left[ {{a}_{ij}} \right]}_{m\times n}}\] or \[A=\left[ {{a}_{ij}} \right]\] .

ELEMENT OF A MATRIX: The numbers a11, a12, etc., in the above matrix are known as the elements of the matrix, generally represented as \[{{a}_{ij}}\], which denotes element in the

\[i\text{th row and }j\text{th column}\text{.}\]

ORDER OF A MATRIX: In above matrix has m rows and n columns, then A is of order

\[m\times n\].

TYPES OF MATRICES:

Row Matrix

Column Matrix

Rectangular Matrix

Horizontal Matrix

Vertical Matrix

Null / Zero Matrix

Square Matrix

Diagonal Matrix

Scalar Matrix

Unit / Identity Matrix

Upper Triangular Matrix

Lower Triangular Matrix

Sub Matrix

Equal Matrices

Principal Diagonal of a Matrix

Singular Matrix

EQUAL MATRICES: Two matrices A and B are said to be equal, if both are having the same order and corresponding elements of both the matrices are equal.

Now, by equating both the given matrices A and B.

\[\begin{align}

& \Rightarrow A=B \\

& \Rightarrow \left( \begin{matrix}

a+b & 3b \\

8 & -6 \\

\end{matrix} \right)=\left( \begin{matrix}

2a+2 & 3b \\

8 & {{b}^{2}}-10 \\

\end{matrix} \right) \\

\end{align}\]

On equating the corresponding elements in both the matrices we get,

\[\begin{align}

& \Rightarrow a+b=2a+2...............\left( 1 \right) \\

& \Rightarrow -6={{b}^{2}}-10...................\left( 2 \right) \\

\end{align}\]

Now, on simplifying the above equation (1) we get,

\[\Rightarrow b=a+2............\left( 3 \right)\]

On simplifying the above equation (2) we get,

\[\begin{align}

& \Rightarrow -6+10={{b}^{2}} \\

& \Rightarrow {{b}^{2}}=4 \\

& \therefore b=2 \\

\end{align}\]

Now, on substituting the value of b in equation (3) we get,

\[\begin{align}

& \Rightarrow b=a+2 \\

& \Rightarrow a=b-2 \\

& \Rightarrow a=2-2\text{ }\left[ \because b=2 \right] \\

& \therefore a=0 \\

\end{align}\]

Similarly, on considering the other possible case,

\[\begin{align}

& \Rightarrow -6+10={{b}^{2}} \\

& \Rightarrow {{b}^{2}}=4 \\

& \therefore b=-2 \\

\end{align}\]

Now, on substituting this value of b we get,

\[\begin{align}

& \Rightarrow b=a+2 \\

& \Rightarrow a=b-2 \\

& \Rightarrow a=-2-2\text{ }\left[ \because b=-2 \right] \\

& \therefore a=-4 \\

\end{align}\]

Note:

It is important to note that when we consider two matrices to be equal then in order to hold the equality every corresponding element in both the matrices should be equal.

In the above simplification if we consider other values of b instead of considering only the positive value then we get an alternate value of a. So, it is important to consider both the positive and negative values of b as it gives different values of a.

Even for the negative values and for the positive values the equality of both the matrices holds true in both the cases.

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