Determinant, Inverse, and Solving Equations by Matrix Method

Write the definition of inverse matrix. In which condition the inverse matrix cannot be defined? Write it.

Inverse matrix: If A and B are two square matrices of the same order, I is an identity matrix of the same order and AB = BA = I then A and B are said to be the inverse of each other. The inverse of A is denoted by A^-1.

If the determinant of a matrix is 0, then inverse matrix is not defined.

If $\begin{bmatrix} 4 \\ 1 \end{bmatrix}$ .C = $\begin{bmatrix} -8 & 12 \\ -2 & 3 \end{bmatrix}$ , find the matrix C.

Let, C = [a, b]

Given,
$\begin{bmatrix} 4 \ 1 \end{bmatrix}$ .C = $\begin{bmatrix} -8 & 12 \ -2 & 3 \end{bmatrix}$ or, $\begin{bmatrix} 4 \ 1 \end{bmatrix}$ .[a, b] = $\begin{bmatrix} -8 & 12 \ -2 & 3 \end{bmatrix}$
or, $\begin{bmatrix} 4a & 4b \ a & b \end{bmatrix}$ = $\begin{bmatrix} -8 & 12 \ -2 & 3 \end{bmatrix}$
Comparing the corresponding element of equal matrix, we get,
a = -2, b = 3

Hence, C = [a, b] = [-2, 3]

If $\begin{bmatrix} 4 \\ 3 \end{bmatrix}$ × p = $\begin{bmatrix} 8 & -4 \\ 6 & -3 \end{bmatrix}$ , find the matrix p.

Let, p = [a, b]

Given,

\begin{bmatrix} 4 \\ 3 \end{bmatrix}

.C =

\begin{bmatrix} 8 & -4 \\ 6 & -3 \end{bmatrix}

or,

\begin{bmatrix} 4 \\ 3 \end{bmatrix}

.[a, b] =

\begin{bmatrix} 8 & -4 \\ 6 & -3 \end{bmatrix}

or,

\begin{bmatrix} 4a & 4b \\ 3a & 3b \end{bmatrix}

\begin{bmatrix} 8 & -4 \\ 6 & -3 \end{bmatrix}

Comparing the corresponding element of equal matrix, we get,
a = 2, b = -1

Hence, p = [a, b] = [2, -1]

If A = $\begin{bmatrix} 1 & 2 & \ \llap{-}3 \\ \ \llap{-}2 & 0 & 5 \\ \end{bmatrix}$ and B = $\begin{bmatrix} 2 & 3 \\ \ \llap{-}4 & 0 \\ 5 & \ \llap{-}1 \\ \end{bmatrix}$ , Find AB.

Given,

A = $\begin{bmatrix} 1 & 2 & \ \llap{-}3 \\ \ \llap{-}2 & 0 & 5 \\ \end{bmatrix}$
B = $\begin{bmatrix} 2 & 3 \\ \ \llap{-}4 & 0 \\ 5 & \ \llap{-}1 \\ \end{bmatrix}$

Now,
AB =

\begin{bmatrix} 1 & 2 & \ \llap{-}3 \\ \ \llap{-}2 & 0 & 5 \\ \end{bmatrix}

\begin{bmatrix} 2 & 3 \\ \ \llap{-}4 & 0 \\ 5 & \ \llap{-}1 \\ \end{bmatrix}

\begin{bmatrix} 2 - 8 - 15 & 3 + 3 \\ -4 - 0 + 25 & -6 - 5 \\ \end{bmatrix}

= <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo fence="true">[</mo><mtable rowspacing="0.16em" columnalign="center center" columnspacing="1em"><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mo>−</mo><mn>21</mn></mrow></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="false"><mn>6</mn></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mn>21</mn></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mo>−</mo><mn>11</mn></mrow></mstyle></mtd></mtr></mtable><mo fence="true">]</mo></mrow><annotation encoding="application/x-tex">\begin{bmatrix}
-21 &amp;  6 \\
21 &amp; -11 \\
\end{bmatrix}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:2.4em;vertical-align:-0.95em;"></span><span class="minner"><span class="mopen delimcenter" style="top:0em;"><span class="delimsizing size3">[</span></span><span class="mord"><span class="mtable"><span class="col-align-c"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.45em;"><span style="top:-3.61em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">−</span><span class="mord">21</span></span></span><span style="top:-2.41em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">21</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.95em;"><span></span></span></span></span></span><span class="arraycolsep" style="width:0.5em;"></span><span class="arraycolsep" style="width:0.5em;"></span><span class="col-align-c"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.45em;"><span style="top:-3.61em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">6</span></span></span><span style="top:-2.41em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">−</span><span class="mord">11</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.95em;"><span></span></span></span></span></span></span></span><span class="mclose delimcenter" style="top:0em;"><span class="delimsizing size3">]</span></span></span></span></span></span>
</span>

If $\begin{vmatrix} 3x & \ \llap{-}6 \\ 4 & 2 \end{vmatrix}$ = 0, find the value of x.

Here,

\begin{vmatrix} 3x & \ \llap{-}6 \\ 4 & 2 \end{vmatrix}

= 0 or, 6x - (-24) = 0
or, 6x = -24
∴ x = -4

If K = $\begin{bmatrix} 2 & \ \llap{-}3 \\ 5 & m \end{bmatrix}$ and |K| = 23 then find the value of m.

Given,
K = $\begin{bmatrix} 2 & \ \llap{-}3 \\ 5 & m \end{bmatrix}$
|K| = 23

Now, $\begin{bmatrix} 2 & \ \llap{-}3 \ 5 & m \end{bmatrix}$ = 23 or, 2m - (-15) = 23
or, 2m + 15 = 23
or, 2m = 8
∴ m = 4

If matrix A = $\begin{bmatrix} 5 & 3 \\ 7 & 4 \\ \end{bmatrix}$ then find A^-1.

Given: A = $\begin{bmatrix} 5 & 3 \\ 7 & 4 \\ \end{bmatrix}$
To find: A^-1

Here, Adjoint of A = $\begin{bmatrix} 4 & -3 \\ -7 & 5 \\ \end{bmatrix}$

Again, for determinant of A,
|A| = $\begin{bmatrix} 5 & 3 \ 7 & 4 \ \end{bmatrix}$ = |20 - 21|
= -1. So, A^-1 exists.

Finally, We Know,
A^-1 = $\frac{1}{\text{|A|}}$ Adjoint of A = $\begin{bmatrix} -4 & 3 \\ 7 & -5 \\ \end{bmatrix}$

Another method

You can also solve it using the formula AA^-1 = I

If the inverse of the matrix $\begin{bmatrix} m & 2 \\ 7 & 3 \\ \end{bmatrix}$ is the matrix $\begin{bmatrix} 3 & -2 \\ -7 & m \\ \end{bmatrix}$ find the value of m.

Given,
A = $\begin{bmatrix} m & 2 \\ 7 & 3 \\ \end{bmatrix}$
A^-1 = $\begin{bmatrix} 3 & -2 \\ -7 & m \\ \end{bmatrix}$
To find: m

Now, We have, A.A^-1 = I or, $\begin{bmatrix} m & 2 \ 7 & 3 \ \end{bmatrix}$ $\begin{bmatrix} 3 & -2 \ -7 & m \ \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \ 0 & 1 \ \end{bmatrix}$
or, $\begin{bmatrix} 3m - 14 & -2m + 2m \ 21 - 21 & -14 + 3m \ \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \ 0 & 1 \ \end{bmatrix}$
Equating the corresponding elements of equal matrices, we get,
or, 3m - 14 = 1
or, 3m = 15
∴ m = 5

Another method

Given,
A =

\begin{bmatrix} m & 2 \\ 7 & 3 \\ \end{bmatrix}

A^-1 =

\begin{bmatrix} 3 & -2 \\ -7 & m \\ \end{bmatrix}

To find: m

Now, Adjoint of A = $\begin{bmatrix} 3 & -2 \\ -7 & m \\ \end{bmatrix}$

Again, for determinant of A,
|A| = $\begin{vmatrix} m & 2 \ 7 & 3 \ \end{vmatrix}$ = 3m - 14

We know, A^-1 = $\frac{1}{\lvert A \rvert}$ Adjoint of A or, $\begin{bmatrix} 3 & -2 \\ -7 & m \\ \end{bmatrix}$ = $\frac{1}{3m - 14}$ $\begin{bmatrix} 3 & -2 \\ -7 & m \\ \end{bmatrix}$
Equating the corresponding elements of equal matrices, we get,
or, 3 = $\frac{3}{3m - 14}$
or, 9m - 42 = 3
or, 9m = 45
∴ m = 5

If matrix A = $\begin{bmatrix} 3 & 1 \\ \ \llap{-}1 & 2 \\ \end{bmatrix}$ , find AA^T.

Given,
A =

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

Now,
A^T = $\begin{bmatrix} 3 & \ \llap{-}1 \\ 1 & 2 \\ \end{bmatrix}$

Finally,
AA^T = $\begin{bmatrix} 3 & 1 \ \ \llap{-}1 & 2 \ \end{bmatrix}$ $\begin{bmatrix} 3 & \ \llap{-}1 \ 1 & 2 \ \end{bmatrix}$ = $\begin{bmatrix} 9 + 1 & -3 + 2 \ -3 + 2 & 1 + 4 \ \end{bmatrix}$
= $\begin{bmatrix} 10 & -1 \ -1 & 5 \ \end{bmatrix}$

If A = $\begin{bmatrix} 2 & 0 \\ \ \llap{-}3 & x \\ \end{bmatrix}$ , B = $\begin{bmatrix} 4 & 0 \\ \ \llap{-}9 & 1 \\ \end{bmatrix}$ and A² = B, find the value of x.

Given,
A = $\begin{bmatrix} 2 & 0 \\ \ \llap{-}3 & x \\ \end{bmatrix}$
B = $\begin{bmatrix} 4 & 0 \\ \ \llap{-}9 & 1 \\ \end{bmatrix}$
A² = B

Now, $\begin{bmatrix} 2 & 0 \ \ \llap{-}3 & x \ \end{bmatrix}$ $\begin{bmatrix} 2 & 0 \ \ \llap{-}3 & x \ \end{bmatrix}$ = $\begin{bmatrix} 4 & 0 \ \ \llap{-}9 & 1 \ \end{bmatrix}$ or, $\begin{bmatrix} 4 - 0 & 0 + 0 \ -6 - 3x & 0 + x^2 \ \end{bmatrix}$ = $\begin{bmatrix} 4 & 0 \ \ \llap{-}9 & 1 \ \end{bmatrix}$
Equating the corresponding elements of equal matrices, we get,
or, -6x - 3x = -9
or, 3 = 3x
∴x = 1

If the inverse of the matrix $\begin{bmatrix} x & 3 \\ 4 & 2 \\ \end{bmatrix}$ can not be defined, find the value of x.

Here,
Let, the given matrix is,
A =

\begin{bmatrix} x & 3 \\ 4 & 2 \\ \end{bmatrix}

According to question, inverse of A cannot be defined, so, |A| = 0 or, $\begin{vmatrix} x & 3 \\ 4 & 2 \\ \end{vmatrix}$
or, 2x - 12 = 0
∴ x = 6

If M = $\begin{bmatrix} 5 & 3 \\ 6 & K \\ \end{bmatrix}$ and |M| = 27 then find the value of K.

Given,
M = $\begin{bmatrix} 5 & 3 \\ 6 & K \\ \end{bmatrix}$
|M| = 27

Now,

\begin{vmatrix} 5 & 3 \\ 6 & K \\ \end{vmatrix}

= 27 or, 5k - 18 = 27
or, 5k = 45
∴ K = 9

If P = $\begin{bmatrix} 2 & 3 \\ 4 & 5 \\ \end{bmatrix}$ and K = $\begin{bmatrix} 1 & 0 \\ 0 & \ \llap{-}1 \\ \end{bmatrix}$ , find PK.

Given,
P = $\begin{bmatrix} 2 & 3 \\ 4 & 5 \\ \end{bmatrix}$
K = $\begin{bmatrix} 1 & 0 \\ 0 & \ \llap{-}1 \\ \end{bmatrix}$

Now,
PK = $\begin{bmatrix} 2 & 3 \ 4 & 5 \ \end{bmatrix}$ $\begin{bmatrix} 1 & 0 \ 0 & \ \llap{-}1 \ \end{bmatrix}$ = $\begin{bmatrix} 2 + 0 & 0 - 3 \ 4 + 0 & 0 - 5 \ \end{bmatrix}$
= $\begin{bmatrix} 2 & -3 \ 4 & -5 \ \end{bmatrix}$

If an inverse matrix of A is $\begin{bmatrix} 1 & 3 \\ 2 & 4 \\ \end{bmatrix}$ , find the matrix A.

Given,
A^-1 =

\begin{bmatrix} 1 & 3 \\ 2 & 4 \\ \end{bmatrix}

Let A =

\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}

<p>
    Now, We know,

A.A^-1 = I
or,

\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}

\begin{bmatrix} 1 & 3 \\ 2 & 4 \\ \end{bmatrix}

\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}

or, <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo fence="true">[</mo><mtable rowspacing="0.16em" columnalign="center center" columnspacing="1em"><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mi>a</mi><mo>+</mo><mn>2</mn><mi>b</mi></mrow></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mn>3</mn><mi>a</mi><mo>+</mo><mn>4</mn><mi>b</mi></mrow></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mi>c</mi><mo>+</mo><mn>2</mn><mi>d</mi></mrow></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="false"><mrow><mn>3</mn><mi>c</mi><mo>+</mo><mn>4</mn><mi>d</mi></mrow></mstyle></mtd></mtr></mtable><mo fence="true">]</mo></mrow><annotation encoding="application/x-tex">\begin{bmatrix}
a + 2b &amp; 3a + 4b \\
c + 2d &amp; 3c + 4d \\
\end{bmatrix}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:2.4em;vertical-align:-0.95em;"></span><span class="minner"><span class="mopen delimcenter" style="top:0em;"><span class="delimsizing size3">[</span></span><span class="mord"><span class="mtable"><span class="col-align-c"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.45em;"><span style="top:-3.61em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord mathnormal">a</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">2</span><span class="mord mathnormal">b</span></span></span><span style="top:-2.41em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord mathnormal">c</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">2</span><span class="mord mathnormal">d</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.95em;"><span></span></span></span></span></span><span class="arraycolsep" style="width:0.5em;"></span><span class="arraycolsep" style="width:0.5em;"></span><span class="col-align-c"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.45em;"><span style="top:-3.61em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">3</span><span class="mord mathnormal">a</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">4</span><span class="mord mathnormal">b</span></span></span><span style="top:-2.41em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">3</span><span class="mord mathnormal">c</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">4</span><span class="mord mathnormal">d</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.95em;"><span></span></span></span></span></span></span></span><span class="mclose delimcenter" style="top:0em;"><span class="delimsizing size3">]</span></span></span></span></span></span> = 
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1 &amp; 0 \\
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</p>

<p>
Equating the corresponding elements of equal matrices, we get,
<span class="left">a + 2b = 1</span>
    or, a = 1 - 2b ---- (i)
</p>

Also, 3a + 4b = 0 or, 3(1 - 2b) + 4b = 0 [From (i)]
or, 3 - 6b + 4b = 0
or, 3 - 2b = 0
∴ b = $\frac{3}{2}$
∴ a = -2 [From (i)]

Similarly, c + 2d = 0 or, c = -2d ---- (ii)

Finally, 3c + 4d = 1 or, -6d + 4d = 1 [From (ii)]
∴ d = $\frac{-1}{2}$
∴ c = 1 [From (ii)]

Hence, A = $\begin{bmatrix} a & b \ c & d \ \end{bmatrix}$ = $\begin{bmatrix} \ \llap{-}2 & 3/2 \ 1 & \ \llap{-}1/2 \ \end{bmatrix}$

If A = $\begin{bmatrix} 3 & 4 \\ 2 & 3 \\ \end{bmatrix}$ and B = $\begin{bmatrix} 3 & \ \llap{-}4 \\ \ \llap{-}2 & 3 \\ \end{bmatrix}$ , then prove that AB is an identity matrix.

Given,
A =

\begin{bmatrix} 3 & 4 \\ 2 & 3 \\ \end{bmatrix}

B =

\begin{bmatrix} 3 & \ \llap{-}4 \\ \ \llap{-}2 & 3 \\ \end{bmatrix}

Now,
AB = $\begin{bmatrix} 3 & 4 \\ 2 & 3 \\ \end{bmatrix}$ $\begin{bmatrix} 3 & \ \llap{-}4 \\ \ \llap{-}2 & 3 \\ \end{bmatrix}$ = $\begin{bmatrix} 9 - 8 & -12 + 12 \\ 6 - 6 & -8 + 9 \\ \end{bmatrix}$
= $\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$

Hence, AB is an identity matrix.

If inverse of matrix A is A^-1 = $\begin{bmatrix} 3 & 2 \\ 5 & 4 \\ \end{bmatrix}$ , find the matrix A.

Given, A^-1 =

\begin{bmatrix} 3 & 2 \\ 5 & 4 \\ \end{bmatrix}

Let, A =

\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}

Now, We Know, A × A^-1 = I or, $\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$ $\begin{bmatrix} 3 & 2 \\ 5 & 4 \\ \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$
or, $\begin{bmatrix} 3a + 5b & 2a + 4b \\ 3c + 5d & 2c + 4d \\ \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$

Equating the corresponding elements of equal matrices, we get, 3a + 5b = 1 or, a = $\frac{1 - 5b}{3}$ ---- (i)

Again, 2a + 4b = 0 or, 2 - 10b + 12b = 0 [From (i)]
or, 2 = -2b
∴ b = -1
∴ a = 2 [From (i)]

Similarly,

3c + 5d = 0

or, c = $\frac{-5d}{3}$ ---- (ii)

Again, 2c + 4d = 1 or, -10d + 12d = 3
or, 2d = 3
∴ d = 3/2
∴c = -15/6 = -5/2 [From (ii)]

Hence, A = $\begin{bmatrix} a & b \ c & d \ \end{bmatrix}$ = $\begin{bmatrix} 2 & -1 \ -5/2 & 3/2 \ \end{bmatrix}$

Define inverse of a matrix. If A = $\begin{bmatrix} 1 & 2 \\ 4 & 5 \\ \end{bmatrix}$ find its inverse.

Inverse matrix: If A and B are two square matrices of same order, I is an identity matrix of same order and AB = BA = I then A and B are said to be the inverse of each other. The inverse of A is denoted by A^-1.

Given,
A =

\begin{bmatrix} 1 & 2 \\ 4 & 5 \\ \end{bmatrix}

Now, Adjoint of A = $\begin{bmatrix} 5 & -2 \\ -4 & 1 \\ \end{bmatrix}$

Again, for determinant of A,
|A| = $\begin{vmatrix} 1 & 2 \ 4 & 5 \ \end{vmatrix}$ = 5 - 8
= -3. So, A^-1 exists.

Finally, We know,
A^-1 = $\frac{1}{\text{|A|}}$ Adjoint of A = $\frac{1}{-3}$ $\begin{bmatrix} 5 & -2 \ -4 & 1 \ \end{bmatrix}$
= $\begin{bmatrix} -5/3 & 2/3 \ 4/5 & -1/3 \ \end{bmatrix}$

If D = $\begin{bmatrix} 1 & 2 \\ 3 & 1 \\ \end{bmatrix}$ , find the value of D².

Given,
D =

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

Now,
D² = $\begin{bmatrix} 1 & 2 \\ 3 & 1 \\ \end{bmatrix}$ $\begin{bmatrix} 1 & 2 \\ 3 & 1 \\ \end{bmatrix}$ = $\begin{bmatrix} 1 + 6 & 2 + 2\\ 3 + 3 & 6 + 1 \\ \end{bmatrix}$
= $\begin{bmatrix} 7 & 4 \\ 6 & 7 \\ \end{bmatrix}$

In which condition can two matrices be multiplied? State it. If P = (1 3 -4) and Q = $\begin{pmatrix} 2 \\ 0 \\ 1 \\ \end{pmatrix}$ , find the matrix QP.

If column of one matrix is equal to the row of another matrix then two matrices can be multiplied. For example: matrix having order [2 × 3][3 × 5] can be multiplied.

Given,
P = (1 3 -4)
Q = $\begin{pmatrix} 2 \ 0 \ 1 \ \end{pmatrix}$

Now,
QP = $\begin{pmatrix} 2 \\ 0 \\ 1 \\ \end{pmatrix}$ (1 3 -4) = $\begin{pmatrix} 2 & 6 & -8 \\ 0 & 0 & 0\\ 1 & 3 & -4\\ \end{pmatrix}$

Find the value of x if the matrix $\begin{bmatrix} x & 8 \\ 4 & 5 \\ \end{bmatrix}$ has no inverse.

Let the given matrix is,
A =

\begin{bmatrix} x & 8 \\ 4 & 5 \\ \end{bmatrix}

If A has no inverse then, |A| = 0 or, $\begin{vmatrix} x & 8 \\ 4 & 5 \\ \end{vmatrix}$ = 0
or, 5x - 32 = 0
∴ x = 32/5

If A = $\begin{bmatrix} 1 & \ \llap{-}1 \\ 3 & \ \llap{-}3 \\ \end{bmatrix}$ and B = $\begin{bmatrix} 2 & \ \llap{-}5 \\ 2 & \ \llap{-}5 \\ \end{bmatrix}$ , show that AB is a null matrix.

Given,
A =

\begin{bmatrix} 1 & \ \llap{-}1 \\ 3 & \ \llap{-}3 \\ \end{bmatrix}

B =

\begin{bmatrix} 2 & \ \llap{-}5 \\ 2 & \ \llap{-}5 \\ \end{bmatrix}

Now,
AB = $\begin{bmatrix} 1 & \ \llap{-}1 \\ 3 & \ \llap{-}3 \\ \end{bmatrix}$ $\begin{bmatrix} 2 & \ \llap{-}5 \\ 2 & \ \llap{-}5 \\ \end{bmatrix}$ = $\begin{bmatrix} 2 - 2 & -5 + 5 \\ 6 - 6 & -15 + 15 \\ \end{bmatrix}$
= $\begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}$

So, AB is a null matrix.

If the determinant of the matrix M = $\begin{bmatrix} \ \llap{-}5 & \ \llap{-}8 \\ p & 12 \\ \end{bmatrix}$ is 20, find the value of p.

Given,
M =

\begin{bmatrix} \ \llap{-}5 & \ \llap{-}8 \\ p & 12 \\ \end{bmatrix}

According to question, |M| = 20 or, $\begin{vmatrix} \ \llap{-}5 & \ \llap{-}8 \\ p & 12 \\ \end{vmatrix}$ = 20
or, -60 + 8p = 20
or, 8p = 80
∴ p = 10

If $\begin{pmatrix} 2 & a \\ b & 4 \\ \end{pmatrix}$ $\begin{pmatrix} \ \llap{-}2 \\ 5 \\ \end{pmatrix}$ = $\begin{pmatrix} 6 \\ 0 \\ \end{pmatrix}$ , find the values of a and b.

Here,

\begin{pmatrix} 2 & a \\ b & 4 \\ \end{pmatrix}

\begin{pmatrix} \ \llap{-}2 \\ 5 \\ \end{pmatrix}

\begin{pmatrix} 6 \\ 0 \\ \end{pmatrix}

or,

\begin{pmatrix} -4 + 5a \\ -2b + 20 \\ \end{pmatrix}

\begin{pmatrix} 6 \\ 0 \\ \end{pmatrix}

Equating the corresponding elements of equal matrices, we get,
or, -4 + 5a = 6
or, 5a = 10
∴ a = 2

Similarly, or, -2b + 20 = 0
∴ b = 10

Hence, the value of a = 2 and b = 10.

If A = $\begin{bmatrix} 2 & 1 \\ 1 & 3 \\ \end{bmatrix}$ , B = $\begin{bmatrix} 2 & 3 \\ 1 & m \\ \end{bmatrix}$ and |BA| = -5, find the value of m.

Given,
A =

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

B =

\begin{bmatrix} 2 & 3 \\ 1 & m \\ \end{bmatrix}

|BA| = -5

Now,
BA = $\begin{bmatrix} 2 & 3 \\ 1 & m \\ \end{bmatrix}$ $\begin{bmatrix} 2 & 1 \\ 1 & 3 \\ \end{bmatrix}$ = $\begin{bmatrix} 4 + 3 & 2 + 9 \\ 2 + m & 1 + 3m \\ \end{bmatrix}$
= $\begin{bmatrix} 7 & 11 \\ 2 + m & 1 + 3m \\ \end{bmatrix}$ --- (i)

Finally, |BA| = -5 or, $\begin{vmatrix} 7 & 11 \ 2 + m & 1 + 3m \ \end{vmatrix}$ = -5 [using (i)]
or, 7 + 21m - 22 - 11m = -5
or, 10m = 10
∴ m = 1

If E = $\begin{bmatrix} -3 & 4 \\ 2 & -2a \\ \end{bmatrix}$ and |E| = 22, find the value of a.

Given,
E =

\begin{bmatrix} -3 & 4 \\ 2 & -2a \\ \end{bmatrix}

|E| = 22

Now, |E| = 22 or, 22 = 6a - 8
or, 30 = 6a
∴ a = 5

If $\begin{pmatrix} -2 & 0 \\ 0 & -5 \\ \end{pmatrix}$ $\begin{pmatrix} x \\ y \\ \end{pmatrix}$ = $\begin{pmatrix} -2 \\ -10 \\ \end{pmatrix}$ , find the value of $\begin{pmatrix} x \\ y \\ \end{pmatrix}$ .

Here,

\begin{pmatrix} -2 & 0 \\ 0 & -5 \\ \end{pmatrix}

\begin{pmatrix} x \\ y \\ \end{pmatrix}

\begin{pmatrix} -2 \\ -10 \\ \end{pmatrix}

or,

\begin{pmatrix} -2x + 0y \\ 0x - 5y \\ \end{pmatrix}

\begin{pmatrix} -2 \\ -10 \\ \end{pmatrix}

Equating the corresponding elements of equal matrices, we get,
or, -2x = -2
∴ x = 1

Similarly, -5y = 10 ∴ y = 2

Hence, $\begin{pmatrix} x \ y \ \end{pmatrix}$ = $\begin{pmatrix} 1 \ 2 \ \end{pmatrix}$

If $\begin{pmatrix} \text{x + 1} \\ \text{y - 2} \\ \end{pmatrix}$ = $\begin{pmatrix} 1 & 0 \\ 0 & \ \llap{-}1 \\ \end{pmatrix}$ $\begin{pmatrix} 2 \\ 3\\ \end{pmatrix}$ find the value of x and y.

Here,

\begin{pmatrix} \text{x + 1} \\ \text{y - 2} \\ \end{pmatrix}

\begin{pmatrix} 1 & 0 \\ 0 & \ \llap{-}1 \\ \end{pmatrix}

\begin{pmatrix} 2 \\ 3\\ \end{pmatrix}

or,

\begin{pmatrix} \text{x + 1} \\ \text{y - 2} \\ \end{pmatrix}

\begin{pmatrix} 2 + 0 \\ 0 - 3 \\ \end{pmatrix}

or,

\begin{pmatrix} \text{x + 1} \\ \text{y - 2} \\ \end{pmatrix}

\begin{pmatrix} 2 \\ -3 \\ \end{pmatrix}

Equating the corresponding elements of equal matrices, we get,
or, x + 1 = 2
∴ x = 1

Again,
or, y - 2 = -3
∴ y = -1

Hence, the value of x = 1 and y = -1.

If the inverse of the matrix $\begin{pmatrix} \ \llap{2}x & 7 \\ 5 & 9 \\ \end{pmatrix}$ is the matrix $\begin{pmatrix} 9 & y \\ \ \llap{-}5 & 4 \\ \end{pmatrix}$ , calculate the values of x and y.

Let, A =

\begin{pmatrix} \ \llap{2}x & 7 \\ 5 & 9 \\ \end{pmatrix}

And, A^-1 =

\begin{pmatrix} 9 & y \\ \ \llap{-}5 & 4 \\ \end{pmatrix}

Now, We have, A.A^-1 = I or, $\begin{pmatrix} \ \llap{2}x & 7 \\ 5 & 9 \\ \end{pmatrix}$ $\begin{pmatrix} 9 & y \\ \ \llap{-}5 & 4 \\ \end{pmatrix}$ = $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$
or, $\begin{pmatrix} 18x - 35 & 2xy + 28 \\ 45 - 45 & 5y + 36 \\ \end{pmatrix}$ = $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$

Equating the corresponding elements of equal matrices, we get,
or, 18x - 35 = 1
or, 18x = 36
∴ x = 2

Similarly,
or, 5y + 36 = 1
or, 5y = -35
∴ y = -7

Hence, the value of x = 2 and y = -7